%Title: QED vertex form factors at two loops
%Author: R. Bonciani (Freiburg U.), P. Mastrolia (Bologna U. and Karlsruhe U.,ITT) and E. Remiddi (Bologna U. and INFN, Bologna) 
%Published: <EM> Nucl.Phys.B </EM> <B> 676 </B> (2004) 399-452. 
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\begin{fmffile}{Fact}

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\begin{titlepage}
\nopagebreak
{\flushright{
        \begin{minipage}{5cm}
        Freiburg-THEP 03/11\\
        TTP03-20\\
        {\tt hep-ph/yymmnnn}\\
        \end{minipage}        }

}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\vspace*{-1.5cm}                        
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\begin{center}
\boldmath
{\Large \bf QED vertex form factors at two loops}\unboldmath
\vskip 1.cm
{\large  R.~Bonciani\footnote{Email: {\tt
Roberto.Bonciani@physik.uni-freiburg.de}}
\footnote{This work was supported by the European Union under
contract HPRN-CT-2000-00149},}
\vskip .2cm
{\it Facult\"at f\"ur Mathematik und Physik, Albert-Ludwigs-Universit\"at
Freiburg, \\ D-79104 Freiburg, Germany} 
\vskip .2cm
{\large P.~Mastrolia\footnote{Email: {\tt Pierpaolo.Mastrolia@bo.infn.it}}},
\vskip .2cm
{\it Dipartimento di Fisica dell'Universit\`a di Bologna, 
I-40126 Bologna, Italy} 
\vskip .1cm
{\it and Institut f\"ur Theoretische Teilchenphysik,
Universit\"at Karlsruhe, \\ D-76128 Karlsruhe, Germany}
\vskip .2cm
{\large E.~Remiddi\footnote{Email: {\tt Ettore.Remiddi@bo.infn.it}}}
\vskip .2cm
{\it Dipartimento di Fisica dell'Universit\`a di Bologna
and INFN Sezione di Bologna, I-40126 Bologna, Italy}
\end{center}
\vskip 1.2cm


\begin{abstract}
We present the closed analytic expression of the form factors of the 
two-loop QED vertex amplitude for on-shell electrons of finite mass $m$ 
and arbitrary momentum transfer $S=-Q^2$. The calculation is carried 
out within the continuous $D$-dimensional regularization scheme, with 
a single continuous parameter $D$, the dimension of the space-time, 
which regularizes at the same time UltraViolet (UV) and InfraRed (IR) 
divergences. The results are expressed in terms of 1-dimensional 
harmonic polylogarithms of maximum weight 4.
\vskip .7cm
{\it Key words}: Feynman diagrams, Multi-loop calculations, Vertex 
diagrams

{\it PACS}: 11.15.Bt, 12.20.Ds
\end{abstract}
\vfill
\end{titlepage}    



\section{Introduction}

This paper is devoted to the evaluation of the 2-loop contributions 
(second order in the expansion in terms of the fine-structure constant) 
of the form factors of the QED vertex amplitude, for arbitrary momentum
transfer $S=-Q^2$ and on-shell external fermion lines, of finite mass 
$m$, in the continuous $D$ regularization scheme. 

The analytic calculation of the imaginary parts of the 
form factors at two-loop level for arbitrary momentum transfer, 
together with the value of the charge slope of the electron, 
were obtained long ago 
in \cite{2loop1} using the Pauli-Villars regularization scheme for the 
ultraviolet (UV) divergences, and giving a small fictitious mass 
$\lambda$ to the photon for the regularization of the soft infrared 
(IR) divergences. The results were given in terms of Nielsen's 
polylogarithms \cite{Nielsen,Kolbig} of maximum weight 3. The properly 
subtracted dispersion relations for the evaluation of the corresponding 
real parts were also written, but their explicit analytic evaluation 
was not carried out in \cite{2loop1}, because the results could not be
expressed in terms of Nielsen's polylogarithms only. 

In \cite{Pie} that analytic integration of the dispersion relations 
could at last be performed, expressing the real parts 
of the 2-loop form factors in terms of the 
1-dimensional harmonic polylogarithms (HPLs), introduced in the 
meanwhile \cite{Polylog,Polylog3}, of maximum weight $w=4$.

In this paper we present the calculation of the real and imaginary parts
of the two-loop form factors within the framework of dimensional
regularization \cite{DimReg}. Both UV and soft IR divergences are 
regularized in terms of the same parameter $D$, the continuous 
dimension of the space-time. 

The Feynman diagrams involved are shown in Fig. \ref{fig1}. 
The fermion lines carry momenta $p_1$ and $p_2$ and are both incoming 
(as in the kinematical case of electron-positron annihilation), 
the outgoing photon has momentum $Q=p_1 + p_2$. 
The electron mass is $m$ and the mass-shell condition for the two 
fermions is $p_1^2 = p_2^2 = -m^2$~. 

%%%%%%%%%%%%%%%%%%%% 2-loop Vertex %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{fmfgraph*}(30,30)
\fmfleft{i1,i2}
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\fmf{fermion,right}{v3,v4}
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%
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\vspace*{8mm}
\caption{\label{fig1} 2-loop vertex diagrams for the QED form factors.
The fermionic external lines are on the mass-shell
$p_{1}^{2}=p_{2}^{2}=-m^2$; the wavy line on the r.h.s. carries 
momentum $Q=p_{1}+p_{2}$, with $Q^{2}=-S$. The arrows label the flow of
the momenta $p_1$ and $p_2$. }
\ec
\efig
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

On Lorentz-invariance grounds, all the vertex diagrams can be expressed 
in terms of at most three factors, corresponding to the three vectors 
proportional to the Dirac matrices $\gamma^{\mu}$, 
to $\sigma^{\mu \nu} Q_{\nu}$, with 
$ \sigma^{\mu \nu} = - \frac{i}{2}[ \gamma^\mu \gamma^\nu 
                                  - \gamma^\nu \gamma^\nu ] $ 
and to $Q^{\mu}$ (in QED the third form factor is of course vanishing, when
the contributions of the various graphs are summed). The on mass-shell 
form factors are functions of the momentum transfer $S=-Q^2$ and of the 
mass of the fermions $m$. Their value, when projected out from the 
Feynman graph amplitudes in the continuous $D$-dimensional regularization 
scheme, is a combination of the scalar integrals associated to the Feynman 
graphs. In \cite{Bon1} all those integrals were separately expressed 
in terms of 17 independent scalar integrals, called Master Integrals 
(MIs), by means of a reduction algorithm based on the integration by 
part identities (IBPs) \cite{Chet}, Lorentz invariance identities (LI) 
\cite{Rem3} and general symmetry relations, implemented for the computer
language FORM \cite{FORM}. The MIs were then calculated in ~\cite{Bon1} 
by the differential equations method \cite{Kot,Rem1,Rem2}.

In this paper we use those results in order to evaluate the explicit 
analytic value of the form factors for any of the (unrenormalized) 
diagrams of Fig. \ref{fig1} and for the full renormalized vertex amplitude 
as well.

The paper is structured as follows.

After a general introduction on the QED form factors, recalled in section 
\ref{Intro}, in section \ref{diagrams} we give the unrenormalized form 
factors for each of the 2-loop Feynman diagrams entering in the calculation
of the 2-loop QED vertex amplitude, within the $D$-dimensional 
regularization scheme. In section \ref{renormalization} the 
subtractions for the renormalization of UV divergences at the second 
order in the fine-structure constant are listed. In section 
\ref{fullyren} we present the full UV-renormalized form factors for the 
2-loop QED vertex amplitude in the space-like region $-S=Q^2>0$; we also 
discuss the analytic continuation to the physical region $S=-Q^2 > 4m^2$,
presenting the imaginary parts of the form factors. In sections 
\ref{Q2>>a} and \ref{Q2<<a} the behaviours of the form factors for large
and small momentum transfer are given (recovering in particular the 
two loop values of the electron $(g-2)$ and of the charge form factor 
slope). In the appendix \ref{app1} we list the definition of the 
propagators used in the explicit calculations, and in appendix 
\ref{app2} we give the 1-loop contributions to the QED vertex form 
factors up to first order in $(D-4)$.


\section{The QED form factors \label{Intro}}

Let us call $V^{\mu}(p_1,p_2)$ the QED vertex amplitude,
corresponding to the annihilation of an electron and a positron, of
momenta $p_1$ and $p_2$, with the two particles on the
mass-shell ($p_{1}^{2}=p_{2}^{2}=-m^2$). Let us define the following
two vectors:
\be
Q^{\mu} = p_{1}^{\mu}+p_{2}^{\mu} \, , \qquad 
\Delta^{\mu} = p_{1}^{\mu}-p_{2}^{\mu} \, , 
\label{b0001}
\ee
such that $Q^{2}=-S$, where $S$ is the c.m. energy squared; in the 
following we will also use the related dimensionless variables 
\be
q^2 = \frac{Q^2}{m^2} \ , \hskip 2cm s = \frac{S}{m^2} \ . 
\label{defq2ands} 
\ee 

In general $V^{\mu}(p_1,p_2)$ can be expressed in terms of three 
dimensionless scalar form factors $F_i(q^2), i=1,2,3$, depending 
only on the dimensionless variable $q^2$ of Eq. (\ref{defq2ands}), 
as follows: 
\begin{eqnarray} 
\hspace*{-0.5cm} 
 V^{\mu}(p_1,p_2) &=& \bar{v}(p_2) \Gamma^{\mu}(p_1,p_2) u(p_1) 
                     \\ 
\hspace*{-0.5cm}  \Gamma^{\mu}(p_1,p_2) &=& \left[ F_{1}(q^2) \ \gamma^{\mu} 
  + \frac{1}{2m} F_{2}(q^2) \ \sigma^{\mu \nu} Q_{\nu} 
  - \frac{i}{m} F_{3}(q^2) \ Q^{\mu} \right] \ ,  \label{b0002} 
\end{eqnarray} 
where $\bar{v}(p_2), u(p_1) $ are the spinor wave functions of the 
positron and the electron, 
$\sigma^{\mu \nu} = - \frac{i}{2}[ \gamma^{\mu},\gamma^{\nu}]$. 
Usually $F_1(q^2)$ is known as the {\it charge} (Dirac) {\it form factor} 
whereas $F_2(q^2)$ as the {\it magnetic} (Pauli) {\it form factor}.
If we consider single Feynman diagrams, any form factors in Eq.
(\ref{b0002}) can be in general different from zero. However, if
$\Gamma^{\mu}(p_1,p_2)$ is the full vertex amplitude, the conservation
of the electromagnetic current forces the third form factor to vanish,
$F_{3}(q^2) = 0$.

The extraction of each form factors $F_i(q^2)$ from Eq. (\ref{b0002}) 
can be carried out by the following general projector operators 
$P_{\mu}^{(i)}$:
\be
\! \! \! \! P_{\mu}^{(i)}(m,p_1,p_2) = 
         \frac{ -i \! \not{\! p_{1}} \! + \! m }{m} 
\biggl[   g_{1}^{(i)} \gamma_{\mu} 
        + \frac{i}{m} g_{2}^{(i)} \Delta_{\mu} 
        - \frac{i}{m} g_{3}^{(i)} Q_{\mu} \biggr] 
         \frac{ i \! \not{\! p_{2}} \! + \! m }{m} ,
\label{b0003} 
\ee 
where the constants $g_{j}^{(i)}, j=1,2,3$, are
properly chosen to have:
\be
{\rm Tr}\left( P_{\mu}^{(i)}(m,p_1,p_2) \Gamma^{\mu}(p_{1},p_{2}) \right) 
  = F_{i}(q^2) .
\label{b0004}
\ee
Let us observe that since we work in a $D$-dimensional space (to 
regularize the divergences arising in the computation) the {\it trace} 
operation is consistently  performed in $D$ dimensions as well.

The explicit values of the constants are:
\bea
g_{1}^{(1)} & = & - \frac{2}{(D-2)}\ \frac{1}{q^2+4} \, ,
\label{b0005} \\
g_{2}^{(1)} & = & - \frac{8(D-1)}{(D-2)}\ \frac{1}{(q^2+4)^2} \, ,
\label{b0006} \\
g_{1}^{(2)} & = & - \frac{8}{(D-2)}\ \frac{1}{q^{2}(q^2+4)} \, ,
\label{b0008} \\
g_{2}^{(2)} & = & - \frac{8}{(D-2)}\ \frac{1}{(q^2+4)^2} \left[
\frac{4}{q^2} + D-2 \right]  \, ,
\label{b0009} \\
g_{3}^{(3)} & = & - \frac{2}{q^{2}(q^2+4)}  \, ,
\label{b00013} 
\eea
and $g_{3}^{(1)} = g_{3}^{(2)} = g_{1}^{(3)} = g_{2}^{(3)} = 0$.

As the spinor traces are in $D$-dimensional space-time, 
in all the above formulas Eqs. (\ref{b0005}-\ref{b00013}), 
all the r.h.s., strictly speaking, should be multiplied by the overall 
constant $(1/4)\ {\rm Tr} {\bf 1}$, where ${\rm Tr} {\bf 1}$ is the trace 
of the unit 
Dirac matrix in $D$-continuous dimensions. The overall constant is in fact 
undetermined for arbitrary $D$, except for its limiting value at $D=4$, 
which is $1$. For simplicity, we will therefore omit systematically 
that overall factor. 

In QED, the form factors are given as an expansion in powers 
of $(\alpha/\pi)$, $\alpha=e^2/4\pi$ being the fine-structure 
constant. Showing explicitly the dependence on the regularizing 
dimension, we write the expansion as 
\bea
F_{1}(D,q^2) & = & 
   1 + \left( \frac{\alpha}{\pi} \right) F^{(1l)}_{1}(D,q^2) 
     + \left( \frac{\alpha}{\pi} \right)^{2} F^{(2l)}_{1}(D,q^2) 
     + {\mathcal O} \Bigg( \left( \frac{\alpha}{\pi} \right) ^{3} \Bigg)
     \, , 
\label{b00020} 
\\
F_{2}(D,q^2) & = & \quad \; \; \,
       \left( \frac{\alpha}{\pi} \right) F^{(1l)}_{2}(D,q^2) 
     + \left( \frac{\alpha}{\pi} \right) ^{2} F^{(2l)}_{2}(D,q^2) 
     + {\mathcal O} \Bigg( \left( \frac{\alpha}{\pi} \right) ^{3} \Bigg) 
     \, , 
\label{b00021} 
\\
F_{3}(D,q^2) & = & \quad \; \; \, 
       \left( \frac{\alpha}{\pi} \right) F^{(1l)}_{3}(D,q^2) 
     + \left( \frac{\alpha}{\pi} \right) ^{2} F^{(2l)}_{3}(D,q^2) 
     + {\mathcal O} \Bigg( \left( \frac{\alpha}{\pi} \right) ^{3} \Bigg)
     \, ,
\label{b00022} 
\eea
where the superscripts ``$1l$'' and ``$2l$'' stand for 1- and 2-loop
contributions, the first term $1$ in $F_{1}(D,q^2)$ is the tree 
approximation and $F_{3}(D,q^2)$ vanishes, as already said, when 
all the graphs are summed up. 

In the following we will focus on the detailed 2-loop computations of  
$F^{(2l)}_1(D,q^2)$, $F^{(2l)}_2(D,q^2)$ and $F^{(2l)}_{3}(D,q^2)$. We 
will present the unrenormalized form factors for each Feynman diagram 
and we will discuss the renormalization procedure, giving the fully 
UV-renormalized result for the QED vertex amplitude. The 1-loop
contributions to the form factors, $F^{(1l)}_{1}(D,q^2)$  and 
$F^{(1l)}_{2}(D,q^2)$, 
%which are needed for the renormalization of the 
%inserted 1-loop subgraphs, 
will be recalled in appendix \ref{app2} for completeness. 

\section{ Unrenormalized contributions \label{diagrams} }

The diagrams which contribute to the order $(\alpha/\pi)^2$ are shown 
in Fig. \ref{fig1}. Following Eq. (\ref{b0002}), we will indicate by 
${\mathcal F}^{(2l,{\tt graph})}_{i}(D,q^2) \ (i=1,2,3),\ 
{\tt graph} \in \{ {\tt a,..,g} \}$ the contribution of each 
unrenormalized graph to the unrenormalized form factors: 
\bea
\! \! \! \! \! \! \left( \underbrace{
\parbox{20mm}{
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{fmfgraph*}(15,15)\fmfkeep{generic}
\fmfforce{0.5w,0.5h}{c1}
\fmfleft{i1,i2}
\fmfright{o1}
\fmfblob{.7w}{c1}
\fmf{plain}{i1,c1}
\fmf{plain}{i2,c1}
\fmf{photon}{c1,o1}
\end{fmfgraph*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
} \hspace*{-.5cm}
}_{{\tt graph} \ \in \ \{ {\tt a,...,g} \} }
\right)^{\mu}
& = & \ \bar{v}(p_2) \Bigg[ {\mathcal F}^{(2l,{\tt graph})}_{1}(D,q^2) 
\, \gamma^{\mu} + \frac{1}{2m} 
{\mathcal F}^{(2l,{\tt graph})}_{2}(D,q^2) \, 
\sigma^{\mu \nu} Q_{\nu} \nn\\
\! \! \! \! \! \! & & \qquad \quad - \frac{i}{m} 
    {\mathcal F}^{(2l,{\tt graph})}_{3}(D,q^2) \ Q^{\mu} \Bigg]u(p_1) 
    \ . 
\label{eq:unrenFF}
\eea

In so doing, each ${\mathcal F}^{(2l,{\tt graph})}_{i}(D,q^2)$ can be 
extracted by the $D$-dimensional projection defined in Eq. 
(\ref{b0004}). After the computation of the {\it trace}, it turns out
that each form factor is expressed in terms of several hundreds of
scalar integrals. According to 
\cite{Bon1}, one can express all those integrals in terms of only 17 
Master Integrals (MIs) {\it via} integration-by-parts identities, 
Lorentz invariance and general symmetry relations (exactly in $D$). 

As an example consider the {\it magnetic} form factor of the diagram 
(g) in Fig. \ref{fig1}. It can be written as a linear combination of 5 
MIs, the coefficients being  ratios of simple polynomials in $D$ 
and $q^2$, the fermion mass squared $m^2$ appearing as a dimensional 
scale factor. (It is to be noted that in the formulas which follow, 
and which are exact in $D$, the dimensionless variables $q^2$ and 
$D$ are never entangled in a same non factorisable polynomial in the 
denominators). 
\bea
\hspace*{-6mm} {\mathcal F}^{(2l,{\tt g})}_{2}(D,q^2) & = & 
\frac{1}{(q^2+4)} 
\Biggl\{
\frac{32(D-4)}{(D-5)} \nn\\
\hspace*{-6mm} & & - \frac{64(D^2-10D+19)}{(D-1)(D-5)} 
\frac{1}{(q^2+4)} 
\Biggr\}
\, \, \parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v3,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v1,v3}
\fmf{plain,tension=0,right=.5}{v2,v1}
\fmf{plain,tension=0,right=.5}{v1,v2}
\end{fmfgraph*} } \nn\\
\hspace*{-6mm} & & - \frac{64(D^3 \! - \! 8D^2 \! + \! 23D \! - \! 
26)}{(D-1)(D-5)} \frac{1}{(q^2 \! + \! 4)^2} \left( \frac{1}{m^2}
\, \, \parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v3,o}
\fmflabel{$\! (p_{2} \! \cdot \! k_{1})$ }{o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v1,v3}
\fmf{plain,tension=0,right=.5}{v2,v1}
\fmf{plain,tension=0,right=.5}{v1,v2}
\end{fmfgraph*} } \hspace*{14mm} \right) \nn\\
\hspace*{-6mm} & & - \frac{1}{(q^2+4)} 
\Biggl\{
\frac{4(D^4-14D^3+59D^2-82D+16)}{(D-1)(D-4)(D-6)} \nn\\
\hspace*{-6mm} & & + \frac{16(3D-8)(D^2-10D+19)}{(D-1)(D-4)(D-5)} 
\frac{1}{(q^2+4)} 
\Biggr\} \left( \frac{1}{m^2} \, \, 
\parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{plain}{i,v1}
\fmf{plain}{v2,o}
\fmf{plain,tension=.15,left}{v1,v2}
\fmf{plain,tension=.15}{v1,v2}
\fmf{plain,tension=.15,right}{v1,v2}
\end{fmfgraph*} } \right) \nn\\
\hspace*{-6mm} & & - \frac{32(D-2)(D^2-10D+19)}{(D-1)(D-5)} 
\frac{1}{(q^2+4)^2}
\, \, \left( \frac{1}{m^2} \, \, 
\parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{photon}{i,v1}
\fmf{photon}{v2,o}
\fmf{plain,tension=.22,left}{v1,v2}
\fmf{plain,tension=.22,right}{v1,v2}
\fmf{plain,right=45}{v2,v2}
\end{fmfgraph*} } \right) \nn\\
\hspace*{-6mm} & & - \frac{2(D - 2)^2}{(q^2 + 4)} 
\Biggl\{
\frac{(3D^4- 53D^3 + 323D^2-795D+642)}{(D-1)(D-3)(D-4)(D-5)(D-6)}
\nn\\
\hspace*{-6mm} & & + 
\frac{4(2D-5)(D^2-10D+19)}{(D-1)(D-3)(D-4)(D-5)} \frac{1}{(q^2+4)} 
\Biggr\} \! \! 
\left( \frac{1}{m^4} \! \! \! \! 
\parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{phantom}{i,v1}
\fmf{phantom}{v1,o}
\fmf{plain,right=45}{v1,v1}
\fmf{plain,left=45}{v1,v1}
\end{fmfgraph*} } \hspace*{-5mm} \right) \! ,
\label{b00023}
\eea
where the MIs depicted on the r.h.s. are those of Fig. 7 of \cite{Bon1}.

Similar formulas hold for the other form factors and graphs, but are too 
lengthy to be reported here. 

Once the form factors are expressed in terms of MIs, one expands the
result around $D=4$ and inserts the values of the MIs, also given in 
\cite{Bon1} as an expansion around $D=4$. In this way one finally obtains 
the required analytic result where 
both UV and soft IR divergences, regulated by the same parameter
$D$, appear as poles in $(D-4)$. In the case of Eq. (\ref{b00023}),
using the Eqs. (88,93,123,125,B.1) of \cite{Bon1}, one will get the 
expression given later in this paper in Eq. (\ref{ex2}).

In this section we will give the contributions 
${\mathcal F}^{(2l,{\tt graph})}_{i}(D,q^2)$ to the 
{\it unrenormalized} form factors from each of the 2-loop 
still unrenormalized graphs (i.e. where the renormalization of the 
inserted 1-loop subgraphs has not yet been carried out). 

The propagators of the graphs are considered in the Feynman gauge and 
the corresponding denominators ${\mathcal D}$'s, which will appear in 
the following formulas, are listed in the appendix \ref{app1}.

The resulting {\it unrenormalized} form factors are given for space-like 
$Q$ ($Q^2>0$ or $S = -Q^2 <0$) and are expressed in terms of 
1-dimensional harmonic polylogarithms \cite{Polylog,Polylog3} of argument 
\be
x = \frac{\sqrt{q^2+4} - \sqrt{q^2} }{\sqrt{q^2+4} + \sqrt{q^2} } 
  = \frac{\sqrt{Q^2+4m^2} - \sqrt{Q^2} }{\sqrt{Q^2+4m^2} + \sqrt{Q^2} } 
\label{b00015}
\ .
\ee

Let us comment shortly here the normalization of the $D$-dimensional 
integrals, or, which is the same, the choice of the loop integration 
measure in $D$ continuous dimensions. With the natural choice 
\be 
\mu_{0}^{(4-D)} \int \frac{d^D k}{(2\pi)^{(D-2)}} \, ,
\label{b00016}
\ee 
where $\mu_0$ is the mass scale, 
one obtains for the simplest loop integral, the 1-loop massive tadpole, 
\be
\mu_{0}^{(4-D)} \int \frac{d^D k}{(2\pi)^{(D-2)}} \frac{1}{k^2+m^2} =
C(D) \, \left( \frac{m^2}{\mu_0^{2}} \right)^{\frac{D-4}{2}} \,
\frac{m^2}{(D-2)(D-4)} \, ,
\label{b00017}
\ee
where $C(D)$ is the following function of the space-time dimension $D$:
\be
C(D) = (4 \pi)^{\frac{(4-D)}{2}} \Gamma \left( 3 - \frac{D}{2} \right) 
\label{b00018} \, ,
\ee
with the limiting value $C(4)=1$ for $D=4$. Note that the explicit 
form of $C(D)$ is essentially irrelevant, as in any physical quantity, 
finite for $D=4$, $C(D)$ can be replaced by 1. The explicit expression 
in Eq. (\ref{b00018}) can matter only in detailed comparisons with 
calculations using a different integration measure. 

To simplify the writing of all the subsequent formulae, we will use 
the following $D$-continuous integration measure 
\be 
\int{\mathfrak D}^Dk = \frac{1}{C(D)} 
\left( \frac{m^2}{\mu_0^{2}} \right)^{\frac{4-D}{2}} 
\int \frac{d^D k}{(2\pi)^{(D-2)}} \, ,
\label{b00016a} 
\ee 
using $m$ as mass scale. With that choice, the 1-loop tadpole simply reads 
\be 
\int{\mathfrak D}^Dk \ \frac{1}{k^2+m^2} = \frac {m^2}{(D-2)(D-4)} \ . 
\label{b00017a} 
\ee 

By using for each loop the integration measure ${\mathcal D}^Dk$ of 
Eq. (\ref{b00016a}) we find the results which follow for the contributions 
to the unrenormalized form factors from the various 2-loop graphs, with 
still unrenormalized 1-loop insertions. 

$\bullet$ The {\it Ladder} graph {\tt(a)} of Fig. \ref{fig1}, defined as
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v1}
\fmf{photon,tension=0}{v4,v3}
\end{fmfgraph*} } & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2 \ 
\frac{{\mathcal N}_{({\tt a})}^{\mu}}{{\mathcal D}_{1} {\mathcal D}_{2} 
{\mathcal D}_{9} {\mathcal D}_{10} {\mathcal D}_{12} 
{\mathcal D}_{13} } \ , 
\label{b1} 
\eea
where 
\bea
\! \! \! \! \! \! {\mathcal N}_{({\tt a})}^{\mu} & = & \bar{v}(p_2)  \gamma_{\sigma}
[i ( \not{\! p_{2}} + \! \! \not{\! k_{1}}) \! + \! m] 
\gamma_{\lambda} 
[i ( \not{\! p_{2}} + \!  \! \not{\! k_{1}} - \!  \! \not{\! k_{2}}) 
\! + \! m] \gamma^{\mu}
[-i ( \not{ \! p_{1}} -  \! \! \not{\! k_{1}} +  \! \! \not{\! k_{2}})
\! + \! m] \times
\nn\\
\! \! \! \! \! \! & & \times
\gamma^{\lambda} 
[-i ( \not{\! p_{1}} -  \! \! \not{\! k_{1}} ) \!  + \! m]
\gamma^{\sigma} u(p_1)  ,
\eea
gives: 
\bea
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt a})}_{1}(D,q^2) & = & 
            \frac{1}{(D-4)^{2}} \Biggl\{ 
            \frac{1}{8} \! 
          + \! \frac{1}{2} \biggl[ 1 \!  
          - \frac{1}{(1\! -\! x)} \! 
   - \frac{1}{(1\! +\! x)} \biggr] H(0;x) \! 
          + \biggl[ 1 \! + \! \frac{1}{(1\! -\! x)^2} 
          \nn\\
\hspace*{-5mm} & &  \hspace*{20mm}     
          - \frac{1}{(1-x)}
          + \frac{1}{(1+x)^2}
          - \frac{1}{(1+x)} \biggr] H(0,0;x)
\Biggr\} \nn\\
\hspace*{-5mm} & & \nn\\
%
%
%
\hspace*{-5mm} & + & \frac{1}{(D-4)} \Biggl\{
          - \frac{5}{32} 
   - \frac{1}{8} \biggl[ 1 - \frac{2}{(1-x)}
   \biggr] H(0;x) - \frac{1}{2} \biggl[  1 - \frac{2}{(1-x)} \nn\\
\hspace*{-5mm} & & \hspace*{18mm}        + \frac{2}{(1 \!  - \!  x)^2}
   \biggr] H(0,0;x) 
          +  \!  \frac{1}{4} \biggl[ 1 \!  
          -  \!  \frac{1}{(1 \!  - \!  x)} \! 
   -  \!  \frac{1}{(1 \!  + \!  x)} \biggr] \bigl[ 
     \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{18mm}     
          - H(0;x) - H(0,0;x) + 2 H(-1,0;x) 
\bigr] \nn\\
\hspace*{-5mm} & &  \hspace*{18mm}       
          + \frac{1}{2} \biggl[ 1  \! +  \! \frac{1}{(1 \! - \! x)^2} 
          -  \! \frac{1}{(1 \! - \! x)}
          +  \! \frac{1}{(1 \! + \! x)^2}  
   -  \! \frac{1}{(1 \! + \! x)} \biggr] \bigl[ \zeta(3)\nn\\
\hspace*{-5mm} & & \hspace*{18mm}      
          - 2 H(0,0;x) - 2 H(0,0,0;x) + 4 H(-1,0,0;x)\nn\\
\hspace*{-5mm} & & \hspace*{18mm}        
          + 4 H(0,1,0;x) \bigr]
\Biggr\}   \nn\\
\hspace*{-5mm} & & \nn\\
\hspace*{-5mm} & & 
%
          + \frac{55}{128}
          + \biggl[ \frac{5}{4(1-x)} 
          - \frac{81}{2(1+x)^4} 
          + \frac{177}{2(1+x)^3} 
          - \frac{123}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &           + \frac{55}{4(1+x)} \! 
          -  \! \frac{23}{8} \biggr] \zeta(2) \! 
          +  \! \biggl[ \frac{3}{5(1-x)^2} \! 
          -  \! \frac{31}{80(1-x)} \! 
          -  \! \frac{51}{2(1+x)^5} \nn\\
\hspace*{-5mm} & &           +  \! \frac{255}{4(1 \! + \! x)^4} \! 
          -  \! \frac{1071}{20(1 \! + \! x)^3} \! 
          +  \! \frac{687}{40(1 \! + \! x)^2} \!
          -  \! \frac{303}{80(1 \! + \! x)} \!
          +  \! \frac{29}{20} \biggr] \zeta^{2}(2) \nn\\
\hspace*{-5mm} & &           - \biggl[ \frac{1}{(1-x)^2}
          - \frac{5}{4(1-x)} 
          - \frac{60}{(1+x)^4}
          + \frac{120}{(1+x)^3}
          - \frac{141}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &           + \frac{41}{4(1+x)} 
   - 2 \zeta(3) \biggr] \zeta(3)
   - \frac{3}{4(1+x)} \biggl[ 1 
   - \frac{1}{(1+x)}  \biggr] \nn\\
\hspace*{-5mm} & &      
%
       +  \biggl[
            \frac{19}{32} 
 - \biggl( \frac{2}{(1-x)^2}
          - \frac{33}{32(1-x)} \!
          + \frac{81}{2(1+x)^5} \!
          - \frac{465}{4(1+x)^4} \nn\\  
\hspace*{-5mm} & &      \hspace*{5mm}
          + \frac{979}{8(1+x)^3} 
   - \frac{865}{16(1+x)^2}
          + \frac{327}{32(1+x)}
          - \frac{9}{8} \biggr) \zeta(2) \nn\\  
\hspace*{-5mm} & &      \hspace*{5mm}
        - \biggl( \frac{1}{4(1-x)^2}  \!
          +  \!\frac{1}{4(1-x)} \!
          -  \!\frac{60}{(1+x)^5} \!
          +  \!\frac{150}{(1+x)^4} \!
          -  \!\frac{126)}{(1+x)^3} \nn\\  
\hspace*{-5mm} & &      \hspace*{5mm}
          + \frac{157}{4(1+x)^2}  
          - \frac{31}{4(1+x)}
          + \frac{9}{4} \biggr) \zeta(3)
          - \frac{1}{(1-x)}
          + \frac{9}{4(1+x)^3} \nn\\  
\hspace*{-5mm} & &      \hspace*{5mm}
          - \frac{27}{8(1+x)^2}
          + \frac{15}{16(1+x)}
          \biggr] H(0;x)  \nn\\
%
\hspace*{-5mm} & &        
     -  \biggl[
           \biggl( \frac{1}{4(1-x)}
          + \frac{1}{4(1+x)} 
          - \frac{1}{4} \biggr) \zeta(2)
          - \biggl( 1
   +  \! \frac{1}{(1 \! - \! x)^2}
          - \frac{1}{(1 \! - \! x)} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}        
          + \frac{1}{(1+x)^2} \!
          - \frac{1}{(1+x)}
          \biggr) \zeta(3)
          \biggr] H(-1;x) \nn\\
%
& &        +  \biggl[
            \frac{27}{8}   \! 
   + \! \biggl( \frac{7}{4(1 \! - \! x)^2}  
   -  \! \frac{13}{8(1 \! - \! x)}  \!
          -  \! \frac{15}{(1 \! + \! x)^5} \!
          + \! \frac{75}{2(1 \! + \! x)^4}   \!
   -  \! \frac{63}{2(1 \!+ \!x)^3}   \nn\\
\hspace*{-5mm} & & \hspace*{5mm}         
          + \frac{23}{2(1+x)^2} 
          - \frac{29}{8(1+x)}  
          + \frac{9}{4} \biggr)  \zeta(2)
          + \frac{17}{8(1-x)^2} 
          - \frac{27}{8(1-x)}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}     
          + \frac{39}{2(1+x)^4}    
   - \frac{93}{2(1+x)^3}  \!
          + \frac{77}{2(1+x)^2}  \! 
          - \frac{47}{4(1+x)}  \!
          \biggr] H(0,0;x)  \nn\\
\hspace*{-5mm} & &  
%
       +  \biggl[ \biggl(
            \frac{1}{2(1-x)}
   - \frac{60}{(1+x)^5}
          + \frac{150}{(1+x)^4}
          - \frac{126}{(1+x)^3}
          + \frac{39}{(1+x)^2} \nn\\
\hspace*{-5mm} & & \hspace*{5mm}      
          - \frac{15}{2(1+x)}  
   + 2 \biggr) \zeta(2)
          \biggr] H(1,0;x) \nn\\
%
\hspace*{-5mm} & &  
       +  \biggl[ \biggl( 1  \! 
          +  \! \frac{1}{2(1 \! - \! x)^2} 
          -  \! \frac{1}{2(1 \! - \! x)}  
          +  \! \frac{1}{2(1 \! + \! x)^2}  
          -  \! \frac{1}{2(1 \! + \! x)}  \biggr) \zeta(2)
          \biggr] \times \nn\\
\hspace*{-5mm} & &  \hspace*{5mm} \times \bigl[ H(0,1;x) 
+ 2 H(-1,0;x) + H(0,-1;x) \bigr] \nn\\
\hspace*{-5mm} & &  
%
       -  \biggl[
            \frac{3}{2} \! 
          -  \! \frac{5}{2(1 \! - \! x)}   \! 
   -  \! \frac{15}{(1 \! + \! x)^3} \! 
          +  \! \frac{45}{2(1 \! + \! x)^2}  \! 
          -  \! \frac{8}{(1 \! + \! x)}
          \biggr] H(-1,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
          \frac{35}{8} \! 
          +  \! \frac{1}{(1 \! - \! x)^2}  \! 
          -  \! \frac{71}{32(1 \! - \! x)} \! 
          - \! \frac{81}{2(1 \! + \! x)^5} \! 
          + \! \frac{525}{4(1 \! + \! x)^4} \!
          - \! \frac{1219}{8(1 \! + \! x)^3} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm}    
          + \frac{1173}{16(1 \! + \! x)^2}
          - \frac{471}{32(1 \! + \! x)} 
          \biggr] H(0,0,0;x) \nn\\
%
& &        -  \biggl[
            \frac{25}{4} \!
          + \! \frac{4}{(1-x)^2}  \!
          - \! \frac{17}{4(1-x)} \!
          + \! \frac{60}{(1+x)^4} \!
          - \! \frac{120}{(1+x)^3} \!
          + \! \frac{73}{(1+x)^2} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm}    
          - \frac{53}{4(1+x)}
          \biggr] H(-1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        +  \frac{1}{2} \biggl[
            1 
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H( -1, -1, 0;x) \nn\\
%
\hspace*{-5mm} & &        +  \biggl[
          \frac{5}{4}
          + \frac{1}{4(1-x)} 
          + \frac{30}{(1+x)^4}
          - \frac{60}{(1+x)^3}
          + \frac{71}{2(1+x)^2} \nn\\
%
\hspace*{-5mm} & &     \hspace*{5mm}    
          - \frac{21}{4(1+x)}
          \biggr] H(0,-1,0;x) \nn\\
%
\hspace*{-5mm} & &        -  \biggl[ 
            \frac{3}{2} \!
          + \frac{2}{(1 \! - \! x)^2} \!
          - \frac{2}{(1 \! - \! x)} \!
          + \frac{1}{(1 \! + \! x)^2} \!
          - \frac{1}{(1 \! + \! x)} \!
          \biggr] H(0,1,0;x)  \nn\\
%
\hspace*{-5mm} & &        +  \biggl[ 
          \frac{3}{2} \!
          + \! \frac{30}{(1 \! + \! x)^4} \!
          - \! \frac{60}{(1 \! + \! x)^3} \!
          + \! \frac{71}{2(1 \! + \! x)^2} \!
          - \! \frac{11}{2(1 \! + \! x)} \!
          \biggr] H(1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        +  \frac{1}{4} \biggl[
          1 
          +  \frac{1}{(1-x)^2} 
          -  \frac{1}{(1-x)} 
          +  \frac{1}{(1+x)^2} 
          -  \frac{1}{(1+x)}  
          \biggr] \times \nn\\
%
\hspace*{-5mm} & &   \hspace*{5mm}  
             \times \bigl[ 5 H(0,0,0,0;x)
          + 16 H(-1, \! -1, \! 0, \! 0;x) 
          - 4  H(-1, \! 0, \! 0, \! 0;x)\nn\\
%
\hspace*{-5mm} & &   \hspace*{5mm}  
   + 2  H(-1, \! 0, \! 1, \! 0;x)  \!
   - \! 16 H(0, \! -1, \! -1, \! 0;x)  \!
   + \! 3  H(0, \! -1, \! 1, \! 0;x) \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}     
          - \! 12 H(0,0,1,0;x) \! 
   + \! 12 H(0,1, \! -1,0;x) \! 
   - \! 4 H(0,1,1,0;x) \bigr] \nn\\  
%
\hspace*{-5mm} & &        +  \biggl[
            \frac{7}{2} \! 
          +  \! \frac{3}{2(1-x)^2}  \! 
          -  \! \frac{1}{(1-x)} \! 
          -  \! \frac{60}{(1+x)^5} \! 
          +  \! \frac{150}{(1+x)^4}    \!        
          -  \! \frac{126}{(1+x)^3}\nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          + \frac{81}{2(1+x)^2} 
          - \frac{9}{(1+x)}
          \biggr] H(0,-1,0,0;x)  \nn\\
%
\hspace*{-5mm} & &        +  \biggl[
            \frac{5}{2} \! 
          +  \! \frac{7}{2(1-x)^2}  \! 
          -  \! \frac{15}{4(1-x)}  \! 
          +  \! \frac{30}{(1+x)^5} \! 
          -  \! \frac{75}{(1+x)^4}   \!         
          +  \! \frac{63}{(1+x)^3} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          - \frac{16}{(1+x)^2}
          + \frac{1}{4(1+x)} 
          \biggr] H(0,0,-1,0;x) 
%
 \nn\\
%
\hspace*{-5mm} & &        -  \biggl[
            3
          + \frac{2}{(1-x)^2}
          - \frac{7}{4(1-x)} 
          - \frac{30}{(1+x)^5}
          + \frac{75}{(1+x)^4}           
          - \frac{63}{(1+x)^3} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm} 
          + \frac{43}{2(1+x)^2} 
          - \frac{23}{4(1+x)} 
          \biggr] H(0,1,0,0;x)  \nn\\
%
\hspace*{-5mm} & &        +  \biggl[
            2 
          + \frac{1}{2(1-x)} 
          - \frac{60}{(1+x)^5}
          + \frac{150}{(1+x)^4}
          - \frac{126}{(1+x)^3}          
          + \frac{39}{(1+x)^2} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          - \frac{15}{2(1+x)} 
          \biggr] H(1,0,0,0;x)  \nn\\
\hspace*{-5mm} & + & 
  {\mathcal O} (D-4) \, , \\
 %
 %
 %
 %
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt a})}_{2}(D,q^2) & = & 
          \frac{1}{(D-4)} \Biggl\{ 
%
        \frac{1}{4} \biggl[
            \frac{1}{(1-x)} \! 
          - \! \frac{1}{(1+x)}
          \biggr] H(0;x) \! 
%
       -  \biggl[
            \frac{1}{(1-x)^2}  \! 
   - \! \frac{1}{(1-x)}  
         \nn\\
\hspace*{-5mm} & &  \hspace*{18mm}          
          - \frac{1}{(1+x)^2}
          + \frac{1}{(1+x)}
          \biggr] H(0,0;x)
\Biggr\}   \nn\\
\hspace*{-5mm} & & \nn\\
\hspace*{-5mm} & & 
%
          +  \biggl[ \frac{7}{8(1-x)^2}  \! 
          -  \! \frac{5}{4(1-x)}  \! 
          +  \! \frac{81}{2(1+x)^4}  \! 
          -  \! \frac{177}{2(1+x)^3}  \! 
   +  \! \frac{465}{8(1+x)^2} \nn\\
\hspace*{-5mm} & &     
          - \frac{39}{4(1+x)} \biggr]  \zeta(2)
          + \biggl[ \frac{17}{160(1-x)} 
          + \frac{51}{2(1+x)^5} 
   - \frac{255}{4(1+x)^4}  \nn\\
\hspace*{-5mm} & &          
          + \frac{2057}{40(1+x)^3} 
          - \frac{1071}{80(1+x)^2} 
          + \frac{17}{160(1+x)}  \biggr]  \zeta^{2}(2)           
          - \biggl[ \frac{5}{2(1-x)^2} \nn\\
\hspace*{-5mm} & & 
          - \frac{5}{2(1-x)} \! 
          + \frac{60}{(1+x)^4} \! 
          - \frac{120}{(1+x)^3} \!
   + \frac{131}{2(1+x)^2}          
          - \frac{11}{2(1 \! + \! x)} \biggr]  \zeta(3) \nn\\
\hspace*{-5mm} & &  
          + \frac{3}{4(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)} \biggr] \nn\\
%
\hspace*{-5mm} & &        +  \biggl[ \biggl(
            \frac{7}{8(1 \! - \! x)^3} \! 
          -  \! \frac{45}{16(1 \! - \! x)^2}  \! 
          +  \! \frac{87}{32(1 \! - \! x)} \!  
          +  \! \frac{81}{2(1 \! + \! x)^5}  \! 
          -  \! \frac{465}{4(1 \! + \! x)^4}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}      
          +   \frac{119}{(1+x)^3} 
          -   \frac{191}{4(1  +  x)^2}  
          +   \frac{119}{32(1  +  x)}  \biggr) \zeta(2) 
          -   \biggl( \frac{1}{4(1  -  x)}  \nn\\ 
\hspace*{-5mm} & & \hspace*{5mm}   
          +   \frac{60}{(1  +   x)^5}  
          - \frac{150}{(1 \! + \! x)^4} \! 
          + \frac{121}{(1 \! + \! x)^3} \!
          - \frac{63}{2(1 \! + \! x)^2}  \! 
          +  \! \frac{1}{4(1 \! + \! x)}  \biggr) \zeta(3) \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
          -  \! \frac{3}{8(1-x)}        
          -  \! \frac{9}{4(1+x)^3} \! 
          +  \! \frac{27}{8(1+x)^2}
   - \! \frac{3}{4(1+x)} 
          \biggr] H(0;x)   \nn\\
\hspace*{-5mm} & &    
%
       +  \biggl[ \biggl(
            \frac{1}{16(1-x)}  \! 
          +  \! \frac{15}{(1+x)^5} \! 
          -  \! \frac{75}{2(1+x)^4} \! 
          +  \! \frac{121}{4(1+x)^3}  \!   
          -  \! \frac{63}{8(1+x)^2}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}       
          + \frac{1}{16(1+x)} \biggr)  \zeta(2)
          + \frac{17}{8(1-x)^2}
          - \frac{7}{4(1-x)} 
          - \frac{39}{2(1+x)^4} \nn\\
\hspace*{-5mm} & & \hspace*{5mm}    
          + \frac{93}{2(1+x)^3} 
          - \frac{295}{8(1+x)^2}
          + \frac{19}{2(1+x)}
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       -  \biggl[ 
            \frac{3}{4(1-x)}
          + \frac{15}{(1+x)^3}
          - \frac{45}{2(1+x)^2}
          + \frac{27}{4(1+x)}
          \biggr] H(-1,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[ \biggl(
            \frac{1}{4(1-x)}
          + \frac{60}{(1+x)^5}
          - \frac{150}{(1+x)^4}
          + \frac{121}{(1+x)^3}
          - \frac{63}{2(1+x)^2}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
          + \frac{1}{4(1+x)} \biggr)  \zeta(2)
          \biggr] H(1,0;x) \nn\\ 
%
\hspace*{-5mm} & &     
       +  \biggl[
            \frac{7}{8(1-x)^3} \! 
          -  \! \frac{29}{16(1-x)^2}  \!  
          +  \! \frac{55}{32(1-x)}   \! 
   +  \! \frac{81}{2(1+x)^5}  \! 
          -  \! \frac{525}{4(1+x)^4}   \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
          + \frac{149}{(1+x)^3}
          - \frac{263}{4(1+x)^2}
   + \frac{215}{32(1+x)} 
          \biggr] H(0,0,0;x) \nn\\ 
%
\hspace*{-5mm} & &      
       +  \biggl[
            \frac{60}{(1+x)^4} \!
          - \frac{120}{(1+x)^3} \!
          + \frac{68}{(1+x)^2} 
          - \frac{8}{(1+x)} \! 
          \biggr] H(-1,0,0;x)  \nn\\ 
%
\hspace*{-5mm} & &  
       -  \biggl[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          + \frac{30}{(1+x)^4} 
          - \frac{60}{(1+x)^3}
          + \frac{33}{(1+x)^2} \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
          - \frac{3}{(1+x)}
          \biggr] \bigl[ H(0,-1,0;x) +  H(1,0,0;x) \bigr] \nn\\ 
%
\hspace*{-5mm} & &  
       -  \biggl[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)^2}
          + \frac{1}{(1+x)}
          \biggr] H(0,1,0;x)   \nn\\ 
%
\hspace*{-5mm} & &        
       -  \biggl[
            \frac{1}{8(1-x)}   
          + \frac{30}{(1+x)^5}  
          - \frac{75}{(1+x)^4}
          + \frac{121}{2(1+x)^3} 
          - \frac{63}{4(1+x)^2} \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
          + \frac{1}{8(1+x)} 
          \biggr] \bigl[ H(0,0,-1,0;x) + H(0,1,0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{25mm} 
   - 2 H(1,0,0,0;x) - 2 H(0,-1,0,0;x) \bigr] \nn\\
\hspace*{-5mm} & +&  {\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt a})}_{3}(D,q^2) & = &  0 \, . \\
\nonumber
\eea

$\bullet$ The {\it Cross} graph {\tt(b)} of Fig. \ref{fig1}, defined as
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v4}
\fmf{photon,tension=0}{v1,v3}
\end{fmfgraph*} 
}  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2 \ 
\frac{{\mathcal N}_{({\tt b})}^{\mu}}{{\mathcal D}_{1} {\mathcal D}_{2} 
{\mathcal D}_{9} {\mathcal D}_{11} {\mathcal D}_{12} 
{\mathcal D}_{13} } \, , 
\label{b2} 
\eea
where 
\bea
\! \! \! \! \! \! {\mathcal N}_{({\tt b})}^{\mu} & = &  \bar{v}(p_2)  \gamma_{\sigma}
[i ( \not{\! p_{2}} - \!  \! \not{\! k_{2}})  \! + \! m] 
\gamma_{\lambda} 
[i ( \not{\! p_{2}} + \!  \! \not{\! k_{1}} - \! \! \not{\! k_{2}})  \! 
+ \! m]
\gamma^{\mu}
[-i ( \not{\! p_{1}} -  \!  \! \not{\! k_{1}} +  \! \! \not{\! k_{2}})  
\! + \! m] \times
\nn\\
\! \! \! \! \! \! & & \times
\gamma^{\sigma} 
[-i ( \not{\! p_{1}} - \!  \! \not{\! k_{1}} )  \! + \! m]
\gamma^{\lambda} u(p_1) ,
\eea
gives: 
\bea
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt b})}_{1}(D,q^2) & = & \frac{1}{(D-4)} 
\Biggl\{   \frac{1}{4}  
          -  \frac{1}{2} \biggl[ 
     1 
   + \frac{1}{(1-x)^2}  
          - \frac{1}{(1-x)}
          + \frac{1}{(1+x)^2}  \nn\\
\hspace*{-5mm} & &   \hspace*{18mm}     
          - \frac{1}{(1+x)} \biggr] \bigl[ \zeta(3)
   - \zeta(2) H(0;x) 
   + 2 H(0,0,0;x) \nn\\
\hspace*{-5mm} & &   \hspace*{38mm} 
   - 2 H(0,-1,0;x) 
   + 2 H(0,1,0;x) \bigr]
%         
\Biggr\}   \nn\\
\hspace*{-5mm} & &
%
          - \frac{1}{2}
          + \frac{1}{2} \biggl[ \frac{1}{(1-x)} 
          - \frac{45}{(1+x)^4} 
          + \frac{99}{(1+x)^3} 
          - \frac{53}{(1+x)^2} \nn\\
\hspace*{-5mm} & &           - \frac{3}{(1+x)} \! 
          +  \! \frac{27}{4} \biggr] \zeta(2) \! 
          +  \! \biggl[ \frac{37}{40(1-x)^2}  \! 
          -  \! \frac{971}{320(1-x)}  \! 
          -  \! \frac{321}{20(1+x)^5} \nn\\
\hspace*{-5mm} & &           
          + \frac{321}{8(1+x)^4} 
          - \frac{2299}{80(1+x)^3} 
          + \frac{125}{32(1+x)^2} 
          - \frac{79}{64(1+x)} \nn\\
\hspace*{-5mm} & &           
          + \frac{119}{40} \biggr]  \zeta^{2}(2)
          -  \biggl[ \frac{6}{(1+x)^4}
          - \frac{12}{(1+x)^3}
          + \frac{7}{(1+x)^2}
          - \frac{1}{(1+x)} \nn\\
\hspace*{-5mm} & &           + 2 \biggr]  \zeta(3)
          - \frac{3}{4 (1+x)} \biggl[ 
     1 
   - \frac{1}{(1+x)} \biggr] \nn\\
%        
\hspace*{-5mm} & &         
          -   \biggl[
            \frac{1}{4}
          - \biggl( \frac{25}{32(1-x)} 
          + \frac{45}{2(1+x)^5}
          - \frac{279}{4(1+x)^4} 
          + \frac{597}{8(1+x)^3}  \nn\\
%        
\hspace*{-5mm} & &      \hspace*{5mm}   
          - \frac{463}{16(1+x)^2}
          + \frac{121}{32(1+x)} 
          - \frac{3}{4} \biggr) \zeta(2)
          - \biggl( \frac{1}{4(1-x)^2} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}  
          - \frac{7}{8(1-x)} \! 
          - \frac{6}{(1+x)^5} \! 
          + \!  \frac{15}{(1+x)^4} \! 
          - \frac{25}{2(1+x)^3} \! 
          + \!  \frac{4}{(1+x)^2} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}  
          -  \! \frac{47}{8(1+x)}  \! 
          +  \! \frac{13}{4} \biggr) \zeta(3) \! 
          -  \! \frac{5}{8(1-x)}  \! 
          -  \! \frac{9}{4(1+x)^3}  \! 
          +  \! \frac{27}{8(1+x)^2}  \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}  
          - \frac{1}{(1+x)}
          \biggr] H(0;x)  \nn\\
%
\hspace*{-5mm} & &        
          + \biggl[
            \frac{45}{(1 \! + \! x)^4} \! 
          -  \! \frac{90}{(1 \! + \! x)^3} \! 
          +  \! \frac{93}{2(1 \! + \! x)^2}  \! 
          -  \! \frac{3}{2(1 \! + \! x)}  \! 
          -  \! \frac{3}{2} 
          \biggr] \zeta(2) H(-1;x) \nn\\
%
\hspace*{-5mm} & &      
          -  \biggl[
            \frac{13}{4}
          + \biggl( \frac{1}{2(1-x)^2} 
          + \frac{49}{32(1-x)} 
          + \frac{15}{2(1+x)^5} 
          - \frac{75}{4(1+x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}          
          +  \! \frac{105}{8(1+x)^3}  \! 
          -  \! \frac{7}{16(1+x)^2}  \! 
          +  \! \frac{17}{32(1+x)}  \! 
          -  \! \frac{3}{2} \biggr)   \zeta(2) \! 
          +  \! \frac{3}{8(1-x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}    
          - \frac{23}{8(1-x)} 
          - \frac{51}{2(1+x)^4} 
          + \frac{129}{2(1+x)^3} 
          - \frac{103}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          + \frac{43}{4(1+x)}  \! 
          \biggr] H(0,0;x) \nn\\
\hspace*{-5mm} & &     
%
       -  \biggl[
            \frac{1}{2(1-x)^2}  \! 
          -  \! \frac{23}{16(1-x)} \! 
          -  \! \frac{45}{(1+x)^5}    \! 
   +  \! \frac{225}{2(1+x)^4}  \! 
          -  \! \frac{351}{4(1+x)^3}  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}     
          + \frac{157}{8(1+x)^2} 
          - \frac{71}{16(1+x)} 
   + \frac{7}{2}
          \biggr] \zeta(2) H(0,-1;x) \nn\\
\hspace*{-5mm} & &       
%
       +  \biggl[
            3
          -  \! \frac{3}{(1 \! - \! x)} \! 
          +  \! \frac{21}{(1 \! + \! x)^3} \! 
          -  \! \frac{63}{2(1 \! + \! x)^2}  \! 
          +  \! \frac{15}{2(1 \! + \! x)}
          \biggr]  \! H(-1,0;x)    \nn\\
\hspace*{-5mm} & &  
       -  \biggl[
            \frac{5}{4} \! 
          -  \! \biggl( \frac{1}{(1 \! - \! x)^2} \! 
          -  \! \frac{5}{2(1 \! - \! x)} \! 
          +  \! \frac{12}{(1 \! + \! x)^5} \! 
          -  \! \frac{30}{(1 \! + \! x)^4} \! 
          +  \! \frac{26}{(1 \! + \! x)^3} \nn\\
\hspace*{-5mm} & &          \hspace*{5mm}     
          -  \! \frac{8}{(1+x)^2} \! 
          +  \! \frac{7}{2(1 \! + \! x)} \biggr) \zeta(2)  \! 
          -  \! \frac{2}{(1-x)} \! 
          +  \! \frac{6}{(1 \! + \! x)^3} \! 
          -  \! \frac{9}{(1 \! + \! x)^2}\nn\\
\hspace*{-5mm} & &         \hspace*{5mm}     
          + \frac{5}{2(1+x)} 
          \biggr] H(1,0;x) \nn\\
%
\hspace*{-5mm} & &        
          -  \biggl[
            1
          - \frac{25}{32(1-x)} 
          - \frac{45}{2(1+x)^5} 
          + \frac{255}{4(1+x)^4} 
          - \frac{501}{8(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}          
          + \frac{359}{16(1+x)^2} 
          - \frac{105}{32(1+x)} 
          \biggr] H(0,0,0;x)  \nn\\
\hspace*{-5mm} & &   
%
       -   \! \biggl[
            3 \! 
          -  \! \frac{42}{(1+x)^4} \! 
   +  \! \frac{84}{(1+x)^3} \! 
          -  \! \frac{41}{(1+x)^2} \! 
          -  \! \frac{1}{(1+x)}
          \biggr] H(0,-1,0;x)  \nn\\
\hspace*{-5mm} & &    
%
       +  \!  \biggl[
            \frac{5}{2} \! 
   -  \! \frac{3}{(1 \! + \! x)^4} \! 
          +  \! \frac{6}{(1 \! + \! x)^3} \! 
          -  \! \frac{3}{2(1 \! + \! x)^2}  \! 
          -  \! \frac{3}{2(1 \! + \! x)} 
          \biggr] H(-1, 0, 0;x) \nn\\
%
\hspace*{-5mm} & &        +   \biggl[
            \frac{3}{2} \! 
          -  \! \frac{12}{(1+x)^4} \! 
          +  \! \frac{24}{(1+x)^3} \! 
          -  \! \frac{11}{(1+x)^2} \! 
          -  \! \frac{1}{(1+x)} \! 
          \biggr]  \! H(0,1,0;x) \nn\\
%
\hspace*{-5mm} & &        -   \biggl[
            1
          - \frac{36}{(1+x)^4}
          + \frac{72}{(1+x)^3}
          - \frac{36}{(1+x)^2}
          \biggr] H(1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        +  \biggl[
            \frac{11}{2}
          + \frac{3}{(1-x)^2}
          - \frac{169}{32(1-x)} 
          - \frac{3}{2(1+x)^5} 
          + \frac{15}{4(1+x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{9}{8(1+x)^3} 
          + \frac{15}{16(1+x)^2} 
          - \frac{153}{32(1+x)} 
          \biggr] H(0,0,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
        -  \biggl[
            7
          + \frac{4}{(1-x)^2}
          - \frac{79}{16(1-x)} 
          + \frac{3}{(1+x)^5}
          - \frac{15}{2(1+x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}    
          + \frac{17}{4(1+x)^3} 
          + \frac{41}{8(1+x)^2} 
          - \frac{159}{16(1+x)} 
          \biggr] H(0,-1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        -  \biggl[
            6 \! 
          +  \! \frac{6}{(1-x)^2} \! 
          -  \! \frac{43}{8(1-x)} \! 
          -  \! \frac{42}{(1+x)^5} \! 
          +  \! \frac{105}{(1+x)^4} \! 
          -  \! \frac{159}{2(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}      
          + \frac{81}{4(1+x)^2}
          - \frac{35}{8(1+x)}
          \biggr] H(0,0,-1,0;x) \nn\\
\hspace*{-5mm} & &          
%
       +  \biggl[
            5
          +  \! \frac{4}{(1-x)^2}   \! 
   -  \! \frac{9}{2(1-x)} \! 
          -  \! \frac{12}{(1+x)^5} \! 
          +  \! \frac{30}{(1+x)^4} \! 
          -  \! \frac{22}{(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm} 
          + \frac{7}{(1+x)^2}
   - \frac{9}{2(1+x)} \! 
          \biggr] H(0, \! 0, \! 1, \! 0, \! x)   \nn\\
\hspace*{-5mm} & & 
       +   \! \biggl[
            4 \! 
          +  \! \frac{2}{(1-x)^2} \! 
          -  \! \frac{17}{4(1-x)} \! 
          + \frac{36}{(1+x)^5}
          - \frac{90}{(1+x)^4} \! 
          +  \! \frac{69}{(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}        
          -  \! \frac{23}{2(1+x)^2} \! 
          -  \! \frac{21}{4(1+x)} \! 
          \biggr] H(0,1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            1 \! 
          +  \! \frac{2}{(1-x)^2} \! 
          -  \! \frac{7}{2(1-x)} \! 
          +  \! \frac{12}{(1+x)^5} \! 
          -  \! \frac{30}{(1+x)^4} \! 
          +  \! \frac{26}{(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{7}{(1+x)^2}
          + \frac{5}{2(1+x)}
          \biggr] H(1,0,0,0;x)  \nn\\
%
\hspace*{-5mm} & &        
          + \frac{1}{2} \biggl[
             1
          +     \frac{1}{(1 \! - \! x)^2}  \! 
          -  \! \frac{1}{(1 \! - \! x)} \! 
          +  \! \frac{1}{(1 \! + \! x)^2} \! 
          -  \! \frac{1}{(1 \! + \! x)} 
          \biggr] \bigl[ 2 \zeta(3) ( H(1;x) \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}
          - H(-1;x) )  
   - 2 \zeta(2) H(-1,0;x) 
   + \zeta(2) H(0,1;x) \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}
   + 4 H( \! -1, \! 0, \! -1, \! 0;x)
   - 4 H( \! -1, \! 0, \! 0, \! 0;x)
   - 4 H( \! -1, \! 0,1, \! 0;x)\nn\\
\hspace*{-5mm} & &       \hspace*{5mm}
   +  \! 10 H(0, \! -1, \! -1, \! 0, \! x) \! 
   -  \! 6 H(0, \! -1, \! 1, \! 0, \! x) \! 
   -  \! 6 H(0, \! 1, \! -1, \! 0, \! x)\nn\\
\hspace*{-5mm} & &       \hspace*{5mm}
   + 2 H(0,1,1,0;x)
   - 4 H(1,0,-1,0;x)
   + 4 H(1,0,1,0;x) \bigr]
   \nn\\
\hspace*{-5mm} &+ &  {\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt b})}_{2}(D,q^2) & = & 
%
        - \biggl[ \frac{1}{8(1-x)^2}  
          + \frac{5}{4(1-x)} 
          - \frac{45}{2(1+x)^4} 
          + \frac{99}{2(1+x)^3} \nn\\
\hspace*{-5mm} & &      
          - \frac{197}{8(1+x)^2} 
          - \frac{15}{4(1+x)} \biggr] \zeta(2) 
        + \biggl[ \frac{39}{80(1-x)^3} 
          - \frac{117}{160(1-x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{337}{320(1-x)} 
          + \frac{321}{20(1+x)^5} 
          - \frac{321}{8(1+x)^4} 
          + \frac{137}{5(1+x)^3} \nn\\
\hspace*{-5mm} & &           
          - \frac{39}{40(1+x)^2} 
          - \frac{337}{320(1+x)}  \biggr]  \zeta^{2}(2)
        + \biggl[ \frac{1}{2(1-x)^2}  
          - \frac{1}{2(1-x)}   \nn\\
\hspace*{-5mm} & &           
          + \frac{6}{(1+x)^4}
          - \frac{12}{(1+x)^3}
          + \frac{13}{2(1+x)^2}  
          - \frac{1}{2(1+x)}  \biggr]  \zeta(3) \nn\\
\hspace*{-5mm} & &           
          - \frac{3}{4(1+x)^2}
          + \frac{3}{4(1+x)} \nn\\
\hspace*{-5mm} & &           
       -  \biggl[ \biggl(
            \frac{1}{8(1 \! - \! x)^3}  \!
          -  \! \frac{13}{16(1 \! - \! x)^2}  \! 
          +  \! \frac{19}{32(1 \! - \! x)}  \! 
   +  \! \frac{45}{2(1 \! + \! x)^5} \! 
          -  \! \frac{279}{4(1 \! + \! x)^4}  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          +  \frac{291}{4(1+x)^3}
          -  \frac{25}{(1+x)^2}
          -  \frac{13}{32(1+x)}  \biggr)  \zeta(2) 
          -  \biggl( \frac{1}{2(1-x)^3}   \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          -  \! \frac{3}{4(1-x)^2} 
          + \frac{1}{8(1-x)}  
          + \frac{6}{(1+x)^5}
          - \frac{15}{(1+x)^4} 
          + \frac{12}{(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{3}{(1+x)^2}
          + \frac{1}{8(1+x)}  \biggr) \zeta(3) 
          + \frac{5}{16(1-x)}
          + \frac{9}{4(1+x)^3}  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{27}{8(1+x)^2}
          + \frac{13}{16(1+x)}
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &          
%
       -  \biggl[
            \frac{3}{4(1-x)^2} 
          - \frac{3}{4(1-x)}
          + \frac{45}{(1+x)^4}
          - \frac{90}{(1+x)^3}
          + \frac{171}{4(1+x)^2}  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          + \frac{9}{4(1+x)} 
          \biggr] \zeta(2) H(-1;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[ \biggl(
            \frac{5}{8(1 \! - \! x)^3} \! 
          -  \! \frac{15}{16(1 \! - \! x)^2}  \! 
          -  \! \frac{15}{32(1 \! - \! x)}  \! 
          +  \! \frac{15}{2(1 \! + \! x)^5} \! 
          -  \! \frac{75}{4(1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          + \frac{25}{2(1+x)^3} 
          - \frac{15}{32(1+x)} \biggr)  \zeta(2)
          - \frac{19}{8(1-x)^2}
          + \frac{21}{4(1-x)} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          -  \! \frac{51}{2(1+x)^4} \! 
          +  \! \frac{129}{2(1+x)^3} \! 
          -  \! \frac{395}{8(1+x)^2} \! 
          +  \! \frac{15}{2(1+x)} \! 
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &        
        -  \biggl[ \biggl(
            \frac{1}{4(1-x)} 
          + \frac{12}{(1+x)^5}
          - \frac{30}{(1+x)^4}
          + \frac{25}{(1+x)^3}
          - \frac{15}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          + \frac{1}{4(1+x)}   \biggr)  \zeta(2)
          - \frac{1}{(1-x)}
          - \frac{6}{(1+x)^3}
          + \frac{9}{(1+x)^2}\nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{2}{(1+x)}
          \biggr] H(1,0;x) \nn\\
%
\hspace*{-5mm} & &       
       -  \biggl[
             \! \frac{19}{4(1-x)} \! 
          +  \! \frac{21}{(1+x)^3} \! 
          -  \! \frac{63}{2(1+x)^2} \! 
          +  \! \frac{23}{4(1+x)} \! 
          \biggr] \!  H(-1,0;x) \nn\\
%
\hspace*{-5mm} & &        
       -  \biggl[
            \frac{3}{4(1-x)^3} \! 
          - \frac{9}{8(1-x)^2}  \! 
          -  \! \frac{21}{16(1-x)}  \! 
          +  \! \frac{45}{(1 \! + \! x)^5} \! 
          -  \! \frac{225}{2(1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}     
          +  \! \frac{84}{(1+x)^3} \! 
          -  \! \frac{27}{2(1+x)^2}  \! 
          -  \! \frac{21}{16(1+x)}  \! 
          \biggr]  \zeta(2) H(0,-1;x) \nn\\
%
\hspace*{-5mm} & &        
      -  \biggl[
            \frac{1}{8(1-x)^3}
          - \frac{13}{16(1-x)^2}
          + \frac{19}{32(1-x)}
          + \frac{45}{2(1+x)^5} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}    
          - \frac{255}{4(1 \! + \! x)^4} \! 
          +  \! \frac{243}{4(1 \! + \! x)^3} \! 
          -  \! \frac{19}{(1 \! + \! x)^2} \! 
          -  \! \frac{13}{32(1 \! + \! x)} \! 
          \biggr]  \! H(0,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{12}{(1+x)^4}
          - \frac{24}{(1+x)^3}
          + \frac{10}{(1+x)^2}
          + \frac{2}{(1+x)}
          \biggr] H(0,1,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          - \frac{36}{(1+x)^4}
          + \frac{72}{(1+x)^3}
          - \frac{33}{(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}    
          - \frac{3}{(1+x)}
          \biggr]  H(1,0,0;x)  \nn\\
%
\hspace*{-5mm} & &        
%
       -  \biggl[
            \frac{3}{4(1-x)^2}
          - \frac{3}{4(1-x)}
          - \frac{3}{(1+x)^4} 
          + \frac{6}{(1+x)^3}
          - \frac{5}{4(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}   
          - \frac{7}{4(1+x)}
          \biggr] H(-1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +   \! \biggl[
            \frac{1}{2(1-x)^2} \! 
          -  \! \frac{1}{2(1-x)} \! 
          -  \! \frac{42}{(1+x)^4} \! 
          +  \! \frac{84}{(1+x)^3} \! 
          -  \! \frac{75}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{9}{2(1+x)}
          \biggr] H(0,-1,0;x)  \nn\\
\hspace*{-5mm} & &           
%
       +  \biggl[
            \frac{5}{8(1 \! - \! x)^3}
          - \frac{15}{16(1 \! - \! x)^2}
          - \frac{11}{32(1 \! - \! x)} \! 
          +  \! \frac{3}{2(1 \! + \! x)^5} \! 
          -  \! \frac{15}{4(1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm} 
          +  \! \frac{1}{(1+x)^3} \! 
          +  \! \frac{9}{4(1+x)^2}
          - \frac{11}{32(1+x)}
          \biggr] H(0,0,0,0;x)  \nn\\
\hspace*{-5mm} & &          
%
       -  \biggl[
            \frac{3}{4(1-x)^3}
          -  \! \frac{9}{8(1-x)^2} 
   +  \! \frac{11}{16(1-x)} \! 
          -  \! \frac{3}{(1 \! + \! x)^5} \! 
          +  \! \frac{15}{2(1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm} 
          -  \! \frac{4}{(1+x)^3} \! 
          -  \! \frac{3}{2(1 \! + \! x)^2}
          + \frac{11}{16(1 \! + \! x)}
          \biggr] H(0,-1,0,0;x) \nn\\
\hspace*{-5mm} & &          
%
       +  \biggl[
            \frac{1}{2(1-x)^3}
          - \frac{3}{4(1-x)^2} 
   + \frac{17}{8(1-x)}
          -  \frac{42}{(1+x)^5}
          +  \frac{105}{(1+x)^4}  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm} 
          -  \! \frac{76}{(1+x)^3} \! 
          +  \! \frac{9}{(1+x)^2}
   + \frac{17}{8(1+x)}
          \biggr] H(0,0,-1,0;x) \nn\\
\hspace*{-5mm} & &            
%
       -  \biggl[
            \frac{3}{4(1-x)}
          - \frac{12}{(1+x)^5}
   + \frac{30}{(1+x)^4}
          -  \frac{21}{(1+x)^3}
          +  \frac{3}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm} 
          +  \frac{3}{4(1+x)}
          \biggr]  \! H(0,0,1,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{1}{(1-x)^3}
          - \frac{3}{2(1-x)^2}
          + \frac{7}{4(1-x)}
          - \frac{36}{(1+x)^5}
          + \frac{90}{(1+x)^4} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}        
          - \frac{66}{(1+x)^3}
          + \frac{9}{(1+x)^2}
          + \frac{7}{4(1+x)}
          \biggr] H(0,1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       -  \biggl[
            \frac{1}{4(1-x)} \! 
          + \frac{12}{(1+x)^5} \! 
          - \frac{30}{(1+x)^4} \! 
          + \frac{25}{(1+x)^3} \! 
          - \frac{15}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}       
          + \frac{1}{4(1+x)}
          \biggr] H(1,0,0,0;x)  \nn\\
\hspace*{-5mm} & + & {\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt b})}_{3}(D,q^2) & = & 0 \, .
\nonumber
\eea


$\bullet$ The {\it Down-Corner} graph {\tt(c)} of Fig. \ref{fig1}, defined as
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2\ 
\frac{{\mathcal N}_{({\tt c})}^{\mu}}{{\mathcal D}_{1} {\mathcal D}_{5} 
{\mathcal D}_{7} {\mathcal D}_{8} {\mathcal D}_{9} 
{\mathcal D}_{10} } \, , 
\label{b3} 
\eea
where 
\bea
{\mathcal N}_{({\tt c})}^{\mu} & = &  \bar{v}(p_2) \gamma_{\sigma} 
[i ( \not{\! k_{2}}) +m] 
\gamma_{\lambda} 
[i ( \not{\! k_{1}} + \! \not{\! k_{2}}) +m] 
\gamma_{\sigma} 
[i ( \not{\! p_{2}} + \! \not{\! k_{1}}) +m]
\gamma^{\mu} \times
\nn\\
& & \times
[-i ( \not{\! p_{1}} - \! \not{\! k_{1}}) +m] 
\gamma^{\lambda} u(p_1)  ,
\eea
gives: 
\bea
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt c})}_{1}(D,q^2) & = & \frac{1}{(D-4)^{2}} 
\Biggl\{
%
         \frac{1}{8}
%
       +  \biggl[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & - & \frac{1}{(D-4)} \Biggl\{
            \frac{3}{32}
   - \frac{7 \zeta(2)}{2}
   - \frac{1}{8} \biggl[
     1
   - \frac{2}{(1+x)} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}  
   - \frac{1}{4} \biggl[
     1 
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
    \biggr] \bigl[ 
           2 \zeta(2)
   - 4 H(0;x)  \nn\\
\hspace*{-5mm} & &    \hspace*{36mm}  
   + H(0,0;x)
   + 2 H(1,0;x) \bigr]
\Biggr\} \nn\\
%
\hspace*{-5mm} & & 
%
          + \frac{17}{128} \! 
       + \biggl[ 
            \frac{23}{16}  \! 
   + \frac{1}{2(1-x)} \! 
          - \frac{15}{4(1+x)^4} \! 
          + \frac{15}{2 (1+x)^3} \! 
          - \frac{23}{8 (1+x)^2}\nn\\
\hspace*{-5mm} & &           
          - \frac{1}{2(1+x)} \biggr] \zeta(2)
       - \biggl[ \frac{9}{(1+x)^2}
          - \frac{9}{(1+x)}
          + \frac{9}{2}  \biggr] \zeta(2) \ln{2}           
       + \biggl[ 
            \frac{1}{20} \nn\\
\hspace*{-5mm} & & 
   -  \! \frac{31}{160(1-x)} \! 
          -  \! \frac{3}{5 (1 \! + \! x)^5} \! 
          +  \! \frac{3}{2  (1 \! + \! x)^4} \! 
          -  \! \frac{71}{40 (1 \! + \! x)^3} \! 
          +  \! \frac{93}{80 (1 \! + \! x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{31}{160 (1 \! + \! x)} \biggr] \zeta^{2}(2)
       - \biggl[ \frac{5}{8 (1 \! - \! x)} \! 
          -  \! \frac{6}{ (1 \! + \! x)^4} \! 
          +  \! \frac{12}{ (1 \! + \! x)^3} \! 
          -  \! \frac{8}{ (1 \! + \! x)^2} \nn\\
\hspace*{-5mm} & &           
          + \frac{11}{8 (1+x)}
          + \frac{1}{8}  \biggr] \zeta(3)
          + \frac{13}{8 (1+x)} \biggr[ 
     1  
          - \frac{1}{(1+x)} \biggr] \nn\\
\hspace*{-5mm} & &           
%
       +  \biggl[
            \frac{49}{32}
       - \biggl( \frac{85}{64 (1-x)}
          + \frac{15}{4 (1+x)^5}
          - \frac{75}{8 (1+x)^4}           
          + \frac{117}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm} 
          +  \! \frac{117}{32 (1+x)^2}  \! 
          -  \! \frac{331}{64 (1+x)} \! 
          + \!  \frac{3}{8} \biggr)  \zeta(2) \! 
       - \biggl( \frac{5}{16(1-x)}   \!    
          +  \! \frac{6}{(1+x)^5} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm} 
          - \frac{15}{(1+x)^4} \! 
          + \frac{45}{4(1+x)^3} \! 
          - \frac{15}{8(1+x)^2} \! 
          + \frac{5}{16(1+x)}   \!   
          - \frac{1}{2}  \biggr) \zeta(3) \nn\\
\hspace*{-5mm} & &  \hspace*{5mm}
          -  \frac{21}{16 (1-x)}   \! 
          -   \!  \frac{39}{8 (1+x)^3}   \! 
          +   \!  \frac{117}{16 (1+x)^2} \! 
          -   \!  \frac{67}{16 (1+x)}  \!  
          \biggr]   H(0;x) \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[
            \frac{1}{4 (1-x)}
          + \frac{1}{4 (1+x)}
          - \frac{1}{4}
          \biggr] \zeta(2) H(1;x) \nn\\
%
\hspace*{-5mm} & &        +  \biggl[
            \frac{21}{2 (1+x)^2}
          - \frac{21}{2 (1+x)}
          + \frac{9}{4}
          \biggr] \zeta(2) H(-1;x) \nn\\
\hspace*{-5mm} & &        
   +  \biggl[
            \frac{35}{8} 
       - \biggl( \frac{1}{16 (1-x)}
          + \frac{3}{ (1+x)^5}
          - \frac{15}{2 (1+x)^4}
          + \frac{19}{4 (1+x)^3} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}        
          +  \frac{3}{8 (1+x)^2} \! 
          -  \! \frac{1}{16 (1+x)} \! 
          -  \! \frac{1}{4} \biggr) \zeta(2) \! 
          +  \! \frac{3}{16 (1-x)^2}  \!          
          -  \! \frac{47}{16 (1-x)} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}        
          -  \! \frac{15}{4 (1+x)^4} \! 
          +  \! \frac{3}{(1+x)^3} \! 
          +  \! \frac{31}{8 (1+x)^2}  \! 
          -  \! \frac{43}{8 (1+x)} \! 
          \biggr] H(0,0;x)  \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[
            \frac{29}{8} \! 
          -  \! \frac{4}{ (1-x)} \! 
          -  \! \frac{6}{ (1+x)^3}  \! 
          +  \! \frac{9 }{(1+x)^2} \! 
          -  \! \frac{25}{4 (1+x)} \! 
          \biggr] H(-1,0;x) \nn\\
\hspace*{-5mm} & &           
%
       + \frac{1}{9} \biggl[
            \frac{3}{(1-x)} \! 
          +  \! \frac{12}{(1+x)^3} \! 
          - \frac{18}{(1+x)^2} \! 
          +  \! \frac{3}{(1+x)} \! 
          \biggr] \zeta(2) H(0,-1;x)  \nn\\
\hspace*{-5mm} & &      
%
       +  \biggl[
            \frac{11}{8}   \!    
       + \biggl( \frac{1}{4(1-x)} \! 
          +  \! \frac{6}{(1 \! + \! x)^5} \! 
          - \frac{15}{(1 \! + \! x)^4} \! 
          +  \! \frac{11}{(1 \! + \! x)^3} \! 
          - \frac{3}{2 (1 \! + \! x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}   
          + \frac{1}{4 (1+x)}
          - \frac{1}{2} \biggr) \zeta(2)
          - \frac{3}{2 (1-x)}
          - \frac{3}{ (1+x)^3}
          + \frac{9}{2 (1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}        
          - \frac{11}{4 (1+x)}
          \biggr] H(1,0;x)  \nn\\
\hspace*{-5mm} & &      
%
       -  \biggl[
            \frac{13}{8}
          + \frac{21}{64 (1-x)}
          + \frac{15}{4 (1+x)^5}
          - \frac{51}{8 (1+x)^4}
          + \frac{21}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}  
          + \frac{197}{32 (1+x)^2}
          - \frac{363}{64 (1+x)}
          \biggr] H(0,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       - \frac{1}{2} \biggl[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] \bigl[
            H(1,1,0;x)
   + 6 H(-1,-1,0;x)  \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}  
          - 3 H(-1,1,0;x)
          - 3 H(1,-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & &        
       +  \biggl[
            3
          - \frac{9}{4 (1-x)}
          + \frac{7}{2 (1+x)^2}
          - \frac{23}{4 (1+x)}
          \biggr] H(-1,0,0;x)  \nn\\
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{19}{4}
          - \frac{9}{4(1-x)}
          + \frac{12}{(1+x)^4}       
          - \frac{24}{(1+x)^3}
          + \frac{17}{ (1+x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          - \frac{29}{4 (1+x)}
          \biggr] H(0,-1,0;x)  \nn\\
\hspace*{-5mm} & &     
%
       -  \biggl[
            \frac{9}{4}      
   - \frac{1}{(1-x)}
          + \frac{6}{ (1+x)^4}
          - \frac{12}{ (1+x)^3}
          + \frac{17}{2 (1+x)^2}  \nn\\
\hspace*{-5mm} & &    \hspace*{5mm} 
   - \frac{7}{2 (1+x)}
          \biggr] H(0,1,0;x)  \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{3}{4}
          + \frac{1}{(1-x)}
          + \frac{12}{ (1+x)^4}
          - \frac{24}{ (1+x)^3}
          + \frac{27}{2 (1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}      
          - \frac{1}{2 (1+x)}
          \biggr]  H(1,0,0;x)   \nn\\
 %
\hspace*{-5mm} & &        
       - \frac{1}{8} \biggl[
            \frac{1}{(1-x)} \! 
          +  \! \frac{4}{(1+x)^3} \! 
          - \frac{6}{(1+x)^2} \! 
          +  \! \frac{1}{(1+x)} \! 
          \biggr]  \bigl[ H(0,0,0,0;x) \nn\\
%
\hspace*{-5mm} & &  \hspace*{5mm}       
          - H(0,-1,0,0;x) \bigr] \nn\\
\hspace*{-5mm} & &        
       -  \biggl[
            1
          - \frac{1}{2 (1-x)} \! 
          - \frac{12}{ (1+x)^5} \! 
          + \frac{30}{ (1+x)^4} \! 
          - \frac{22}{ (1+x)^3} \! 
          + \frac{3}{ (1+x)^2} \nn\\
\hspace*{-5mm} & &  \hspace*{5mm}           
          - \frac{1}{2(1+x)}
          \biggr] H(0,0,-1,0;x)  \nn\\
\hspace*{-5mm} & &        
%
       +  \biggl[
            \frac{1}{2} \! 
          - \frac{1}{4(1-x)} \! 
          - \frac{6}{(1+x)^5}  \! 
          +  \! \frac{15}{(1+x)^4} \! 
          - \frac{11}{(1+x)^3} \! 
          +  \! \frac{3}{2(1+x)^2}\nn\\
\hspace*{-5mm} & &   \hspace*{5mm}     
          - \frac{1}{4(1+x)}
          \biggr] H(0,0,1,0;x) \nn\\
%
\hspace*{-5mm} & &        
        -  \biggl[
            1 \! 
          -  \! \frac{3}{8(1-x)} \! 
          -  \! \frac{12}{(1+x)^5} \! 
          +  \! \frac{30}{(1+x)^4} \! 
          -  \! \frac{43}{2(1+x)^3}    \!      
          +  \! \frac{9}{4(1+x)^2} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}    
          - \frac{3}{8(1+x)}
          \biggr] H(0,1,0,0;x)  \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[
            \frac{1}{2} \! 
          -  \! \frac{1}{4(1-x)} \! 
          -  \! \frac{6}{(1+x)^5} \! 
          +  \! \frac{15}{(1+x)^4} \! 
          -  \! \frac{11}{(1+x)^3} \! 
          +  \! \frac{3}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &           - \frac{1}{4(1+x)}
          \biggr]  H(1,0,0,0;x)  \nn\\
\hspace*{-5mm} & + & {\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt c})}_{2}(D,q^2) & = & \frac{1}{(D-4)} 
\Biggl\{ 
       \frac{1}{4} \biggl[
          \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
%
\hspace*{-5mm} & & 
       + \biggl[ \frac{1}{16 (1-x)^2} \! 
          +  \! \frac{1}{16 (1-x)}  \! 
          +  \! \frac{15}{4 (1 \! + \! x)^4} \! 
          -  \! \frac{15}{2 (1 \! + \! x)^3} \! 
          +  \! \frac{89}{16 (1 \! + \! x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{31}{16 (1+x)} \biggr]  \zeta(2)
   - \frac{6 \zeta(2) \ln{2}}{ (1+x)} \biggl[
     1 
   - \frac{1}{ (1+x)} \biggr] 
       - \biggl[ \frac{27}{40 (1-x)^3}  \nn\\
\hspace*{-5mm} & &           
          -  \! \frac{81}{80 (1-x)^2} \! 
          +  \! \frac{3}{80 (1-x)} \! 
          -  \! \frac{3}{5 (1 \! + \! x)^5} \! 
          +  \! \frac{3}{2 (1 \! + \! x)^4} \! 
          -  \! \frac{69}{40 (1 \! + \! x)^3} \nn\\
\hspace*{-5mm} & &           
          +  \! \frac{87}{80 (1 \! + \! x)^2} \! 
          +  \! \frac{3}{80 (1 \! + \! x)} \biggr]  \zeta^{2}(2) \! 
       -  \! \biggl[ \frac{1}{4 (1 \! - \! x)^2} \! 
          -  \! \frac{1}{4 (1 \! - \! x)} \! 
          -  \! \frac{6}{ (1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &           
          +  \! \frac{12}{ (1+x)^3} \! 
          - \frac{27}{4 (1+x)^2} \! 
          +  \! \frac{3}{4 (1 \! + \! x)} \biggr]  \zeta(3) \! 
          - \frac{13}{8 (1 \! + \! x)} \biggl[
     1  \! 
   - \frac{1}{(1 \! + \! x)} \biggr] 
%
           \nn\\
\hspace*{-5mm} & &        
       +  \biggl[ \biggl(
            \frac{1}{16  (1 \! - \! x)^3} \! 
          -  \! \frac{27}{32 (1 \! - \! x)^2} \! 
          +  \! \frac{57}{64 (1 \! - \! x)} \! 
          +  \! \frac{15}{4 (1 \! + \! x)^5} \! 
          -  \! \frac{75}{8 (1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}       
          + \frac{7}{(1+x)^3}
          + \frac{21}{8 (1+x)^2}
          - \frac{263}{64 (1+x)} \biggr) \zeta(2) 
          - \biggl( \frac{1}{4  (1-x)^3} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          + \frac{3}{8 (1-x)^2}
          - \frac{3}{8 (1-x)}
          + \frac{6}{ (1+x)^5}
          - \frac{15}{ (1+x)^4}
          + \frac{43}{4 (1+x)^3} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          - \frac{9}{8 (1+x)^2}
          - \frac{3}{8 (1+x)} \biggr) \zeta(3) 
          - \frac{27}{32 (1-x)}
          + \frac{39}{8 (1+x)^3} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          - \frac{117}{16 (1+x)^2}
          + \frac{105}{32 (1+x)}
          \biggr] H(0;x) \nn\\
%
\hspace*{-5mm} & &        
       + \frac{3}{2} \biggl[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          - \frac{5}{(1+x)^2}
          + \frac{5}{(1+x)}
          \biggr] \zeta(2) H(-1;x) \nn\\
%
\hspace*{-5mm} & &        
       -  \biggl[ \biggl(
            \frac{3}{4 (1-x)^3} \! 
          -  \! \frac{9}{8 (1-x)^2} \! 
          -  \! \frac{3}{16 (1-x)} \! 
          +  \! \frac{3}{(1 \! + \! x)^5}   \!      
          -  \! \frac{15}{2 (1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}
          + \frac{9}{2 (1+x)^3}
          + \frac{3}{4 (1+x)^2}
          - \frac{3}{16 (1+x)}  \biggr) \zeta(2)     
          + \frac{13}{16 (1-x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}
          + \frac{1}{16 (1-x)}
          - \frac{15}{4 (1+x)^4}
          + \frac{3}{(1+x)^3}
          + \frac{51}{16 (1+x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}
          - \frac{53}{16 (1+x)}
          \biggr] H(0,0;x)  \nn\\
%
\hspace*{-5mm} & &       
%
       +  \biggl[
            \frac{5}{4 (1-x)}
          - \frac{6}{ (1+x)^3}   
          + \frac{9}{ (1+x)^2}
          - \frac{17}{4 (1+x)}
          \biggr] H(-1,0;x) \nn\\
%
\hspace*{-5mm} & &    
%
       +  \biggl[
            \frac{3 \zeta(2)}{8 (1-x)}
          - \frac{6 \zeta(2)}{ (1+x)^5}
          + \frac{15 \zeta(2)}{ (1+x)^4}
          - \frac{21 \zeta(2)}{2 (1+x)^3}
          + \frac{3 \zeta(2)}{4 (1+x)^2} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}    
          + \frac{3 \zeta(2)}{8 (1+x)}
          - \frac{1}{2(1-x)}
          + \frac{3}{(1+x)^3}
          - \frac{9}{2 (1+x)^2} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}   
          + \frac{2}{(1+x)}
          \biggr] H(1,0;x)  \nn\\
%
\hspace*{-5mm} & &       
%
       +  \biggl[
            \frac{1}{16 (1 \! - \! x)^3} \! 
          -  \! \frac{19}{32 (1 \! - \! x)^2} \! 
          +  \! \frac{41}{64 (1 \! - \! x)} \! 
          +  \! \frac{15}{4 (1 \! + \! x)^5} \! 
          -  \! \frac{51}{8 (1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}
          + \frac{1}{(1+x)^3}
          + \frac{39}{8 (1+x)^2}
          - \frac{215}{64 (1+x)}
          \biggr] H(0,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       +  \biggl[
            \frac{6}{ (1+x)^4} \! 
          -  \! \frac{12}{ (1+x)^3} \! 
          +  \! \frac{7}{ (1+x)^2} \! 
          -  \! \frac{1}{(1+x)} \! 
          \biggr] H(0,1,0;x)  \nn\\
%
\hspace*{-5mm} & &        
          +   \! \frac{1}{2} \! \biggl[
            \frac{1}{(1-x)^2} \! 
          -  \! \frac{1}{(1-x)} \! 
          -  \! \frac{5}{(1+x)^2} \! 
          +  \! \frac{5}{(1+x)} \! 
          \biggr]  \! H(-1,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       -  \!  \biggl[
            \frac{12}{(1+x)^4} \! 
          -  \! \frac{24}{(1+x)^3} \! 
          +  \! \frac{14}{(1+x)^2} \! 
          -  \! \frac{2}{(1+x)} \! 
          \biggr]  \! H(0, \! -1,0;x)  \nn\\
%
\hspace*{-5mm} & &        
       -  \! \biggl[
            \frac{1}{2 (1-x)^2} \! 
          -  \! \frac{1}{2 (1-x)} \! 
          +  \! \frac{12}{ (1+x)^4} \! 
          -  \! \frac{24}{ (1+x)^3} \! 
          +  \! \frac{23}{2 (1+x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}  
          +  \! \frac{1}{2 (1 \! + \! x)} \! 
          \biggr] \!  H(1,0,0;x)   \nn\\
%
\hspace*{-5mm} & &   
%
       +  \frac{1}{2} \biggl[
            \frac{1}{(1-x)^3}
          - \frac{3}{2 (1-x)^2}
          - \frac{1}{(1+x)^3}
          + \frac{3}{2 (1+x)^2}
          \biggr] \times \nn\\
\hspace*{-5mm} & &      \hspace*{8mm}
    \times \bigl[ 3 \zeta(2) H(0,-1;x) 
   - H(0,0,0,0;x) 
   + H(0,-1,0,0;x) \bigr] \nn\\
\hspace*{-5mm} & &           
%
       +  \biggl[
            \frac{3}{4 (1-x)}
   - \frac{12}{(1+x)^5}
          + \frac{30}{ (1+x)^4}
          - \frac{21}{ (1+x)^3}
          + \frac{3}{2 (1+x)^2}  \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}
   + \frac{3}{4 (1+x)}
          \biggr]  H(0,0,-1,0;x)  \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[
            \frac{3}{8 (1-x)}
          - \frac{6}{(1+x)^5} 
   + \frac{15}{(1+x)^4}
   - \frac{21}{2 (1+x)^3}
          + \frac{3}{4 (1+x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}
          + \frac{3}{8 (1+x)}
          \biggr] H(0,0,1,0;x) \nn\\
%
\hspace*{-5mm} & &        -  \biggl[
            \frac{1}{2 (1-x)^3}
          - \frac{3}{4 (1-x)^2}
          - \frac{3}{4 (1-x)}
          + \frac{12}{ (1+x)^5}
          - \frac{30}{ (1+x)^4} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}       
          + \frac{41}{2 (1+x)^3}
          - \frac{3}{4 (1+x)^2}
          - \frac{3}{4 (1+x)} 
          \biggr] H(0,1,0,0;x) \nn\\
%
\hspace*{-5mm} & & 
       +  \biggl[
            \frac{3}{8 (1-x)} \! 
          -  \! \frac{6}{ (1+x)^5} \! 
          +  \! \frac{15}{ (1+x)^4} \! 
          -  \! \frac{21}{2 (1+x)^3} \! 
          +  \! \frac{3}{4 (1+x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm}    
          + \frac{3}{8 (1+x)}
          \biggr] H(1,0,0,0;x)  \nn\\
\hspace*{-5mm} & + & 
{\mathcal O} (D-4) , \\
%
%
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt c})}_{3}(D,q^2) & = & 
%
      - \biggl[ \frac{3}{16 (1-x)^2}
          - \frac{3}{16 (1-x)}
          + \frac{21}{16 (1+x)^2}
          - \frac{21}{16 (1+x)} \biggr]  \zeta(2) \nn\\
\hspace*{-5mm} & &           
          - \frac{3 \zeta(2) \ln{2}}{ (1-x)}
 \biggl[ 
     1
   - \frac{1}{ (1-x)} \biggr] 
      + \biggl[ \frac{27}{80 (1-x)^3}
          - \frac{81}{160 (1-x)^2} \nn\\
%
\hspace*{-5mm} & &        
          - \frac{27}{80 (1+x)^3}
          + \frac{81}{160 (1+x)^2} \biggr] \zeta^{2}(2)
      - \biggl[ \frac{1}{8 (1-x)^2}
          - \frac{1}{8 (1-x)} \nn\\
\hspace*{-5mm} & &           
          + \frac{1}{8 (1+x)^2}
          - \frac{1}{8 (1+x)} \biggr] \zeta(3)
          - \frac{1}{16 (1-x)^2}
          + \frac{1}{16 (1-x)} \nn\\
%
\hspace*{-5mm} & &        
       -  \biggl[ \biggl( 
            \frac{3}{16 (1-x)^3}
          - \frac{69}{32 (1-x)^2}
          + \frac{33}{16 (1-x)}
          + \frac{9}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &           \hspace*{5mm}   
          - \frac{15}{32 (1+x)^2}
          - \frac{3}{16 (1+x)} \biggl)  \zeta(2)
       - \biggl( \frac{1}{8 (1-x)^3}
          + \frac{3}{16 (1-x)^2} \nn\\
\hspace*{-5mm} & &           \hspace*{5mm}   
          + \frac{1}{8 (1+x)^3}
          - \frac{3}{16 (1+x)^2}  \biggl)  \zeta(3)
          + \frac{1}{8 (1-x)^3}
          - \frac{3}{16 (1-x)^2} \nn\\
\hspace*{-5mm} & &          \hspace*{5mm}    
          - \frac{11}{32 (1-x)} \! 
          + \frac{13}{32 (1+x)} \! 
          \biggr] H(0;x)  \nn\\
%
\hspace*{-5mm} & &       
%
       +   \biggl[  \biggl(
            \frac{3}{8 (1-x)^3}
          - \frac{9}{16 (1-x)^2}
          -  \frac{3}{8 (1+x)^3}
          + \frac{9}{16 (1+x)^2} \biggr)  \zeta(2) \nn\\
\hspace*{-5mm} & &          \hspace*{5mm} 
          -  \! \frac{1}{8 (1-x)^4} \! 
          +  \! \frac{1}{4 (1-x)^3} \! 
          +  \! \frac{5}{8 (1-x)^2} \! 
          -  \! \frac{3}{4 (1-x)} \! 
          -  \! \frac{13}{16 (1 \! + \! x)^2}\nn\\
\hspace*{-5mm} & &          \hspace*{5mm} 
          + \frac{13}{16 (1+x)}
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &    
      -  \biggl[
            \frac{3}{16 (1-x)^3}
          - \frac{49}{32 (1-x)^2}
          + \frac{23}{16 (1-x)}
          + \frac{9}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &         \hspace*{5mm}  
          - \frac{19}{32 (1+x)^2}
          - \frac{1}{16 (1+x)}
          \biggr] H(0,0,0;x) \nn\\
%
\hspace*{-5mm} & &        
%
       - \frac{1}{4}  \biggl[
            \frac{5}{(1 \! - \! x)^2}  \! 
          -  \! \frac{5}{(1 \! - \! x)} \! 
          -  \! \frac{1}{(1 \! + \! x)^2} \! 
          +  \! \frac{1}{(1 \! + \! x)} \! 
          \biggr] \bigl[  3 \zeta(2) H(-1;x)   \nn\\
\hspace*{-5mm} & &         \hspace*{5mm}  
          + H(-1,0,0;x) 
   - H(1,0,0;x) \bigr] \nn\\
%
\hspace*{-5mm} & &        
       - \frac{1}{4}  \biggl[
            \frac{1}{(1-x)^3}
          - \frac{3}{2 (1-x)^2}
          - \frac{1}{(1+x)^3}
          + \frac{3}{2 (1+x)^2}
          \biggr]  \times  \nn\\
\hspace*{-5mm} & &         \hspace*{8mm}  
        \times \bigl[ 3 \zeta(2) H(0,-1;x)
   + H(0,0,0,0;x)
   - H(0,-1,0,0;x)  \nn\\
\hspace*{-5mm} & &         \hspace*{13mm}  
   + H(0,1,0,0;x) \bigr] \nn\\
%
\hspace*{-5mm} & + & {\mathcal O} (D-4) \, .
\eea


$\bullet$ The {\it Up-Corner} graph {\tt(d)} of Fig. \ref{fig1}, defined as
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2\ 
\frac{{\mathcal N}_{({\tt d})}^{\mu}}{{\mathcal D}_{3} {\mathcal D}_{4} 
{\mathcal D}_{6} {\mathcal D}_{7} {\mathcal D}_{12} 
{\mathcal D}_{13} } \, , 
\label{b4} 
\eea
where 
\bea
{\mathcal N}_{({\tt d})}^{\mu} & = & \bar{v}(p_2) \gamma_{\sigma} 
[i ( \not{\! p_{2}} + \! \not{\! k_{1}} - \not{\! k_{2}}) +m] 
\gamma^{\mu} 
[-i ( \not{\! p_{1}} - \! \not{\! k_{1}} + \not{\! k_{2}}) +m] 
\gamma_{\lambda}   \times \nn\\
& & \times [-i ( \not{\! k_{2}}) +m] 
\gamma_{\sigma} 
[-i (\not{\! k_{1}}) +m]
\gamma^{\lambda} u(p_1) ,
\eea
due to reasons of symmetry gives:
\bea
{\mathcal F}^{(2l,{\tt d})}_{1}(D,q^2) & = &  {\mathcal F}^{(2l,{\tt c})}_{1}(D,q^2) 
\, , \\
{\mathcal F}^{(2l,{\tt d})}_{2}(D,q^2) & = &  {\mathcal F}^{(2l,{\tt c})}_{2}(D,q^2) 
\, , \\
{\mathcal F}^{(2l,{\tt d})}_{3}(D,q^2) & = & - {\mathcal F}^{(2l,{\tt c})}_{3}(D,q^2) 
\, .
\eea


$\bullet$ The {\it Up-Self-Mass-insertion} graph {\tt(e)} of Fig. \ref{fig1}, 
defined as 
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,right}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} }  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2\ 
\frac{{\mathcal N}_{({\tt e})}^{\mu}}{{\mathcal D}_{1} {\mathcal D}_{2} 
{\mathcal D}^{2}_{9} {\mathcal D}_{10} {\mathcal D}_{12}} 
\label{b5} 
\eea
where 
\bea
\! \! \! \! \! \! {\mathcal N}_{({\tt e})}^{\mu} & = &  \bar{v}(p_2)  \gamma_{\sigma}
[i ( \not{\! p_{2}} +  \!  \! \not{\! k_{1}}) \!  + \! m] 
\gamma_{\mu} 
[-i ( \not{\! p_{1}} -  \! \! \not{\! k_{1}})  \! + \! m] 
\gamma_{\lambda} 
[-i ( \not{\! p_{1}} - \!  \! \not{\! k_{1}} + \!  \! \not{\! k_{2}})  
\! + \! m]
\gamma^{\lambda} \times
\nn\\
\! \! \! \! \! \! & & \times
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] 
\gamma^{\sigma} u(p_1),
\eea
gives: 
\bea
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt e})}_{1}(D,q^2) & = & \frac{1}{(D-4)^{2}} 
\Biggl\{ 
%
            \frac{11}{8}
          + \frac{3}{(1+x)^2}
          - \frac{3}{(1+x)}
%
        -  \biggl[
            1
          + \frac{1}{2(1-x)}  \nn\\
\hspace*{-5mm} & &  \hspace*{20mm}     
          - \frac{3}{(1+x)^3}
          + \frac{9}{2(1+x)^2}
          - \frac{4}{(1+x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & + & \frac{1}{(D-4)} 
\Biggl\{ 
%
            \frac{19}{32}
          - \frac{2}{(1+x)^2}
          + \frac{2}{(1+x)}
       - \biggl( \frac{1}{4(1-x)}
          + \frac{3}{2(1+x)^3} \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}         
          -  \! \frac{9}{4(1 \! + \! x)^2} \! 
          +  \! \frac{2}{(1 \! + \! x)}  \! 
          -  \! \frac{1}{2} \biggr) \zeta(2)  \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}     
%
       +  \biggl[
            \frac{7}{8}
          + \frac{3}{4(1-x)}
          - \frac{5}{(1+x)^3}       
          + \frac{15}{2(1+x)^2}\nn\\
\hspace*{-5mm} & &    \hspace*{23mm} 
          - \frac{5}{(1+x)}
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}     
%
       -  \biggl[
            \frac{1}{4}
          - \frac{1}{(1-x)}
          + \frac{3}{2(1+x)^3}   
          - \frac{9}{4(1+x)^2}\nn\\
\hspace*{-5mm} & &    \hspace*{23mm} 
          + \frac{5}{4(1+x)}
          \biggr] H(0,0;x)  \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}     
%
       - \frac{3}{2} \biggl[
            \frac{1}{(1 \! - \! x)} \! 
          -  \! \frac{2}{(1 \! + \! x)^3}    \!  
          +  \! \frac{3}{(1 \! + \! x)^2} \! 
          -  \! \frac{2}{(1 \! + \! x)} \! 
          \biggr] H( \! -1, \! 0, \! x)  \nn\\
\hspace*{-5mm} & &    \hspace*{18mm}     
%
       -  \biggl[
            \frac{1}{2}
          - \frac{1}{2(1-x)}        
          - \frac{1}{2(1+x)}
          \biggr] H(1,0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
%
          + \frac{179}{128}
       + \biggl[ \frac{3}{8(1-x)}
          + \frac{21}{4(1+x)^4}
          - \frac{13}{(1+x)^3}
          + \frac{105}{8(1+x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{53}{8(1+x)}
          + \frac{31}{16} \biggr]  \zeta(2)
       + \biggl[ \frac{11}{8(1-x)}
          - \frac{3}{2(1+x)^3}
          + \frac{9}{4(1+x)^2} \nn\\
\hspace*{-5mm} & &           
          -  \! \frac{7}{8(1 \! + \! x)} \! 
          -  \! \frac{5}{8} \biggr]  \zeta(3)
          +  \! \frac{59}{8(1 \! + \! x)^2} \! 
          -  \! \frac{59}{8(1 \! + \! x)} \nn\\
\hspace*{-5mm} & &   
       -  \biggl[
            \frac{45}{32} \! 
       + \biggl( \frac{17}{8} \! 
          -  \! \frac{75}{64(1-x)}  \! 
   -  \! \frac{21}{4(1 \! + \! x)^5} \! 
          +  \! \frac{105}{8(1 \! + \! x)^4} \! 
          -  \! \frac{215}{16(1 \! + \! x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}     
          + \frac{225}{32(1+x)^2}
          - \frac{291}{64(1+x)} \biggr)  \zeta(2)
          + \frac{21}{16(1-x)}
          - \frac{89}{8(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}  
          + \frac{267}{16(1+x)^2}
          - \frac{155}{16(1+x)}
          \biggr] H(0;x)  \nn\\
\hspace*{-5mm} & &    
%
       -  \biggl[
             \frac{3}{4(1 \! - \! x)} \! 
          -  \! \frac{3}{2(1 \! + \! x)^3} \! 
          +  \! \frac{9}{4(1 \! + \! x)^2} \! 
          -  \! \frac{3}{2(1 \! + \! x)} \! 
          \biggr] \zeta(2) H(-1;x)  \nn\\
\hspace*{-5mm} & &           
%
       -  \frac{1}{4} \biggl[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] \zeta(2) H(1;x) \nn\\
%
\hspace*{-5mm} & &         
       +  \biggl[
            \frac{13}{4} \! 
          + \frac{15}{16(1-x)^2} \! 
          - \frac{57}{16(1-x)} \! 
          + \frac{21}{4(1+x)^4} \! 
          - \frac{8}{(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}        
          + \frac{45}{8(1+x)^2}
          - \frac{25}{8(1+x)}
          \biggr] H(0,0;x)  \nn\\
%
\hspace*{-5mm} & &         
%
       -  \biggl[
            \frac{13}{8} \! 
          -  \! \frac{15}{4(1-x)}  \! 
          +  \! \frac{5}{(1+x)^3} \! 
          -  \! \frac{15}{2(1+x)^2} \! 
          +  \! \frac{3}{(1+x)} \! 
          \biggr] H(-1,0;x)  \nn\\
%
\hspace*{-5mm} & &         
%
       +  \biggl[
            \frac{5}{4} 
          - \frac{3}{2(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H(1,0;x)  \nn\\
%
\hspace*{-5mm} & &         
%
       -  \biggl[
            \frac{9}{8}
          + \frac{37}{64(1-x)} 
          - \frac{21}{4(1+x)^5}
          + \frac{105}{8(1+x)^4}
          - \frac{239}{16(1+x)^3} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}     
          + \frac{297}{32(1+x)^2} 
          - \frac{323}{64(1+x)}
          \biggr] H(0,0,0;x)  \nn\\
%
\hspace*{-5mm} & &         
%
       +  \biggl[
            3
          - \frac{9}{2(1-x)}
          + \frac{3}{(1+x)^3} 
          - \frac{9}{2(1+x)^2}
          \biggr] H(-1,-1,0;x)  \nn\\
%
\hspace*{-5mm} & &         
%
       -   \biggl[
             \frac{9}{4} \! 
          -  \! \frac{3}{(1 \!- \!x)} \! 
          +  \! \frac{3}{2(1 \!+ \!x)^3}  \! 
   -  \!\frac{9}{4(1 \!+ \!x)^2} \!
          -  \!\frac{3}{4(1 \!+ \!x)} \!
          \biggr] H(-1,0,0;x) \nn\\
\hspace*{-5mm} & &         
%
       -  \biggl[
            \frac{9}{4} \!
          -  \! \frac{3}{(1 \! - \! x)} \!
          +  \! \frac{3}{2(1 \! + \! x)^3}  \!
          -  \! \frac{9}{4(1 \! + \! x)^2} \!
          -  \! \frac{3}{4(1 \! + \! x)} \!
          \biggr] H(0,-1,0;x)\nn\\
\hspace*{-5mm} & &            
%
       - \frac{1}{4}  \biggl[
            1
          - \frac{1}{(1-x)}
   - \frac{1}{(1+x)}
          \biggr] \bigl[   \zeta(2) H(1;x)   
   + 6 H(-1,1,0;x) \nn\\
\hspace*{-5mm} & &    \hspace*{10mm}     
   - 4 H(0,1,0;x)  
          + 6 H(1,-1,0;x) 
   - 4 H(1,0,0;x)  \nn\\
\hspace*{-5mm} & &    \hspace*{10mm}   
   - 4 H(1,1,0;x) \bigr]\nn\\
\hspace*{-5mm} & + & 
%
{\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt e})}_{2}(D,q^2) & = & \frac{1}{(D-4)^{2}} 
\Biggl\{ 
%
            \frac{3}{(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)} \biggr]
%
       + \frac{3}{4} \biggl[
            \frac{1}{(1-x)}
          - \frac{4}{(1+x)^3} \nn\\
\hspace*{-5mm} & &    \hspace*{20mm}             
          + \frac{6}{(1+x)^2}
          - \frac{3}{(1+x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & + & \frac{1}{(D-4)} 
\Biggl\{        
          - \frac{5}{(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)} \biggr] \nn\\
\hspace*{-5mm} & &     \hspace*{18mm}  
       -  \biggl[
            \frac{3}{4(1-x)} \! 
          -  \! \frac{8}{(1 \! + \! x)^3} \! 
          +  \! \frac{12}{(1 \! + \! x)^2}  \! 
          -  \! \frac{19}{4(1 \! + \! x)} \! 
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &     \hspace*{18mm}
       + \frac{3}{8} \biggl[
            \frac{1}{(1-x)} \! 
          -  \! \frac{4}{(1+x)^3}   \!          
          +  \! \frac{6}{(1+x)^2} \! 
          -  \! \frac{3}{(1+x)} \! 
          \biggr] \bigl[ \zeta(2) \nn\\
\hspace*{-5mm} & &     \hspace*{25mm}
         -  H(0,0;x) 
  + 2 H(-1,0;x) \bigr]
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
%
       - \biggl[ \frac{7}{16(1-x)^2}
          - \frac{1}{16(1-x)}
          + \frac{21}{4(1+x)^4}
          - \frac{29}{2(1+x)^3} \nn\\
\hspace*{-5mm} & &           
          + \frac{239}{16(1+x)^2}
          - \frac{97}{16(1+x)} \biggr]  \zeta(2)
          - \frac{51}{8(1+x)^2}
          + \frac{51}{8(1+x)} \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[ \biggl(
            \frac{7}{16(1-x)^3} 
          - \frac{21}{32(1-x)^2}
          - \frac{25}{64(1-x)}
          + \frac{21}{4(1+x)^5} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm}         
          - \frac{105}{8(1+x)^4}
          + \frac{13}{(1+x)^3}
          - \frac{51}{8(1+x)^2}
          + \frac{119}{64(1+x)} \biggl)  \zeta(2) \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          -  \! \frac{53}{32(1-x)}   \! 
          +  \! \frac{105}{8(1+x)^3} \! 
          -  \! \frac{315}{16(1+x)^2} \! 
          +  \! \frac{263}{32(1+x)} \! 
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &        
%
       +  \biggl[
            \frac{19}{16(1-x)^2}
          - \frac{25}{16(1-x)} 
          - \frac{21}{4(1+x)^4}
          + \frac{13}{2(1+x)^3} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          - \frac{47}{16(1+x)^2}
          + \frac{33}{16(1+x)}
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &        
          - \frac{1}{2} \biggl[
            \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H(1,0;x)  \nn\\
%
\hspace*{-5mm} & &        
%
       +  \!  \biggl[
            \frac{1}{4(1-x)}
          + \frac{8}{(1+x)^3} 
          - \frac{12}{(1+x)^2}
          + \frac{15}{4(1+x)}
          \biggr] H(-1,0;x)  \nn\\
%
\hspace*{-5mm} & &  
       -  \biggl[
            \frac{7}{16(1 \! - \! x)^3} \! 
          -  \! \frac{21}{32(1 \! - \! x)^2}  \! 
          -  \! \frac{49}{64(1 \! - \! x)} \! 
          +  \! \frac{21}{4(1 \! + \! x)^5} \! 
          -  \! \frac{105}{8(1 \! + \! x)^4} \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          + \frac{29}{2(1+x)^3}
          - \frac{69}{8(1+x)^2}
          + \frac{191}{64(1+x)}
          \biggr] H(0,0,0;x)  \nn\\
%
\hspace*{-5mm} & &  
%
       - \frac{3}{8} \biggl[
            \frac{1}{(1-x)} \! 
          -  \! \frac{12}{(1+x)^3} \! 
          +  \! \frac{6}{(1+x)^2}  \! 
          -  \! \frac{3}{(1+x)} \! 
          \biggr] \bigl[ \zeta(3) \! 
   - H(-1;x) \nn\\
\hspace*{-5mm} & &   \hspace*{10mm} 
   - 2 H(-1,-1,0;x) 
   + H(-1,0,0;x) 
   + H(0,-1,0;x) \bigr] \nn\\
%
\hspace*{-5mm} & + &  
%
{\mathcal O} (D-4) \, , \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt e})}_{3}(D,q^2) & = & \frac{1}{(D-4)^{2}} 
\Biggl\{ 
%
          -  \frac{3}{2 (1-x)} \biggl[ 
     1
   -  \! \frac{1}{(1-x)} \biggr] \! 
%
       +  \! \frac{3}{2} \biggl[
            \frac{1}{(1-x)^3} \! 
          -  \! \frac{3}{2 (1-x)^2} \nn\\
\hspace*{-5mm} & &        \hspace*{20mm}     
          + \frac{3}{4 (1-x)}
          - \frac{1}{4 (1+x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & + & \frac{1}{(D-4)} 
\Biggl\{          
          -  \frac{1}{2 (1-x)} \biggl[ 
     1
   - \frac{1}{(1-x)} \biggr]
%
       + \frac{3}{4} \biggl[  
            \frac{1}{(1-x)^3} \! 
          - \frac{3}{2(1-x)^2} \nn\\
\hspace*{-5mm} & &        \hspace*{18mm}   
          +  \! \frac{3}{4(1-x)} \! 
          - \frac{1}{4(1+x)} \biggr] \biggl[ \zeta(2) \! 
   +  \! \frac{2}{3} H(0;x) \! 
   - H(0,0;x) \nn\\
\hspace*{-5mm} & &        \hspace*{18mm}  
   + 2 H(-1,0;x) \biggr]
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
       + \biggl[ \frac{1}{4 (1-x)^3}
          + \frac{9}{16 (1-x)^2}
          - \frac{9}{8 (1-x)}
          - \frac{7}{16 (1+x)^2} \nn\\
\hspace*{-5mm} & & 
          + \frac{3}{4 (1+x)} \biggr] \zeta(2)
          + \frac{19}{16 (1-x)} \biggl[ 
     1
   - \frac{1}{(1-x)} \biggr]\nn\\
\hspace*{-5mm} & & 
%
       +  \biggl[ \biggl(
            \frac{9}{16 (1-x)^3}
          - \frac{27}{32 (1-x)^2}
          + \frac{19}{32 (1-x)}
          - \frac{7}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          + \frac{21}{32 (1+x)^2}  
          - \frac{17}{32 (1+x)} \biggr) \zeta(2)
          - \frac{13}{8 (1-x)^3}
          + \frac{39}{16 (1-x)^2} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          + \frac{21}{32 (1+x)^2} 
          + \frac{9}{32 (1-x)}
          - \frac{35}{32 (1+x)}
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &         
%
       -  \biggl[
            \frac{7}{8 (1-x)^4}
          - \frac{3}{2 (1-x)^3}
          + \frac{3}{8 (1-x)^2}
          - \frac{1}{16 (1-x)} \nn\\
\hspace*{-5mm} & &      \hspace*{5mm} 
          + \frac{7}{16 (1+x)^2}
          - \frac{1}{8 (1+x)}
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &        
      + \frac{1}{2}  \biggl[
            \frac{1}{(1-x)^3} \! 
          -  \! \frac{3}{2(1-x)^2} \! 
          -  \! \frac{3}{4(1-x)} \! 
          +  \! \frac{5}{4(1+x)} \! 
          \biggr] H(-1,0;x) \nn\\
%
\hspace*{-5mm} & &        
      +  \biggl[
            \frac{21}{16 (1-x)^3}
          - \frac{63}{32 (1-x)^2}
          + \frac{37}{32 (1-x)}
          - \frac{7}{16 (1+x)^3} \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}     
      + \frac{21}{32 (1+x)^2}
          - \frac{23}{32 (1+x)}
          \biggr] H(0,0,0;x) \nn\\
\hspace*{-5mm} & &   
      - \frac{3}{4}  \biggl[
            \frac{1}{(1-x)^3} \! 
          -  \! \frac{3}{2(1-x)^2} \! 
          +  \! \frac{3}{4(1-x)} \! 
          -  \! \frac{1}{4(1+x)} \! 
          \biggr] \bigr[ \zeta(3) \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}    
   - \zeta(2) H(-1;x)
   - 2 H(-1,-1,0;x) 
   + H(-1,0,0;x) \nn\\
\hspace*{-5mm} & &       \hspace*{5mm}    
   + H(0,-1,0;x) \bigr] \nn\\
%
\hspace*{-5mm} & + &  {\mathcal O} (D-4) \, .
\eea


$\bullet$ The {\it Down-Self-Mass-insertion} graph {\tt(f)} of 
Fig. \ref{fig1}, defined as 
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,left}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} }  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2\ 
\frac{{\mathcal N}_{({\tt f})}^{\mu}}{{\mathcal D}_{1} {\mathcal D}_{2} 
{\mathcal D}_{9} {\mathcal D}^{2}_{10} {\mathcal D}_{13}} \, , 
\label{b6} 
\eea
where 
\bea
\! \! \! \! \! \! {\mathcal N}_{({\tt f})}^{\mu} & = & \bar{v}(p_2) \gamma_{\sigma}
[i ( \not{\! p_{2}} +  \!  \! \not{\! k_{1}}) \!  + \! m] 
\gamma_{\lambda} 
[i ( \not{\! p_{2}} + \!  \! \not{\! k_{1}} - \!  \! \not{\! k_{2}}) \! 
 + \! m] 
\gamma^{\lambda} 
[i ( \not{\! p_{2}} +   \! \! \not{\! k_{1}})  \! + \! m] 
\gamma_{\mu} \times
\nn\\
\! \! \! \! \! \! & & \times
[-i ( \not{\! p_{1}} -  \! \! \not{\! k_{1}})  \! + \! m] 
\gamma^{\sigma} u(p_1) ,
\eea
due to reasons of symmetry, gives: 
\bea
{\mathcal F}^{(2l,{\tt f})}_{1}(D,q^2) & = & 
{\mathcal F}^{(2l,{\tt e})}_{1}(D,q^2) \, ,
\\
{\mathcal F}^{(2l,{\tt f})}_{2}(D,q^2) & = & 
{\mathcal F}^{(2l,{\tt e})}_{2}(D,q^2) \, ,
\\
{\mathcal F}^{(2l,{\tt f})}_{3}(D,q^2) & = & 
- {\mathcal F}^{(2l,{\tt e})}_{3}(D,q^2) \, .
\eea


$\bullet$ The {\it Vacuum-Polarization-insertion} graph {\tt(g)} 
of Fig. \ref{fig1}, defined as 
\bea
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.3h}{v3}
\fmfforce{0.2w,0.7h}{v4}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v3}
\fmf{photon}{v2,v4}
\fmf{plain}{v1,v5}
\fmf{plain,left}{v3,v4}
\fmf{plain,right}{v3,v4}
\end{fmfgraph*} }  & = & 
\int {\mathfrak{D}}^Dk_1\;{\mathfrak{D}}^Dk_2\ 
\frac{{\mathcal N}_{({\tt g})}^{\mu}}{{\mathcal D}^{2}_{1} {\mathcal D}_{7} 
{\mathcal D}_{8} {\mathcal D}_{9} {\mathcal D}_{10}} \, , 
\label{b7} 
\eea
where 
\bea
{\mathcal N}_{({\tt g})}^{\mu} & = & - \bar{v}(p_2) \gamma_{\sigma} 
[i ( \not{\! p_{2}} +  \!  \! \not{\! k_{1}})  \! + \! m] 
\gamma_{\mu} 
[-i ( \not{\! p_{1}} - \!  \! \not{\! k_{1}})  \! + \! m] 
\gamma_{\lambda} 
\gamma_{\lambda} 
[i ( \not{\! k_{1}} + \!  \! \not{\! k_{2}})  \! + \! m] \times
\nn\\
& & \times
\gamma^{\sigma} 
[i ( \not{\! k_{2}})  \! + \! m] u(p_1),
\eea
gives: 
\bea
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt g})}_{1}(D,q^2) & = &\frac{1}{(D-4)^{2}} 
\Biggl\{ 
%
          - \frac{1}{3}
%
       - \frac{2}{3} \biggl[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] H(0;x)  
%
\Biggr\}   \nn\\
\hspace*{-5mm} & + & 
\frac{1}{(D-4)} \Biggl\{ 
%
            \frac{1}{8}
   + \frac{1}{6} \biggl[
     1
   - \frac{2}{(1-x)} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &      \hspace*{18mm}
       - \frac{1}{3} \biggl[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \biggr] \bigl[ \zeta(2)
   - H(0;x) \nn\\
\hspace*{-5mm} & &      \hspace*{26mm}
   - H(0,0;x)
   + 2 H(-1,0;x) \bigr]
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
%
          + \frac{223}{864}
      + \biggl[ 2 \! 
          - \frac{1}{3(1-x)} \! 
          +  \! \frac{98}{3(1+x)^4} \! 
          - \frac{196}{3(1+x)^3} \! 
          +  \! \frac{229}{6(1+x)^2} \nn\\
\hspace*{-5mm} & &           
          - \frac{17}{3(1+x)} \biggr]  \zeta(2)
          - \frac{49}{9(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)}  \biggr]\nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[
            \frac{409}{216} \! 
       - \biggl( \frac{1}{4(1-x)} \! 
          +  \! \frac{6}{(1 \! + \! x)^5} \! 
          -  \! \frac{15}{(1 \! + \! x)^4} \! 
          +  \! \frac{11}{(1 \! + \! x)^3} \! 
          -  \! \frac{3}{2(1 \! + \! x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{1}{4(1+x)} \biggr)  \zeta(2)
          - \frac{77}{27(1-x)}
          - \frac{89}{9(1+x)^3}
          + \frac{89}{6(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}       
          -  \! \frac{635}{108(1+x)}
          \biggr] H(0;x)  \nn\\
\hspace*{-5mm} & &   
%
       +  \biggl[
            \frac{5}{18}
          + \frac{1}{3(1-x)} 
          + \frac{62}{9(1+x)^4}
          - \frac{124}{9(1+x)^3}
          + \frac{163}{18(1+x)^2} \nn\\
\hspace*{-5mm} & &     \hspace*{5mm}
          - \frac{2}{(1+x)}
          \biggr] H(0,0;x) \nn\\
%
\hspace*{-5mm} & &        
       + \frac{1}{6} \biggl[
            1
          - \frac{2}{(1-x)}
          \biggr] H(-1,0;x)  \nn\\
%
\hspace*{-5mm} & &        
%
       -  \biggl[
            \frac{1}{3}
          - \frac{7}{12(1-x)}
          + \frac{6}{(1+x)^5}
          - \frac{15}{(1+x)^4}
          + \frac{11}{(1+x)^3}  \nn\\
\hspace*{-5mm} & &   \hspace*{5mm} 
          - \frac{3}{2(1+x)^2}       
   - \frac{7}{12(1+x)}
          \biggr] H(0,0,0;x) \nn\\
%
\hspace*{-5mm} & & 
       + \frac{1}{3} \biggl[
            1 \! 
          - \frac{1}{(1-x)} \! 
          - \frac{1}{(1+x)} \! 
          \biggr]  \bigl[
     \zeta(3) \! 
   -  \! \zeta(2) H(-1;x)  \! 
   + \!  H(-1,0;x) \nn\\
\hspace*{-5mm} & &   \hspace*{10mm} 
   - 2 H(-1,-1,0;x) 
   + H(-1,0,0;x)
   + H(0,-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & + &  {\mathcal O} (D-4) \, , 
\label{ex1} \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt g})}_{2}(D,q^2) & = & \frac{1}{(D-4)} \Biggl\{ 
%
       \frac{1}{3} \biggl[
            \frac{1}{(1+x)}
          - \frac{1}{(1-x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
%
       - \biggl[ \frac{1}{6(1-x)} \! 
          -  \! \frac{34}{(1+x)^4} \! 
          +  \! \frac{68}{(1+x)^3} \! 
          -  \! \frac{33}{(1+x)^2} \! 
          -  \! \frac{5}{6(1+x)} \biggr] \zeta(2) \nn\\
\hspace*{-5mm} & &  
          + \frac{17}{3(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)} \biggr] \nn\\
\hspace*{-5mm} & &           
%
       -  \biggl[ \biggl(
            \frac{3}{8(1-x)} \! 
          - \frac{6}{(1+x)^5} \! 
          +  \! \frac{15}{(1+x)^4} \! 
          - \frac{21}{2(1+x)^3} \! 
          +  \! \frac{3}{4(1+x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}       
          + \frac{3}{8(1+x)} \biggr)  \zeta(2)
          + \frac{11}{9(1-x)}
          + \frac{31}{3(1+x)^3}
          - \frac{31}{2(1+x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}     
          + \frac{71}{18(1+x)}
          \biggr] H(0;x) \nn\\
\hspace*{-5mm} & &           
%
       +  \biggl[
            \frac{1}{6(1-x)}
          - \frac{22}{3(1+x)^4}
          + \frac{44}{3(1+x)^3}
          - \frac{23}{3(1+x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm} 
          + \frac{1}{6(1+x)}
          \biggr] H(0,0;x)  \nn\\
\hspace*{-5mm} & &       
%
       + \frac{1}{3}  \biggl[
            \frac{1}{(1+x)}
          - \frac{1}{(1-x)}
          \biggr] H(-1,0;x)  \nn\\
\hspace*{-5mm} & &       
%
       -  \biggl[
            \frac{3}{8(1-x)}
          - \frac{6}{(1+x)^5}
          + \frac{15}{(1+x)^4}
          - \frac{21}{2(1+x)^3}
          + \frac{3}{4(1+x)^2} \nn\\
\hspace*{-5mm} & &    \hspace*{5mm}    
          + \frac{3}{8(1+x)}
          \biggr] H(0,0,0;x)  \nn\\
\hspace*{-5mm} & + & {\mathcal O} (D-4) \, , 
\label{ex2} \\
\hspace*{-5mm} {\mathcal F}^{(2l,{\tt g})}_{3}(D,q^2) & = & 0 \, .
\label{ex3}
\eea

Note the structure of the contribution to the third {\it 
unrenormalized} form factor ${\mathcal F}^{(2l)}_{3}(D,x)$ of the seven 
diagrams given above: the contributions of the {\it Ladder}, {\it Cross} 
and {\it Vacuum-Polarization-insertion} vertices, 
the diagrams (a), (b) and (g) of Fig. \ref{fig1}, vanish 
separately, while those of the 
{\it Corner} and {\it Self-energy-insertion} diagrams, 
Fig. \ref{fig1} (c), (d), (e) and (f), cancel pairwise. 
That leads to the vanishing of the third form factor 
\be
{\mathcal F}^{(2l)}_{3}(D,q^2) = 
\sum_{{\tt graph}} {\mathcal F}^{(2l,{\tt graph})}_{3}(D,x) = 0 \ ,
\label{EFFE3}
\ee
as expected (and already repeatedly anticipated) 
from the conservation of the electromagnetic current.



\section{Renormalization subtractions \label{renormalization}}

%%%%%%%%%%%%%%%%%%%% Counter-terms Vertex %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bfig
\bc
%
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.64w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v5}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=75,l.d=.15w}{v5}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.31w,0.79h}{v22}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v2}
\fmfv{label=$\otimes$}{v22}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=60,l.d=.1w}{v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.31w,0.21h}{v22}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v2}
\fmfv{label=$\otimes$}{v22}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=-50,l.d=.08w}{v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.52w,0.55h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexagram,d.f=full}{v3}
%\fmfv{d.s=circle,d.f=empty,label=$K$,label.d=0}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m^{(1l)}$,l.a=60,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \\
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.52w,0.45h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexagram,d.f=full}{v3}
%\fmfv{d.s=circle,d.f=empty,label=$K$,label.d=0}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m^{(1l)}$,l.a=-60,l.d=.1w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.52w,0.55h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}^{(1l)}$,l.a=50,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
%\hspace{2mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.52w,0.45h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}^{(1l)}$,l.a=-50,l.d=.12w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
\hspace{5mm}
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.35w,0.5h}{v33}
%\fmfdot{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexagram,d.f=full}{v3}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{3}^{(1l)}$,l.a=150,l.d=.1w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \\
%
%%%%%%%%%%%%%%%%%%%%%%%
%
\subfigure[]{
\begin{fmfgraph*}(25,25) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.64w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v5}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}^{(2l)}$,l.a=75,l.d=.15w}{v5}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{phantom}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }
%
%%%%%%%%%%%%%%%%%%%%%%%
%
\vspace*{8mm}
\caption{\label{fig2} Subtraction terms for the renormalization at 
2 loops. } 
\ec
\efig
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a next step we have to renormalize the 1-loop insertions in the 
above 2-loop graphs, when present (i.e. in all the graphs, with the 
exception of the {\it Cross} graph {\tt b} of Fig. \ref{fig1}). 
That will be done by subtracting from each graphs suitable 
contributions, which will be specified in the next subsection, 
proportional to the 1-loop renormalization constants 
$Z_1^{(1l)}(D)$ (charge), $\ Z_2^{(1l)}(D)$ (electron wave function; 
in QED one has $Z_2^{(1l)}(D) = - Z_1^{(1l)}(D)$, due to 
the Ward identity), $\ Z_3^{(1l)}(D)$ (photon wave function) and 
$\delta m^{(1l)}(D,m)$ (electron mass). 

Those subtractions are sufficient to obtain the 2-loop renormalized form 
factor $F_2^{(2l)}(D,q^2)$. To obtain also the renormalized value 
of $F_1^{(2l)}(D,q^2)$, we have to subtract from the unrenormalized 
charge form factor (obtained after the subtractions due to the renormalization 
of the 1-loop insertions) its value at $q^2=0$, 
$Z_{1}^{(2l)}(D) = {\mathcal{F}}_1^{(2l)}(D,q^2=0)$. 


\subsection{1-loop renormalization constants times 1-loop subdiagrams
\label{1Lcnt}}

We give here the values of the subtractions to the 2-loop graphs due 
to the renormalization of the 1-loop insertions shown in 
Fig. \ref{fig2} {\tt(a-h)}. 

The 1-loop renormalization constants, $Z_{1}^{(1l)}(D)$, $\ Z_{2}^{(1l)}(D)$, 
$\ Z_{3}^{(1l)}(D)$ and $\delta m^{(1l)}(D,m)$, calculated according to the 
{\it On-Shell} renormalization prescription, have the following 
expressions exact in $D$: 
\bea
Z_{1}^{(1l)}(D) & = & 
- \frac{1}{2} \frac{(D-1)}{(D-3)(D-4)} 
\label{c001} 
 \ , \\
 & & \nonumber \\
Z_{2}^{(1l)}(D)  &=& - Z_{1}^{(1l)}(D) \qquad (\mbox{\rm Ward identity}) \ ,\\ 
 & & \nonumber \\
Z_{3}^{(1l)}(D) & = & 
\frac{2}{3(D-4)}  
\label{c002} 
 \ , \\
\delta m^{(1l)}(D,m) & = &  
m \ \frac{1}{2} \frac{(D-1)}{(D-3)(D-4)} \ ,  
\label{c003} 
\eea

The subtractions graphs of Fig. \ref{fig2} {\tt(a-h)} are defined by the
following relations:

$\bullet$ graph (a) in Fig. (\ref{fig2})

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.55w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v5}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=55,l.d=.1w}{v5}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \stackrel{def}{=} 
 \ Z_{1}^{(1l)}(D) \ \times 
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \, ;
\label{c1} 
\ee


$\bullet$ graphs (b) and (c) in Fig. (\ref{fig2})

\vspace*{3mm}

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.39w,0.71h}{v22}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v2}
\fmfv{label=$\otimes$}{v22}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=40,l.d=.1w}{v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  
 = 
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.38w,0.28h}{v22}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v2}
\fmfv{label=$\otimes$}{v22}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=-45,l.d=.08w}{v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }   \stackrel{def}{=} 
 \ Z_{1}^{(1l)}(D) \ \times 
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \, ;
\label{c2} 
\ee

\vspace*{5mm}

equal to the graph (a) of Fig. (\ref{fig2}).


$\bullet$ graph (d) in Fig. (\ref{fig2})

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m^{(1l)}$,l.a=60,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }   \stackrel{def}{=} 
% \ \delta m^{(1l)}(D,m) \ \times
\ \frac{\delta m^{(1l)}(D,m)}{m} \ \times \left( m 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \right) ;
\label{c3} 
\ee

$\bullet$ graph (e) in Fig. (\ref{fig2})

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m^{(1l)}$,l.a=-60,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }   \stackrel{def}{=} 
\ \frac{\delta m^{(1l)}(D,m)}{m} \ \times \left( m 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \right) ;
\label{c4} 
\ee

\vspace*{5mm}


$\bullet$ graphs (f) and (g) in Fig. (\ref{fig2})


\vspace*{2mm}

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}^{(1l)}$,l.a=60,l.d=.12w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  =
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}^{(1l)}$,l.a=-60,l.d=.12w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } 
\stackrel{def}{=} 
\ Z_{2}^{(1l)}(D) \ \times 
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \, ;
\label{c5} 
\ee

\vspace*{2mm}

$\bullet$ graph (h) in Fig. (\ref{fig2})

\be
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.55h}{v3}
\fmfforce{0.42w,0.5h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexagram,d.f=full}{v3}
%\fmfv{d.s=hexacross}{v3}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{3}^{(1l)}$,l.a=180,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  
\stackrel{def}{=} 
\ Z_{3}^{(1l)}(D) \ \times 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.42w,0.5h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexagram,d.f=full}{v3}
%\fmfv{d.s=hexacross}{v3}
%\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \, .
\label{c6} 
\ee

\vspace*{4mm}

From the {\it r.h.s} of Eqs. (\ref{c1}-\ref{c6}), it turns out that only
three 1-loop vertex subdiagrams appear in the calculation, two of them 
being in fact equal for symmetry reasons. Each of the subdiagrams can be 
written in terms of its vertex form factors by using an expression analogous 
to Eq. (\ref{b0002}) and Eq. (\ref{eq:unrenFF}). 

The first subdiagram, occurring in Eqs. 
(\ref{c1},\ref{c2},\ref{c5},\ref{c6}), is exactly the 1-loop QED vertex:
 
\be
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } =
\int {\mathfrak{D}}^Dk_1 \ 
\frac{{\mathcal N}^{\mu}}
{{\mathcal D}_{1} {\mathcal D}_{9} {\mathcal D}_{10} } \ , 
\label{c_c2} 
\ee
where 
\bea
\! \! \! \! \! \! \! \! \! \! {\mathcal N}^{\mu} & = &  
\bar{v}(p_2)  \gamma_{\sigma} 
[i ( \not{\! p_{2}} + \! \! \not{\! k_{1}}) \! +\! m] 
\gamma^{\mu} 
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] 
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] 
\gamma_{\sigma} u(p_1) \ . 
\eea
The corresponding form factors are 
\bea
\hspace*{-5mm} {\mathcal F}_1^{(1l)}(D,q^2) & = &  \frac{1}{(D-4)} \Biggl\{ 
%
          - \frac{1}{2}
       - \biggl[
            1
          - \frac{1}{(1+x)}
          - \frac{1}{(1-x)}
          \biggr] H(0;x) 
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
       + \frac{1}{4} \biggl[
            1
          - \frac{2}{(1-x)} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & & 
       - \frac{1}{2} \biggl[
            1
          - \frac{1}{(1+x)}
          - \frac{1}{(1-x)}
          \biggr] \bigl[ \zeta(2)
   - H(0;x) 
   - H(0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{5mm} 
   + 2 H(-1,0;x) \bigr]  \nn\\
\hspace*{-5mm} & + & 
(D-4) \Biggl\{ \frac{1}{8} \biggl[
            1
          - \frac{2}{(1-x)} \biggr] \bigl[ \zeta(2)
   - H(0,0;x) 
   + 2 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
        + \frac{1}{4} \biggl[
            1
          - \frac{1}{(1+x)}
          - \frac{1}{(1-x)}
          \biggr] \bigl[ \zeta(2)
   + 2 \zeta(3) \nn\\
\hspace*{-5mm} & & \hspace*{25mm} 
          - (4 \! -  \! \zeta(2)) H(0;x)  \! 
   - \!  2 \zeta(2) H(-1;x) \! 
   - \!  H(0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{25mm} 
   + 2 H(-1,0;x)
   - H(0,0,0;x)
   + 2 H(-1,0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{25mm} 
   + 2 H(0,-1,0;x)
   - 4 H(-1,-1,0;x) \bigr] \Biggr\} \nn\\
\hspace*{-5mm} & + & {\mathcal O} \left( (D-4)^2 \right) \, , 
\label{1loopF1} \\
%
\hspace*{-5mm} {\mathcal F}_2^{(1l)}(D,q^2) & = & - \frac{1}{2} 
          \biggl[ 
     \frac{1}{(1-x)}
   - \frac{1}{(1+x)} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & - & 
(D-4) \Biggl\{ \frac{1}{4} 
          \biggl[ 
     \frac{1}{(1-x)}
   - \frac{1}{(1+x)} \biggr] \bigl[ \zeta(2)
   - 4 H(0;x)
          - H(0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
   + 2 H(-1,0;x) \bigr] \Biggr\} \nn\\
\hspace*{-5mm} & + & {\mathcal O} \left( (D-4)^2 \right) \, , 
\label{1loopF2}\\
\hspace*{-5mm} {\mathcal F}_3^{(1l)}(D,q^2) & = & 0 
\label{c7} \, .
\eea
Let us observe that these are exactly the 1-loop {\it unrenormalized} 
form factors, ${\mathcal F}_i^{(1l)}(D,q^2)$ ($i=1,2,3$). 
As in the 2-loop counterterms they are multiplied by the 1-loop 
renormalization constants Eqs. (\ref{c001}-\ref{c003}), which behave 
for $D\to4$ as $1/(D-4)$, 
the 1-loop form factors must be evaluated up to the first order 
term in $(D-4)$ included; similarly, as the 1-loop unrenormalized form 
factors can develop poles in $(D-4)$ (that is the case of the charge 
form factor), the 1-loop renormalization constants are also needed 
up to first order in $(D-4)$ (a requirement trivially fulfilled by 
Eqs. (\ref{c001},\ref{c002},\ref{c003}), which are exact in $D$). 

The second subdiagram occurring in the {\it r.h.s.} of Eq. (\ref{c3}) 
is:
\be
m \parbox{20mm} {\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } =
m \int {\mathfrak{D}}^Dk_1 \ 
\frac{{\mathcal U}^{\mu}}
{{\mathcal D}_{1} {\mathcal D}^{2}_{9} {\mathcal D}_{10} } \ , 
\label{c_c1} 
\ee
where 
\bea
\! \! \! \! \! \! \! \! \! \! {\mathcal U}^{\mu} & = &  
\bar{v}(p_2)  \gamma_{\sigma} 
[i ( \not{\! p_{2}} + \! \! \not{\! k_{1}}) \! +\! m] 
\gamma^{\mu} 
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] i
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] 
\gamma_{\sigma} u(p_1) \ ; 
\eea
the corresponding (dimensionless) form factors are:
\bea
\hspace*{-5mm} {\mathcal F}_1^{(\otimes,up)}(D,q^2) & = &  \frac{1}{(D-4)} 
\Biggl\{ 
            1
          - \frac{2}{(1+x)}
          - \frac{2}{(1+x)^2}
 - \biggl[ 
            \frac{1}{(1-x)}
          - \frac{2}{(1+x)} \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
          + \frac{3}{(1+x)^2}
          - \frac{2}{(1+x)^3}
          \biggr] H(0;x)
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
          + 1 
   - \biggl[ \frac{1}{2(1+x)}
   - \frac{3}{2(1+x)^2}
   + \frac{1}{(1+x)^3} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & & 
      - \frac{1}{2} \biggl[ 
            \frac{1}{(1-x)} \! 
          -  \! \frac{2}{(1 \! + \! x)}  \! 
          +  \! \frac{3}{(1 \! + \! x)^2} \! 
          -  \! \frac{2}{(1 \! + \! x)^3} \biggr]  \! 
   \bigl[ \zeta(2) \! 
   - H(0;x)  \nn\\
\hspace*{-5mm} & & \hspace*{10mm} 
   - H(0,0;x)
   + 2 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & + & 
(D-4) \Biggl\{ 
     1
          - \frac{3}{(1+x)}
          + \frac{3}{(1+x)^2}   \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
       + \!  \biggl[ \frac{1}{4} \biggl(
            \frac{1}{(1-x)} \! 
          -  \! \frac{2}{(1 \! + \! x)}  \! 
          +  \! \frac{3}{(1 \! + \! x)^2} \! 
          -  \! \frac{2}{(1 \! + \! x)^3} \biggr) \zeta(2)   \nn\\
\hspace*{-5mm} & & \hspace*{23mm}  
       -  \frac{1}{2(1-x)}
          + \frac{3}{(1+x)} 
          - \frac{9}{2(1+x)^2}  \nn\\
\hspace*{-5mm} & & \hspace*{23mm}  
          + \frac{1}{(1+x)^3} \biggr] H(0;x)  \nn\\
\hspace*{-5mm} & & \hspace*{18mm}  
       + \!  \biggl[ 
            \frac{1}{2(1-x)} \! 
          -  \! \frac{1}{(1 \! + \! x)}  \! 
          +  \! \frac{3}{2(1 \! + \! x)^2} \! 
          -  \! \frac{1}{(1 \! + \! x)^3} \biggr] \bigl[ \zeta(3)  \nn\\
\hspace*{-5mm} & & \hspace*{27mm}  
          - H(-1;x) \bigr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm}  
       +  \frac{1}{4} \biggl[ 
            \frac{1}{(1-x)} \! 
          -  \! \frac{3}{(1 \! + \! x)}  \! 
          +  \! \frac{6}{(1 \! + \! x)^2} \! 
          -  \! \frac{4}{(1 \! + \! x)^3} \biggr]  \bigl[ \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{27mm}  
          - H(0,0;x)
          + 2 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm}  
       -  \frac{1}{4} \biggl[ 
            \frac{1}{(1-x)} \! 
          -  \! \frac{2}{(1+x)}  \! 
          +  \! \frac{3}{(1+x)^2}  \! 
          -  \! \frac{2}{(1+x)^3} \biggr] \times  \nn\\
\hspace*{-5mm} & & \hspace*{27mm} 
        \times \bigl[ 
     H(0,0,0;x)
   + 4 H(-1,-1,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{27mm}  
   - 2 H(-1,0,0;x)
   - 2 H(0,-1,0;x) \bigr]
\Biggr\}   \nn\\
\hspace*{-5mm} & + & {\mathcal O} \left( (D-4)^2 \right) \, , \\
%
%
\hspace*{-5mm} {\mathcal F}_2^{(\otimes,up)}(D,q^2) & = &  \frac{1}{(D-4)} 
\Biggl\{ 
            \frac{2}{(1+x)} \biggl[
     1
          - \frac{1}{(1+x)} \biggr]
 + \frac{1}{2} \biggl[ 
            \frac{1}{(1-x)}
          - \frac{3}{(1+x)} \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
          + \frac{6}{(1+x)^2}
          - \frac{4}{(1+x)^3}
          \biggr] H(0;x)
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
       - \frac{2}{(1+x)} \biggl[
     1
          - \frac{1}{(1+x)} \biggr]\nn\\
\hspace*{-5mm} & & 
       + \frac{2}{(1+x)} \biggl[ 
            1
   - \frac{3}{(1+x)}
   + \frac{2}{(1+x)^2} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & & 
      + \frac{1}{4} \biggl[ 
            \frac{1}{(1-x)}
          - \frac{3}{(1+x)} 
          + \frac{6}{(1+x)^2}
          - \frac{4}{(1+x)^3} \biggr] 
   \bigl[ \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{10mm} 
   - H(0,0;x)
   + 2 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & + & 
(D-4) \Biggl\{ \frac{1}{(1+x)} \biggl[ 1 - \frac{1}{(1+x)} \biggr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm}
       + \frac{1}{2(1+x)} \biggl[ 
             1 
   - \frac{3}{(1+x)}
   + \frac{2}{(1+x)} \biggr] 
   \bigl[ 2 \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
          - 3 H(0;x)
   - 2 H(0,0;x) 
   + 4 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm}
       - \frac{1}{8} \biggl[ 
            \frac{1}{(1 \! - \! x)} \! 
          -  \! \frac{3}{(1 \! + \! x)}  \! 
          +  \! \frac{6}{(1 \! + \! x)^2} \! 
          -  \! \frac{4}{(1 \! + \! x)^3} \biggr] 
   \bigl[ 2 \zeta(3) \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
          + \zeta(2) ( H(0;x)
   - 2 H(-1;x) ) 
   - H(0,0,0;x)  \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
         - 4 H(-1,-1,0;x)
  + 2 H(-1,0,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
  + 2 H(0,-1,0;x)\bigr]  
%
\Biggr\}  \nn\\
\hspace*{-5mm} & + & {\mathcal O} \left( (D-4)^2 \right) \, , \\
%
%
\hspace*{-5mm} {\mathcal F}_3^{(\otimes,up)}(D,q^2) & = &  \frac{1}{(D-4)} 
\Biggl\{ 
         - \frac{1}{(1-x)} \biggl[
     1
          - \frac{1}{(1-x)} \biggr]
 - \frac{1}{4} \biggl[ 
            \frac{1}{(1+x)}
          - \frac{3}{(1-x)} \nn\\
\hspace*{-5mm} & & \hspace*{18mm} 
          + \frac{6}{(1-x)^2}
          - \frac{4}{(1-x)^3}
          \biggr] H(0;x)
%
\Biggr\}   \nn\\
\hspace*{-5mm} & & 
       - \frac{1}{(1-x)} \biggl[
     1
          - \frac{1}{(1-x)} \biggr]\nn\\
\hspace*{-5mm} & & 
       + \frac{1}{4} \biggl[ 
            \frac{1}{(1+x)}
          + \frac{1}{(1-x)}
   - \frac{6}{(1-x)^2}
   + \frac{4}{(1-x)^3} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & & 
      - \frac{1}{8} \biggl[ 
            \frac{1}{(1-x)}
          - \frac{3}{(1+x)} 
          + \frac{6}{(1+x)^2}
          - \frac{4}{(1+x)^3} \biggr] 
   \bigl[ \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{10mm} 
   - H(0,0;x)
   + 2 H(-1,0;x) \bigr] \nn\\
\hspace*{-5mm} & + & 
(D-4) \Biggl\{ \frac{1}{2(1-x)} \biggl[ 1 - \frac{1}{(1-x)} \biggr] \nn\\
\hspace*{-5mm} & & \hspace*{18mm}
       - \frac{1}{4} \biggl[ 
            \frac{1}{(1+x)} 
   - \frac{3}{(1-x)^2}
   + \frac{2}{(1-x)^3} \biggr] H(0;x) \nn\\
\hspace*{-5mm} & & \hspace*{18mm}
       + \frac{1}{8} \biggl[ 
            \frac{1}{(1 \! + \! x)} \! 
          +  \! \frac{1}{(1 \! - \! x)}  \! 
          -  \! \frac{6}{(1 \! - \! x)^2} \! 
          +  \! \frac{4}{(1 \! - \! x)^3} \biggr] 
   \bigl[ \zeta(2) \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
          - H(0,0;x)
   + 2 H(-1,0;x) \bigr]  \nn\\
\hspace*{-5mm} & & \hspace*{18mm}
       + \frac{1}{16} \biggl[ 
            \frac{1}{(1 \! + \! x)} \! 
          -  \! \frac{3}{(1 \! - \! x)}  \! 
          +  \! \frac{6}{(1 \! - \! x)^2} \! 
          -  \! \frac{4}{(1 \! - \! x)^3} \biggr] \times \nn\\
\hspace*{-5mm} & & \hspace*{23mm}
  \times  \bigl[ 2 \zeta(3)
          + \zeta(2) ( H(0;x)
   - 2 H(-1;x) )\nn\\
\hspace*{-5mm} & & \hspace*{28mm}
          - H(0,0,0;x)
         - 4 H(-1,-1,0;x) \nn\\
\hspace*{-5mm} & & \hspace*{28mm}
  + 2 H(-1,0,0;x)
  + 2 H(0,-1,0;x) \bigr]  
%
\Biggr\}  \nn\\
\hspace*{-5mm} & + & {\mathcal O} \left( (D-4)^2 \right)
\label{c8} \, .
\eea

The last subdiagram, occurring in the {\it r.h.s.} of Eq. (\ref{c4}) is:
\be
m \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.3h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } =
m \int {\mathfrak{D}}^Dk_1 \ 
\frac{{\mathcal V}^{\mu}}
{{\mathcal D}_{1} {\mathcal D}_{9} {\mathcal D}^{2}_{10} } \ , 
\label{c_c3} 
\ee
where 
\bea
\! \! \! \! \! \! \! \! \! \! {\mathcal V}^{\mu} & = &  
\bar{v}(p_2)  \gamma_{\sigma} 
[i ( \not{\! p_{2}} + \! \! \not{\! k_{1}}) \! +\! m] i
[i ( \not{\! p_{2}} + \! \! \not{\! k_{1}}) \! +\! m]
\gamma^{\mu} 
[-i ( \not{\! p_{1}} - \! \! \not{\! k_{1}}) \! +\! m] 
\gamma_{\sigma} u(p_1) \ ; 
\eea
its form factors are:
\bea
\! \! \! \! \! \! {\mathcal F}_1^{(\otimes,down)}(D,q^2) & = & 
{\mathcal F}_1^{(\otimes,up)}(D,q^2)\\ 
\! \! \! \! \! \! {\mathcal F}_2^{(\otimes,down)}(D,q^2) & = & 
{\mathcal F}_2^{(\otimes,up)}(D,q^2) \\
\! \! \! \! \! \! {\mathcal F}_3^{(\otimes,down)}(D,q^2) & = & 
- {\mathcal F}_3^{(\otimes,up)}(D,q^2) 
\label{c9} \, .
\eea

We will write the contributions to the 2-loop form factors from the 
subtraction graphs with the 1-loop renormalization counter-terms 
of Fig. \ref{fig2} as 
${\mathcal F}_i^{(C,{\tt cnt})}(D,q^2)$ ($i=1,2,3$), where the $C$ in 
the superscript 
stands for ``counter-term'' and the ${\tt cnt}$ (${\tt cnt} \in \{
{\tt a,...,e} \}$) refers to the corresponding graphs of Fig. \ref{fig2}.
According to Eqs. (\ref{c1}-\ref{c5}), we have:
\bea
 & {\mathcal F}_i^{(C,{\tt a})}(D,q^2) \, = & 
Z_{1}^{(1l)}(D) \times {\mathcal F}_i^{(1l)}(D,q^2) \, , \\
 & {\mathcal F}_i^{(C,{\tt b})}(D,q^2)  \,= &  
{\mathcal F}_i^{(C,{\tt c})}(D,q^2)  \,  = \, Z_{1}^{(1l)}(D) 
\times {\mathcal F}_i^{(1l)}(D,q^2) \, , \\
& {\mathcal F}_i^{(C,{\tt d})}(D,q^2)  \, = & \, 
                     \frac{1}{m}\ \delta m^{(1l)}(D,m) 
\times {\mathcal F}_i^{(\otimes,up)}(D,q^2) \, , \\
 & {\mathcal F}_i^{(C,{\tt e})}(D,q^2)  \, = & \, 
                     \frac{1}{m}\ \delta m^{(1l)}(D,m) 
\times {\mathcal F}_i^{(\otimes,down)}(D,q^2) \, , \\
 & {\mathcal F}_i^{(C,{\tt f})}(D,q^2)  \, = &  
{\mathcal F}_i^{(C,{\tt g})}(D,q^2)  \, = \, Z_{2}^{(1l)}(D) \times 
{\mathcal F}_i^{(1l)}(D,q^2) \, , \\
& {\mathcal F}_i^{(C,{\tt h})}(D,q^2)  \, = & \, Z_{3}^{(1l)}(D) 
\times {\mathcal F}_i^{(1l)}(D,q^2) \, , 
\eea
where $i=1,2,3$.

As can be seen from Eqs. (\ref{c7},\ref{c8},\ref{c9}), the 
total contribution to the third form factor vanishes:
\be
{\mathcal F}^{(C)}_{3}(D,q^2) = 
\sum_{{\tt cnt}} {\mathcal F}^{(C,{\tt cnt})}_{3}(D,q^2) = 0 \ .
\label{EFFE3CT}
\ee

\subsection{2-loop charge renormalization constant. \label{2Lcnt}}

The only subtraction from the 2-loop renormalization counter-terms, in
our case, is given by the product of the charge renormalization constant
at 2-loop times the tree-level vertex, corresponding to the diagram
{\tt(i)} of Fig. \ref{fig2}:
\be
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.56w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v5}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}^{(2l)}$,l.a=75,l.d=.15w}{v5}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{phantom}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  \stackrel{def}{=}  Z_{1}^{(2l)}(D) \, \times \,
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{phantom}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, .
\ee

The 2-loop charge renormalization constant $Z_1^{(2l)}(D)$ is given 
by the value at $q^2=0$ of the unrenormalized charge form factor 
at two loops, 
\begin{equation} 
{\mathcal F}^{(2l)}_{1}(D,q^2) = 
   \sum_{{\tt graph}} {\mathcal F}^{(2l,{\tt graph})}_{1}(D,q^2) 
   - \sum_{{\tt cnt}} {\mathcal F}^{(C,{\tt cnt})}_{1}(D,q^2) \, ,
\end{equation} 
where ${\tt graph} \in \{ {\tt a,...,g} \}$ runs over the 
diagrams of Fig. \ref{fig1} and ${\tt cnt} \in \{ {\tt a,...,h} \}$ runs
over the subtraction graphs of Fig. \ref{fig2}. 

As the evaluation of the value at $q^2=0$ is somewhat simpler than the 
value for arbitrary $q^2$, we give here its expression exact in $D$: 
\bea
Z_{1}^{(2l)}(D) & = & {\mathcal{F}}^{(2l)}_{1}(D,q^2=0) \nonumber\\ 
    &=&  \frac{(D-6)}{4(D-5)(D-4)(D-3)} 
            \Big( 360-650 D+470 D^2              \nn \\
    & & \hspace*{35mm} - 167 D^3 + 29 D^4 - 2 D^5 \Big) 
\left( \frac{1}{m^2} \, \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{plain}{i,v1}
\fmf{plain}{v2,o}
\fmf{photon,tension=.15,left}{v1,v2}
\fmf{plain,tension=.15}{v1,v2}
\fmf{photon,tension=.15,right}{v1,v2}
\end{fmfgraph*} }  \hspace*{-3mm} \right)
\nn \\
& + &  \frac{(D + 4)}{8 (D - 6) (D - 4)^2} 
           \Big( 736 - 1348 D + 800 D^2 
            \nn \\
& & \hspace*{35mm} 
            - 207 D^3 + 24 D^4 - D^5 \Big) 
\left( \frac{1}{m^2} \, \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{plain}{i,v1}
\fmf{plain}{v2,o}
\fmf{plain,tension=.15,left}{v1,v2}
\fmf{plain,tension=.15}{v1,v2}
\fmf{plain,tension=.15,right}{v1,v2}
\end{fmfgraph*} } \hspace*{-3mm} \right)
\nn \\
& + & \frac{(D-2)}{48(D-7)(D-6)(D-5)^2(D-4)^2(D-3)^2} 
           \times \nn \\
%& & \
%             \nn \\
& & \hspace*{15mm}
           \Big(  7131744 - 9801144 D + 1271956 D^2 + 5512286 D^3 
             \nn \\
& & \hspace*{18mm}
             - 4884843 D^4 + 2058126 D^5 - 514065 D^6 + 79836 D^7 
             \nn \\
& & \hspace*{18mm}
             - 7567 D^8 + 400 D^9 - 9 D^{10} \Big) 
 \left( \frac{1}{m^4} \! \! 
\parbox{15mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i}
\fmfright{o}
\fmf{phantom}{i,v1}
\fmf{phantom}{v1,o}
\fmf{plain,right=45}{v1,v1}
\fmf{plain,left=45}{v1,v1}
\end{fmfgraph*} } \hspace*{-3mm} \right) \, , 
\eea
where the MIs depicted in the r.h.s. are those of Fig. 7 of \cite{Bon1}.

The corresponding expansion in $(D-4)$ is:
\bea
Z_{1}^{(2l)}(D) & = &
       - \frac{9}{8}   \frac{1}{(D-4)^2} 
       + \frac{55}{32} \frac{1}{(D-4)}  
       - \frac{7685}{1152}
       \nn \\
       & &
       + \frac{55}{8} \zeta(2)
       - 6 \zeta(2) \ln2
       + \frac{3}{2} \zeta(3)
       + {\mathcal O} (D-4) \, .
\eea
Let us recall once more that this counter-term is required for the 
renormalization of the {\it charge} form factor 
${\mathcal F}^{(2l)}_{1}(D,q^2)$ only. 


\section{The renormalized form factors \label{fullyren}}

As a first step the renormalization procedure requires the proper 
subtraction of the 8 graphs with the counter-terms at one loop of Fig. 
\ref{fig2} 
from the 7 unrenormalized 2-loop graphs of section \ref{diagrams}. 
That is carried out according to the following scheme: 
\bea
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v1}
\fmf{photon,tension=0}{v4,v3}
\end{fmfgraph*} } \right| _{ren} & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v1}
\fmf{photon,tension=0}{v4,v3}
\end{fmfgraph*} } - 
\parbox{20mm}{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.55w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}$,l.a=55,l.d=.1w}{v5}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, , \\
 & & \nonumber \\
%Cross
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v4}
\fmf{photon,tension=0}{v1,v3}
\end{fmfgraph*} 
}  \right| _{ren} & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=.3}{v2,v3}
\fmf{plain,tension=.3}{v3,v5}
\fmf{plain,tension=.3}{v1,v4}
\fmf{plain,tension=.3}{v4,v5}
\fmf{photon,tension=0}{v2,v4}
\fmf{photon,tension=0}{v1,v3}
\end{fmfgraph*} 
} \, , \\
 & & \nonumber \\
%Up-Corner
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }    \right| _{ren} & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }  - 
\parbox{20mm}
{\begin{fmfgraph*}(15,15) 
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.4w,0.70h}{v22}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
%\fmfv{d.s=hexacross}{v2}
\fmfv{label=$\otimes$}{v22}
\fmfv{l=$Z_{1}$,l.a=60,l.d=.1w}{v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, , \\
 & & \nonumber \\
%Down-Corner
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }   \right| _{ren} & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain,tension=0}{v2,v5}
\fmf{plain,tension=0}{v3,v4}
\fmf{photon,tension=.4}{v1,v4}
\fmf{plain,tension=.4}{v4,v5}
\fmf{plain,tension=0}{v1,v3}
\fmf{photon,tension=0}{v2,v3}
\end{fmfgraph*} }  - 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.4w,0.30h}{v11}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v11}
\fmfv{l=$Z_{1}$,l.a=-10,l.d=.2w}{v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, , \\
 & & \nonumber \\
%Up-SM
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,right}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} }   \right| _{ren} & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,right}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} }  
- \parbox{20mm}
{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m$,l.a=60,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  
- \parbox{20mm}
{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.51h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}$,l.a=80,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, , \\
 & & \nonumber \\
%Down-SM
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,left}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} }  \right| _{ren}  & = & \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v1}
\fmfforce{0.2w,0.07h}{v2}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v2}
\fmf{plain}{i2,v1}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v3}
\fmf{photon,tension=.25,left}{v3,v4}
\fmf{plain,tension=.25}{v3,v4}
\fmf{plain}{v4,v5}
\fmf{plain}{v1,v5}
\fmf{photon}{v1,v2}
\end{fmfgraph*} } 
- \parbox{20mm}
{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$\delta m$,l.a=-60,l.d=.4w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} }  
- \parbox{20mm}
{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.5w,0.7h}{v3}
\fmfforce{0.5w,0.49h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{2}$,l.a=-60,l.d=.4w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, , \\
 & & \nonumber \\
%Vac-Pol
\left. \parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.3h}{v3}
\fmfforce{0.2w,0.7h}{v4}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v3}
\fmf{photon}{v2,v4}
\fmf{plain}{v1,v5}
\fmf{plain,left}{v3,v4}
\fmf{plain,right}{v3,v4}
\end{fmfgraph*} } \right| _{ren}  & = & 
\parbox{20mm}{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.3h}{v3}
\fmfforce{0.2w,0.7h}{v4}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v3}
\fmf{photon}{v2,v4}
\fmf{plain}{v1,v5}
\fmf{plain,left}{v3,v4}
\fmf{plain,right}{v3,v4}
\end{fmfgraph*} }
- \hspace*{0.5cm}
\parbox{20mm}
{\begin{fmfgraph*}(15,15)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.2w,0.5h}{v3}
\fmfforce{0.42w,0.5h}{v33}
\fmfforce{0.8w,0.5h}{v5}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmfv{label=$\otimes$}{v33}
\fmfv{l=$Z_{3}$,l.a=180,l.d=.15w}{v3}
\fmf{photon}{v5,o}
\fmf{plain}{v2,v5}
\fmf{photon}{v1,v2}
\fmf{plain}{v1,v5}
\end{fmfgraph*} } \, .
\eea
As already anticipated in the previous section, we will indicate by 
${\mathcal F}_i^{(2l)}(D,q^2)$ the sum of the 
contributions to the form factors from all the above graphs: 
\begin{equation} 
{\mathcal F}^{(2l)}_{i}(D,q^2) = 
   \sum_{{\tt graph}} {\mathcal F}^{(2l,{\tt graph})}_{i}(D,q^2) 
   - \sum_{{\tt cnt}} {\mathcal F}^{(C,{\tt cnt})}_{i}(D,q^2) \, ,
\end{equation} 
where ${\tt graph} \in \{ {\tt a,...,g} \}$ runs over the 
diagrams of Fig. \ref{fig1} and ${\tt cnt} \in \{ {\tt a,...,h} \}$ runs
over the counter-terms of Fig. \ref{fig2}. 
Note that, as $ Z_2^{(1l)}(D) = - Z_1^{(1l)}(D) $ (the Ward identity), 
all the terms with $ Z_2^{(1l)}(D) $ are canceled by corresponding 
terms with $ Z_1^{(1l)}(D) $, so that 
only a single term proportional to $ Z_1^{(1l)}(D) $ remains in 
the sum of the counter-terms. 

For $i=3$ the renormalized form factor vanishes, as expected 
\be 
  F^{(2l)}_{3}(D,q^2) = 0 \, . 
\ee 

To obtain the 2-loop fully renormalized form factors $ F^{(2l)}_{i}(D,q^2) $ 
we have to subtract from the first form factor 
${\mathcal F}^{(2l)}_{1}(D,q^2)$ its value at $q^2=0$ (2-loop charge 
renormalization); no renormalization is needed for the second 
form factor, and we have finally 
\begin{eqnarray} 
 F^{(2l)}_{1}(D,q^2) &=& {\mathcal F}^{(2l)}_{1}(D,q^2) 
                      -  {\mathcal F}^{(2l)}_{1}(D,0)
   \label{2lFiren} \nn\\ 
                     &=& {\mathcal F}^{(2l)}_{1}(D,q^2) 
                      -   Z_{1}^{(2l)}(D) \, , \\
 F^{(2l)}_{2}(D,q^2) &=& {\mathcal F}^{(2l)}_{2}(D,q^2) \ .
\end{eqnarray} 

We report now the analytic expression of the UV-renormalized form factors 
$ F^{(2l)}_{i}(D,q^2) $ in the space-like region $-s = q^2> 0$, 
in terms of HPLs of the variable $x$ already introduced in Eq. (\ref{b00015}), 
$$ x = \frac{\sqrt{q^2+4} - \sqrt{q^2} }{\sqrt{q^2+4} + \sqrt{q^2} } \ , $$ 
up to the finite term in the expansion in $(D-4)$: 
\bea
\hspace*{-5mm} F^{(2l)}_{1}(D,q^2) & = & 
       \frac{1}{(D-4)^2} \Bigg\{
          \frac{1}{2}
%
       -  \Bigg[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
          \Bigg] H(0;x) \nn \\
%
\hspace*{-5mm} & & \hspace*{20mm}
       +  \Bigg[
            1
          + \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          + \frac{1}{(1+x)^2} \nn \\
%
\hspace*{-5mm} & & \hspace*{25mm}
          - \frac{1}{(1+x)}
          \Bigg] H(0,0;x) 
       \Bigg\} \nn \\
%
\hspace*{-5mm} & - &
        \frac{1}{(D-4)} \Bigg\{
            1
          + \frac{1}{2} \Bigg[
                    1  
                  - \frac{1}{(1-x)}
                  - \frac{1}{(1+x)}
             \Bigg] \bigl[ \zeta(2)
                           + 2 H(-1,0;x) \bigr] \nn \\
%
\hspace*{-5mm} & & \hspace*{18mm}
%
       -  \Bigg[
            \frac{7}{4}
          - \frac{2}{(1-x)}
          - \frac{3}{2(1+x)}
          \Bigg] H(0;x) \nn \\
%
       \hspace*{-5mm} & & \hspace*{18mm}
%
       + \Bigg[
            1
          + \frac{2}{(1-x)^2}
          - \frac{3}{2(1-x)}
          + \frac{1}{(1+x)^2} \nn \\
       \hspace*{-5mm} & & \hspace*{23mm}
          - \frac{1}{2(1+x)}
         \Bigg] H(0,0;x) \nn \\
%
       \hspace*{-5mm} & & \hspace*{18mm}
       -  \frac{1}{2}  \Bigg[
            1 \! 
          +  \! \frac{1}{(1-x)^2} \! 
          -  \! \frac{1}{(1-x)} \! 
          +  \! \frac{1}{(1+x)^2} \! 
          -  \! \frac{1}{(1+x)} \! 
          \Bigg] \times \nn \\
       \hspace*{-5mm} & & \hspace*{23mm}
          \times \bigl[  H(0;x) \zeta(2) 
                 + 4 H(-1,0,0;x)
                 + H(0,-1,0;x)  \nn \\
       \hspace*{-5mm} & & \hspace*{28mm}
                 - 3 H(0,0,0;x)
          \bigr]
%
      \Bigg\} \nn \\
  \hspace*{-5mm} & &
          +  \! \frac{1387}{216} \! 
          - \frac{49}{9(1+x)} \biggl[ 
     1 \! 
   - \frac{1}{(1+x)} \biggr] \! 
       + \!  \Bigg[
            \frac{51}{16}  \! 
          +  \! \frac{1}{2(1-x)} \! 
          - \frac{82}{3(1+x)^4}  \nn \\
       \hspace*{-5mm} & &
          + \frac{200}{3(1+x)^3}
          - \frac{221}{6(1+x)^2}
          - \frac{33}{8(1+x)}
            \Bigg] \zeta(2)
       - \Bigg[
            3 
          + \frac{18}{(1+x)^2}  \nn \\
       \hspace*{-5mm} & &
          - \frac{18}{(1+x)}
            \Bigg] \zeta(2) \ln2
       - \Bigg[
            \frac{7}{4} 
          +  \frac{1}{(1-x)^2}
          - \frac{1}{2(1-x)}
          - \frac{42}{(1+x)^4} \nn \\
       \hspace*{-5mm} & &
          + \frac{84}{(1+x)^3}
          - \frac{95}{2(1+x)^2}
          + \frac{6}{(1+x)}
            \Bigg] \zeta(3)
      + \Bigg[
            \frac{181}{40}
          + \frac{61}{40(1-x)^2} \nn \\
       \hspace*{-5mm} & &
          -  \! \frac{1219}{320(1-x)} \! 
          -  \! \frac{171}{4(1+x)^5} \! 
          +  \! \frac{855}{8(1 \! + \! x)^4} \! 
          -  \! \frac{6867}{80(1 \! + \! x)^3} \! 
          +  \! \frac{749}{32(1 \! + \! x)^2} \! 
          \nn \\
          \hspace*{-5mm} & &
          - \frac{1731}{320(1+x)}
            \Bigg] \zeta^2(2) \nn \\
%
 \hspace*{-5mm} & &
%
       -  \Bigg[
            \frac{3355}{864} \! 
          -  \! \frac{985}{216(1-x)} \! 
          -  \! \frac{89}{9(1+x)^3} \! 
          + \!  \frac{89}{9(1 \! + \! x)^2} \! 
          -  \! \frac{3521}{432(1 \! + \! x)} \! 
          \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
      + \Bigg(
            \frac{53}{12} 
          + \frac{2}{(1-x)^2}
          - \frac{19}{24(1-x)}
          + \frac{21}{(1+x)^5}
          - \frac{54}{(1+x)^4}
          \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{45}{(1+x)^3}
          - \frac{3}{(1+x)^2}
          - \frac{349}{24(1+x)}
             \Bigg) \zeta(2)
      - \Bigg(
            2 
          - \frac{7}{4(1-x)}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{42}{(1+x)^5}
          - \frac{105}{(1+x)^4}
          + \frac{91}{(1+x)^3}
          - \frac{63}{2(1+x)^2}  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{5}{4(1+x)}
             \Bigg) \zeta(3)
          \Bigg] H(0;x)  \nn \\
%
 \hspace*{-5mm} & &
       +  \Bigg[
            \frac{5}{2} 
          + \frac{1}{2(1-x)}
          + \frac{45}{(1+x)^4}
          - \frac{90}{(1+x)^3}
          + \frac{135}{2(1+x)^2} \nn \\
%
 \hspace*{-5mm} & & \hspace*{5mm} 
          - \frac{22}{(1+x)}
          \Bigg]  \zeta(2) H(-1;x) \nn \\
 \hspace*{-5mm} & &
       +  \Bigg[
            \frac{2137}{144} \! 
          +  \! \frac{4}{(1 \! - \! x)^2} \! 
          -  \! \frac{21}{2(1 \! - \! x)} \! 
          +  \! \frac{494}{9(1 \! + \! x)^4} \! 
          -  \! \frac{1258}{9(1 \! + \! x)^3} \! 
          -  \! \frac{1258}{9(1 \! + \! x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm} 
          + \frac{1130}{9(1+x)^2}
          - \frac{1081}{24(1+x)}
      + \Bigg(
            \frac{13}{4}
          + \frac{5}{4(1-x)^2}
          - \frac{105}{32(1-x)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm} 
          - \frac{33}{2(1+x)^5}
          + \frac{165}{4(1+x)^4}
          - \frac{281}{8(1+x)^3}
          + \frac{203}{16(1+x)^2}
          \nn \\
\hspace*{-5mm} & & \hspace*{5mm} 
          - \frac{137}{32(1+x)}
            \Bigg) \zeta(2)
          \Bigg] H(0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       -  \Bigg[
            \frac{55}{8} \! 
          -  \! \frac{9}{(1-x)} \! 
          -  \! \frac{48}{(1+x)^3} \! 
          +  \! \frac{72}{(1+x)^2} \! 
          -  \! \frac{115}{4(1+x)} \! 
          \Bigg] H(-1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       -  \Bigg[
            \frac{5}{2}  \! 
          -  \! \frac{1}{2(1-x)^2} \! 
          -  \! \frac{19}{16(1-x)} \! 
          -  \! \frac{45}{(1 \! + \! x)^5} \! 
          +  \! \frac{225}{2(1 \! + \! x)^4} \! 
          -  \! \frac{363}{4(1 \! + \! x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm} 
          + \frac{185}{8(1+x)^2}
          - \frac{67}{16(1+x)}
          \Bigg] \zeta(2) H(0,-1;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            4 \! 
          -  \! \frac{4}{(1-x)} \! 
          -  \! \frac{12}{(1+x)^3} \! 
          +  \! \frac{18}{(1+x)^2} \! 
          -  \! \frac{10}{(1+x)} \! 
       + \!  \Bigg(
            1 \! 
          +  \!  \frac{1}{(1-x)^2} \nn\\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{3}{2(1-x)}
          - \frac{36}{(1+x)^5}
          + \frac{90}{(1+x)^4}
          - \frac{78}{(1+x)^3}
          + \frac{28}{(1+x)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{7}{2(1+x)}
            \Bigg)  \zeta(2)
         \Bigg] H(1,0;x) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            1
          - \frac{1}{(1-x)}
          - \frac{1}{(1+x)}
         \Bigg]  H(-1,-1,0;x) \nn \\
%
\hspace*{-5mm} & & 
       - \Bigg[
            \frac{3}{2} \! 
          +  \! \frac{4}{(1-x)^2} \! 
          -  \! \frac{7}{2(1-x)} \! 
          +  \! \frac{63}{(1+x)^4} \! 
          -  \! \frac{126}{(1+x)^3} \! 
          +  \! \frac{135}{2(1+x)^2}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{4}{(1+x)}
         \Bigg] H(-1,0,0;x) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            4
          - \frac{1}{2(1-x)}
          + \frac{96}{(1+x)^4}
          - \frac{192}{(1+x)^3}
          + \frac{221}{2(1+x)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{15}{(1+x)}
         \Bigg] H(0,-1,0;x) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            \frac{8}{3} \! 
          -  \! \frac{1}{(1-x)^2} \! 
          +  \! \frac{41}{24(1-x)} \! 
          +  \! \frac{21}{(1+x)^5} \! 
          -  \! \frac{69}{(1+x)^4} \! 
          +  \! \frac{75}{(1+x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{95}{4(1+x)^2}
          - \frac{223}{24(1+x)}
         \Bigg] H(0,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{5}{2}
          + \frac{2}{(1-x)^2}
          - \frac{2}{(1-x)}
          + \frac{24}{(1+x)^4}
          - \frac{49}{(1+x)^3}
          + \frac{29}{(1+x)^2} \nn\\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{5}{(1+x)}
         \Bigg] H(0,1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            4 \! 
          +  \! \frac{90}{(1+x)^4} \! 
          -  \! \frac{180}{(1+x)^3} \! 
          +  \! \frac{197}{2(1+x)^2} \! 
          -  \! \frac{17}{2(1+x)} \! 
         \Bigg] H(1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{7}{2} \! 
          +  \! \frac{5}{2(1-x)^2} \! 
          -  \! \frac{67}{16(1-x)} \! 
          +  \! \frac{63}{(1 \! + \! x)^5} \! 
          -  \! \frac{315}{2(1 \! + \! x)^4} \! 
          +  \! \frac{517}{4(1 \! + \! x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{271}{8(1+x)^2} 
          - \frac{19}{16(1+x)}
         \Bigg] H(0,-1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{11}{2} \! 
          +  \! \frac{5}{2(1-x)^2} \! 
          -  \! \frac{21}{8(1-x)} \! 
          -  \! \frac{96}{(1 \! + \! x)^5} \! 
          +  \! \frac{240}{(1 \! + \! x)^4} \! 
          -  \! \frac{373}{2(1 \! + \! x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{169}{4(1+x)^2}
          - \frac{45}{8(1+x)}
         \Bigg] H(0,0,-1,0;x)  \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            \frac{27}{4} \! 
          +  \! \frac{17}{4(1 \! - \! x)^2} \! 
          -  \! \frac{217}{32(1 \! - \! x)} \! 
          -  \! \frac{3}{2(1 \! + \! x)^5} \! 
          +  \! \frac{15}{4(1 \! + \! x)^4} \! 
          -  \! \frac{17}{8(1 \! + \! x)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{59}{16(1+x)^2}
          - \frac{201}{32(1+x)}
         \Bigg] H(0,0,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            3
          + \frac{1}{(1-x)^2}
          - \frac{2}{(1-x)}
          - \frac{24}{(1+x)^5}
          + \frac{60}{(1+x)^4}
          - \frac{44}{(1+x)^3}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{7}{(1+x)^2}
          - \frac{2}{(1+x)}
         \Bigg] H(0,0,1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            1 \! 
          +  \! \frac{7}{4(1-x)} \! 
          -  \! \frac{90}{(1+x)^5} \! 
          +  \! \frac{225}{(1+x)^4} \! 
          -  \! \frac{175}{(1+x)^3} \! 
          +  \! \frac{75}{2(1+x)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{5}{4(1+x)}
         \Bigg] H(0,1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            2
          + \frac{2}{(1-x)^2}
          - \frac{5}{2(1-x)}
          - \frac{36}{(1+x)^5}
          + \frac{90}{(1+x)^4}
          - \frac{78}{(1+x)^3}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{29}{(1+x)^2}
          - \frac{9}{2(1+x)}
         \Bigg] H(1,0,0,0;x)  \nn \\
%
 \hspace*{-5mm} & &
       +  \Bigg[
            1
          + \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          + \frac{1}{(1+x)^2}
          - \frac{1}{(1+x)}
          \Bigg]  \big[ \zeta(3) H(1;x)  \nn\\
\hspace*{-5mm} & & \hspace*{5mm}
                        +  \! \zeta(2) H(-1,0;x) \! 
                        +  \! 4 H( \! -1, \! -1,0,0;x) \! 
                        +  \! 2 H( \! -1,0, \! -1,0;x) \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
                        -  \! 3 H(-1,0,0,0;x)  \! 
                        +  \! H(0,-1, \! -1,0;x) \! 
                        -  \! 2 H(1,0, \! -1,0;x) \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
                        + 2 H(1,0,1,0;x)
                 \big] \nn \\
%
\hspace*{-5mm} & + & {\mathcal O}(D-4)
\label{FF1fr} \ , \\
\hspace*{-5mm} F^{(2l)}_{2}(D,q^2) & = &  
        \frac{1}{(D-4)}  \Bigg\{
          \frac{1}{2} \left[ \frac{1}{(1+x)} - \frac{1}{(1-x)} 
                     \right] H(0;x) 
       + \Bigg[
             \frac{1}{(1+x)^2}
             \nn \\
\hspace*{-5mm} & & \hspace*{18mm}
          -  \frac{1}{(1+x)}
          -  \frac{1}{(1-x)^2}
          +  \frac{1}{(1-x)}
          \Bigg] H(0,0;x) 
          \Bigg\} \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{17}{8(1-x)} \! 
          -   \!  \frac{26}{(1+x)^4} \! 
          +  \!   \frac{64}{(1+x)^3} \! 
          -   \!  \frac{43}{(1+x)^2} \! 
          +  \! \frac{23}{8(1+x)} \! 
         \Bigg] \zeta(2)  \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{12}{(1+x)^2}
          - \frac{12}{(1+x)}
         \Bigg] \zeta(2) \ln2
       - \Bigg[
            \frac{69}{80(1-x)^3}
          - \frac{207}{160(1-x)^2} \nn \\
\hspace*{-5mm} & &
          + \frac{327}{320(1-x)}
          - \frac{171}{4(1+x)^5}
          + \frac{855}{8(1+x)^4}
          - \frac{3291}{40(1+x)^3} \nn \\
\hspace*{-5mm} & &
          + \frac{1323}{80(1+x)^2}
          + \frac{327}{320(1+x)}
         \Bigg] \zeta^2(2)
       - \Bigg[
            \frac{5}{2(1-x)^2}
          - \frac{5}{2(1-x)} \nn \\
\hspace*{-5mm} & &
          +   \frac{42}{(1+x)^4}
          -   \frac{84}{(1+x)^3}
          + \frac{91}{2(1+x)^2}
          - \frac{7}{2(1+x)}
         \Bigg] \zeta(3) \nn \\
%
\hspace*{-5mm} & &
          + \frac{17}{3(1+x)} \biggl[ 
     1
   - \frac{1}{(1+x)} \biggr] \nn \\
%
 \hspace*{-5mm} & & 
       -  \Bigg[
             \Bigg(
            \frac{19}{8(1-x)^2} \! 
          -  \! \frac{75}{16(1-x)} \! 
          -  \! \frac{21}{(1+x)^5} \! 
          +  \! \frac{54}{(1+x)^4} \! 
          -  \! \frac{173}{4(1+x)^3}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          +  \! \frac{13}{4(1 \! + \! x)^2} \! 
          +  \! \frac{149}{16(1 \! + \! x)} \! 
             \Bigg) \zeta(2) \! 
      +  \! \Bigg(
            \frac{7}{8(1-x)} \! 
          +  \!   \frac{42}{(1 \! + \! x)^5} \! 
          -  \!  \frac{105}{(1 \! + \! x)^4} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{175}{2(1+x)^3} 
          - \frac{105}{4(1+x)^2}
          + \frac{7}{8(1+x)}
             \Bigg) \zeta(3)
          - \frac{7}{144(1-x)}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{31}{3(1+x)^3}
          - \frac{31}{2(1+x)^2}
          + \frac{751}{144(1+x)}
          \Bigg] H(0;x) \nn \\
%
 \hspace*{-5mm} & &
       +  \Bigg[
            \frac{9}{4(1-x)^2}
          - \frac{9}{4(1-x)}
          - \frac{45}{(1+x)^4}
          + \frac{90}{(1+x)^3}
          - \frac{231}{4(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{51}{4(1+x)}
          \Bigg] \zeta(2) H(-1;x) \nn \\
%
 \hspace*{-5mm} & &
       -  \Bigg[
            \frac{9}{4(1-x)}
          + \frac{48}{(1+x)^3}
          - \frac{72}{(1+x)^2}
          + \frac{87}{4(1+x)}
          \Bigg]  H(-1,0;x) \nn \\
%
\hspace*{-5mm} & &
       +  \Bigg[
            \frac{9}{4(1-x)^3} \! 
          - \frac{27}{8(1-x)^2} \! 
          + \frac{21}{16(1-x)} \! 
          - \frac{45}{(1+x)^5} \! 
          + \frac{225}{2(1+x)^4} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{87}{(1+x)^3}
          + \frac{18}{(1+x)^2}
          + \frac{21}{16(1+x)}
          \Bigg] \zeta(2) H(0,-1;x) \nn \\
%
 \hspace*{-5mm} & &
       -  \Bigg[
             \Bigg(
            \frac{7}{8(1 \! - \! x)^3} \! 
          -  \! \frac{21}{16(1 \! - \! x)^2} \! 
          +  \! \frac{1}{32(1 \! - \! x)} \! 
          -  \! \frac{33}{2(1 \! + \! x)^5} \! 
          +  \! \frac{165}{4(1 \! + \! x)^4}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{135}{4(1+x)^3}
          + \frac{75}{8(1+x)^2}
          + \frac{1}{32(1+x)}
             \Bigg) \zeta(2)
          - \frac{1}{2(1-x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{1}{8(1-x)}
          + \frac{166}{3(1+x)^4}
          - \frac{422}{3(1+x)^3}
          + \frac{709}{6(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{773}{24(1+x)}
          \Bigg] H(0,0;x) \nn \\
%
\hspace*{-5mm} & &
       +  \Bigg[
             \Bigg(
            \frac{3}{4(1-x)}
          + \frac{36}{(1+x)^5}
          - \frac{90}{(1+x)^4}
          + \frac{75}{(1+x)^3}
          - \frac{45}{2(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{3}{4(1+x)}
             \Bigg) \zeta(2)
          - \frac{1}{(1-x)}
          + \frac{12}{(1+x)^3}
          - \frac{18}{(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{7}{(1+x)} 
          \Bigg]  H(1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       +  \Bigg[
            \frac{1}{4(1-x)^2}
          - \frac{1}{4(1-x)}
          + \frac{63}{(1+x)^4}
          - \frac{126}{(1+x)^3}
          + \frac{257}{4(1+x)^2}  \nn \\
       \hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{5}{4(1+x)}
          \Bigg] H(-1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       -  \Bigg[
            \frac{1}{2(1-x)^2}
          - \frac{1}{2(1-x)}
          +  \frac{96}{(1+x)^4}
          -  \frac{192}{(1+x)^3}
          + \frac{197}{2(1+x)^2}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{5}{2(1+x)}
          \Bigg] H(0,-1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{7}{8(1-x)^2}
          - \frac{51}{16(1-x)}
          -  \frac{21}{(1+x)^5}
          +  \frac{69}{(1+x)^4} 
          - \frac{293}{4(1+x)^3}\nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{91}{4(1+x)^2}
          + \frac{77}{16(1+x)}
          \Bigg] H(0,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          -  \frac{24}{(1+x)^4}
          +  \frac{48}{(1+x)^3}
          -  \frac{25}{(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{1}{(1+x)}
         \Bigg] H(0,1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{1}{(1-x)^2}
          - \frac{1}{(1-x)}
          +  \frac{90}{(1+x)^4}
          -  \frac{180}{(1+x)^3}
          + \frac{89}{(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{1}{(1+x)}
         \Bigg] H(1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            \frac{1}{4(1-x)^3} \! 
          - \frac{3}{8(1-x)^2} \! 
          - \frac{7}{16(1-x)} \! 
          +  \! \frac{63}{(1+x)^5} \! 
          - \frac{315}{2(1+x)^4}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{124}{(1+x)^3}
          - \frac{57}{2(1+x)^2}
          - \frac{7}{16(1+x)}
         \Bigg] H(0,-1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            \frac{1}{2(1-x)^3}
          - \frac{3}{4(1-x)^2}
          + \frac{7}{2(1-x)}
          - \frac{96}{(1+x)^5}
          + \frac{240}{(1+x)^4}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{357}{2(1+x)^3}
          + \frac{111}{4(1+x)^2} 
          + \frac{7}{2(1+x)}
         \Bigg] H(0,0,-1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{3}{8(1-x)^3} \! 
          -  \! \frac{9}{16(1-x)^2} \! 
          +  \! \frac{11}{32(1-x)} \! 
          -  \! \frac{3}{2(1 \! + \! x)^5}  \! 
          +  \! \frac{15}{4(1 \! + \! x)^4} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          - \frac{2}{(1+x)^3}
          - \frac{3}{4(1+x)^2}
          + \frac{11}{32(1+x)}
         \Bigg] H(0,0,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       - \Bigg[
            \frac{3}{2(1-x)}
          - \frac{24}{(1+x)^5}
          + \frac{60}{(1+x)^4}
          - \frac{42}{(1+x)^3}
          + \frac{3}{(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{3}{2(1+x)}
         \Bigg] H(0,0,1,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            \frac{25}{8(1-x)}
          -  \frac{90}{(1+x)^5}
          +  \frac{225}{(1+x)^4}
          - \frac{335}{2(1+x)^3}
          + \frac{105}{4(1+x)^2} \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{25}{8(1+x)}
         \Bigg] H(0,1,0,0;x) \nn \\
%
 \hspace*{-5mm} & &
       + \Bigg[
            \frac{3}{4(1-x)}
          +  \frac{36}{(1+x)^5}
          -  \frac{90}{(1+x)^4}
          +  \frac{75}{(1+x)^3}
          - \frac{45}{2(1+x)^2}  \nn \\
\hspace*{-5mm} & &  \hspace*{5mm}
          + \frac{3}{4(1+x)}
         \Bigg] H(1,0,0,0;x)  \nn \\
\hspace*{-5mm} & + & {\mathcal O}(D-4) \ .
\label{FF2fr}
\eea

Even after the full renormalization has been carried out, the on-shell 
renormalized form factors still develop polar singularities in $(D-4)$, 
due to soft IR divergences. These divergences are not physical and are 
removed in any physical process by the corresponding divergences due 
to soft real emission.

As it is easy to check explicitly, the IR divergences are the same as 
in~\cite{2loop1,Pie}, provided the following formal replacement is done 
\be
\log \left( \frac{\lambda}{m} \right) = - \frac{1}{(D-4)} \, , 
\ee 
so that they follow the general structure already pointed out in~\cite{YFS}. 
The finite parts are however different in the $D$-dimensional 
and in the $\lambda$-mass regularization schemes. 


\subsection{Continuation to time-like momentum transfer and imaginary 
parts \label{im}}

The expressions of the UV-renormalized form factors given in the
previous section, Eqs. (\ref{FF1fr},\ref{FF2fr}), can be analytically 
continued to the time-like region, $S = -Q^2 > 0$, and in particular 
above the physical threshold $S>4m^2$, where the form factors 
develop an imaginary part. 

The analytic continuation to $S+i\epsilon$ for $S>4m^2$ is performed 
with the substitution 
\be
x = - y + i \epsilon \, ,
\label{analyticY}
\ee
where now:
\be
y = \frac{\sqrt{s} - \sqrt{s-4}    }{\sqrt{s} + \sqrt{s-4} } \,
  = \frac{\sqrt{S} - \sqrt{S-4m^2} }{\sqrt{S} + \sqrt{S-4m^2} } \, , 
\label{Yins}
\ee 
with $S=m^2 s$ according to Eq. (\ref{defq2ands}). 

The imaginary part of the form factors is originated by the imaginary parts
developed by the polylogarithms with rightmost index equal to 0 (for
details see \cite{Pie,Polylog,Polylog3}).

In the kinematical region above threshold the form factors can be written as 
\bea
F^{(2l)}_{1}(D,-s-i\epsilon) & = & \Re \, F^{(2l)}_{1}(D,-s) 
                               + i \pi \, \Im \, F^{(2l)}_{1}(D,-s) \, , \\
F^{(2l)}_{2}(D,-s-i\epsilon) & = & \Re \, F^{(2l)}_{2}(D,-s) 
                               + i \pi \, \Im \, F^{(2l)}_{2}(D,-s) \, ; 
\eea
for short, we give the explicit expressions of the imaginary parts only: 
\bea
\hspace*{-5mm} \Im \, F^{(2l)}_{1}(D,-s) & = & 
     \frac{1}{(D-4)^2} 
       \Bigg\{
          - 1
          + \frac{1}{(1-y)}
          + \frac{1}{(1+y)} 
          + \Bigg[
              1
            + \frac{1}{(1-y)^2}
          \nn \\ 
\hspace*{-5mm} & & \hspace*{20mm} 
            - \frac{1}{(1-y)}
            + \frac{1}{(1+y)^2}
            - \frac{1}{(1+y)}
          \Bigg] H(0;y)
       \Bigg\} \nn \\
\hspace*{-5mm} & + &
       \frac{1}{(D-4)}
       \Bigg\{
            \frac{7}{4}
          - \frac{3}{2(1-y)}
          - \frac{2}{(1+y)}
          - \Bigg[
            1
          + \frac{1}{(1-y)^2}
          \nn \\ 
\hspace*{-5mm} & & \hspace*{18mm} 
          - \frac{1}{2(1-y)}
          + \frac{2}{(1+y)^2}
          - \frac{3}{2(1+y)}
          \Bigg] H(0;y)
          \nn \\ 
\hspace*{-5mm} & & \hspace*{18mm} 
          + \Bigg[
            1
          - \frac{1}{(1-y)}
          - \frac{1}{(1+y)}
          \Bigg] H(1;y)
          \nn \\ 
\hspace*{-5mm} & & \hspace*{18mm} 
          + \frac{1}{2} \Bigg[
            1 \! 
          +  \! \frac{1}{(1 \! - \! y)^2} \! 
          -  \! \frac{1}{(1 \! - \! y)} \! 
          +  \! \frac{1}{(1 \! + \! y)^2} \! 
          -  \! \frac{1}{(1 \! + \! y)} \! 
          \Bigg]  \! \times \nn \\ 
\hspace*{-5mm} & & \hspace*{25mm} 
           \times \big[
            4 \zeta(2)
          -  3 H(0,0;y)
          -  2 H(0,1;y) \nn \\ 
\hspace*{-5mm} & & \hspace*{25mm} 
          -  4 H(1,0;y)
          \big]
       \Bigg\} \nn \\
\hspace*{-5mm} & &
          -  \! \frac{3355}{864} \! 
          +  \! \frac{89}{9(1-y)^3} \! 
          -  \! \frac{89}{6(1-y)^2} \! 
          +  \! \frac{3521}{432(1-y)} \! 
          +  \! \frac{985}{216(1 \! + \! y)} \nn \\ 
\hspace*{-5mm} & & 
       - \Bigg(
            \frac{7}{4}  \! 
          +  \!  \frac{15}{(1-y)^4} \! 
          -  \!  \frac{30}{(1-y)^3} \! 
          +  \! \frac{83}{4(1-y)^2} \! 
          -  \! \frac{21}{4(1-y)} \! 
          +  \! \frac{3}{(1 \! + \! y)^2} \nn \\ 
\hspace*{-5mm} & & 
          - \frac{5}{2(1+y)}
            \Bigg) \zeta(2) 
       + \Bigg(
            2 
          +  \frac{42}{(1-y)^5}
          -  \frac{105}{(1-y)^4}
          +  \frac{91}{(1-y)^3} \nn \\ 
\hspace*{-5mm} & & 
          - \frac{63}{2(1-y)^2}
          + \frac{5}{4(1-y)}
          - \frac{7}{4(1+y)}
            \Bigg) \zeta(3) \nn \\ 
\hspace*{-5mm} & & 
%
       + \Bigg[
            \frac{2137}{144}
          + \frac{494}{9(1-y)^4}
          - \frac{1258}{9(1-y)^3}
          + \frac{1130}{9(1-y)^2}
          - \frac{1081}{24(1-y)}\nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{4}{(1+y)^2} 
          - \frac{21}{2(1+y)}
          - \Bigg(
            \frac{7}{2} 
          + \frac{15}{(1-y)^5}
          - \frac{75}{2(1-y)^4} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{33}{(1-y)^3}
          - \frac{9}{(1-y)^2}
          - \frac{2}{(1-y)}
          + \frac{3}{(1+y)^2}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{7}{2(1+y)}
            \Bigg) \zeta(2)
          \Bigg] H(0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            4
          - \frac{12}{(1-y)^3}
          + \frac{18}{(1-y)^2}
          - \frac{10}{(1-y)}
          - \frac{4}{(1+y)}
         \Bigg] H(-1;y) \nn \\
\hspace*{-5mm} & &
       + \Bigg[
            \frac{55}{8} \! 
          -  \frac{48}{(1-y)^3} \! 
          +  \!  \frac{72}{(1-y)^2} \! 
          - \frac{115}{4(1-y)} \! 
          - \frac{9}{(1+y)} \! 
         \Bigg] H(1;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            4 \! 
          +  \!  \frac{90}{(1-y)^4} \! 
          -  \!  \frac{180}{(1-y)^3} \! 
          +  \! \frac{197}{2(1 \! - \! y)^2} \! 
          -  \! \frac{17}{2(1 \! - \! y)} \! 
         \Bigg] H(-1,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{5}{2} \! 
          +  \!  \frac{24}{(1-y)^4} \! 
          -  \!  \frac{48}{(1-y)^3} \! 
          +  \!  \frac{29}{(1-y)^2} \! 
          -  \!  \frac{5}{(1-y)} \! 
          +  \!  \frac{2}{(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          -  \frac{2}{(1+y)}
         \Bigg]  H(0,-1;y) \nn \\
%
\hspace*{-5mm} & & 
       - \Bigg[
            \frac{8}{3} \! 
          +  \!  \frac{21}{(1 \! - \! y)^5} \! 
          -  \!  \frac{69}{(1 \! - \! y)^4} \! 
          +  \!  \frac{75}{(1 \! - \! y)^3} \! 
          -  \! \frac{95}{4(1 \! - \! y)^2} \! 
          -  \! \frac{223}{24(1 \! - \! y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{(1+y)^2} 
          + \frac{41}{24(1+y)}
         \Bigg] H(0,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            4
          +  \frac{96}{(1-y)^4}
          -  \frac{192}{(1-y)^3}
          + \frac{221}{2(1-y)^2} 
          -  \frac{15}{(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{2(1+y)}
         \Bigg] H(0,1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{3}{2} \! 
          +   \! \frac{63}{(1-y)^4} \! 
          -  \!  \frac{126}{(1-y)^3} \! 
          +  \! \frac{135}{2(1-y)^2} \! 
          -  \! \frac{4}{(1-y)} \! 
          +  \! \frac{4}{(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{7}{2(1+y)}
         \Bigg] H(1,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            1
          - \frac{1}{(1-y)}
          - \frac{1}{(1+y)}
         \Bigg] H(1,1;y) \nn \\
%
\hspace*{-5mm} & & 
       - \Bigg[
            2 \! 
          -  \!  \frac{36}{(1-y)^5} \! 
          +  \!  \frac{90}{(1-y)^4} \! 
          -  \!  \frac{78}{(1-y)^3}  \! 
          +  \! \frac{29}{(1-y)^2} \! 
          -  \! \frac{9}{2(1-y)}  \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{2}{(1+y)^2}
          - \frac{5}{2(1+y)}
         \Bigg]  H(-1,0,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            1
          -  \frac{90}{(1-y)^5}
          +  \frac{225}{(1-y)^4}
          -  \frac{175}{(1-y)^3}
          + \frac{75}{2(1-y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{5}{4(1-y)}
          + \frac{7}{4(1+y)}
         \Bigg] H(0,-1,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            3 \! 
          -   \! \frac{24}{(1-y)^5} \! 
          +   \! \frac{60}{(1-y)^4} \! 
          -  \!  \frac{44}{(1-y)^3}  \! 
          +  \! \frac{7}{(1-y)^2} \! 
          -  \! \frac{2}{(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{1}{(1+y)^2} 
          - \frac{2}{(1+y)}
         \Bigg] H(0,0,-1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{27}{4}
          - \frac{3}{2(1-y)^5}
          + \frac{15}{4(1-y)^4}
          - \frac{17}{8(1-y)^3} 
          + \frac{59}{16(1-y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{201}{32(1-y)}
          + \frac{17}{4(1+y)^2} 
          - \frac{217}{32(1+y)}
         \Bigg] H(0,0,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{11}{2}
          - \frac{96}{(1-y)^5}
          + \frac{240}{(1-y)^4}
          - \frac{373}{2(1-y)^3}
          + \frac{169}{4(1-y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{45}{8(1-y)}
          + \frac{5}{2(1+y)^2}
          - \frac{21}{8(1+y)}
         \Bigg] H(0,0,1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{7}{2}
          + \frac{63}{(1-y)^5}
          - \frac{315}{2(1-y)^4}
          + \frac{517}{4(1-y)^3} 
          - \frac{271}{8(1-y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{19}{16(1-y)}
          + \frac{5}{2(1+y)^2} 
          - \frac{67}{16(1+y)}
         \Bigg] H(0,1,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            1 
          + \frac{1}{(1-y)^2}
          - \frac{1}{(1-y)}
          + \frac{1}{(1+y)^2}
          - \frac{1}{(1+y)}
         \Bigg] \times \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
         \times \big[ 
             \zeta(2) H(-1;y)
           - 4 \zeta(2) H(1;y)
           + 2 H(-1,0,-1;y)    \nn \\
\hspace*{-5mm} & & \hspace*{10mm}
           - 2 H(-1,0,1;y)
           + H(0,1,1;y)
           + 3 H(1,0,0;y) \nn \\
\hspace*{-5mm} & & \hspace*{10mm}
           + 2 H(1,0,1;y)
           + 4 H(1,1,0;y)
         \big] \nn \\
%
\hspace*{-5mm} & + & {\mathcal O}(D-4) \, , \\
\hspace*{-5mm} \Im \, F^{(2l)}_{2}(D,-s) & = & 
      \frac{1}{(D-4)}
       \Bigg\{
            \frac{1}{2(1-y)}
          - \frac{1}{2(1+y)}
          + \Bigg[
            \frac{1}{(1-y)^2}
          - \frac{1}{(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{18mm}
          - \frac{1}{(1+y)^2}
          + \frac{1}{(1+y)}
            \Bigg] H(0;y) 
       \Bigg\} \nn \\
%
\hspace*{-5mm} & &     
          - \frac{31}{3(1-y)^3}
          + \frac{31}{2(1-y)^2}
          - \frac{751}{144(1-y)}
          + \frac{7}{144(1+y)} \nn \\
\hspace*{-5mm} & & 
         + \Bigg[
            \frac{15}{(1-y)^4} \! 
          -  \! \frac{30}{(1-y)^3} \! 
          +  \! \frac{39}{2(1-y)^2} \! 
          -  \! \frac{9}{2(1-y)} \! 
          -  \!  \frac{3}{2(1+y)^2} \nn \\
\hspace*{-5mm} & &
          +  \frac{3}{2(1+y)}
           \Bigg] \zeta(2)
         - \Bigg[
             \frac{42}{(1-y)^5}
          -  \frac{105}{(1-y)^4}
          + \frac{175}{2(1-y)^3}  \nn \\
\hspace*{-5mm} & & 
          - \frac{105}{4(1-y)^2} 
          + \frac{7}{8(1-y)}
          + \frac{7}{8(1+y)}
           \Bigg] \zeta(3) \nn \\
\hspace*{-5mm} & &
       - \Bigg[
            \frac{12}{(1 \! - \! y)^3} \! 
          - \frac{18}{(1 \! - \! y)^2} \! 
          + \frac{7}{(1 \! - \! y)} \! 
          - \frac{1}{(1 \! + \! y)} \! 
         \Bigg] H(-1;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            \frac{166}{3(1 \! - \! y)^4} \! 
          - \!  \frac{422}{3(1 \! - \! y)^3} \! 
          +  \! \frac{709}{6(1 \! - \! y)^2} \! 
          -  \! \frac{773}{24(1 \! - \! y)} \! 
          -  \! \frac{1}{2(1 \! + \! y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{8(1+y)}
          - \Bigg(
            \frac{15}{(1-y)^5}
          - \frac{75}{2(1-y)^4}
          + \frac{127}{4(1-y)^3} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{81}{8(1-y)^2} 
          + \frac{5}{16(1-y)}
          - \frac{1}{2(1+y)^3}
          + \frac{3}{4(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{5}{16(1+y)}
            \Bigg) \zeta(2)
         \Bigg] H(0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{48}{(1-y)^3}
          - \frac{72}{(1-y)^2}
          + \frac{87}{4(1-y)} 
          + \frac{9}{4(1+y)}
         \Bigg] H(1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
             \frac{90}{(1-y)^4}
          -  \frac{180}{(1-y)^3}
          +  \frac{89}{(1-y)^2}
          + \frac{1}{(1-y)}
          + \frac{1}{(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{(1+y)}
         \Bigg]  H(-1,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
             \frac{24}{(1-y)^4}
          -  \frac{48}{(1-y)^3}
          +  \frac{25}{(1-y)^2}
          - \frac{1}{(1-y)}
          - \frac{1}{(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{1}{(1+y)}
         \Bigg] H(0,-1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[ 
            \frac{21}{(1-y)^5} \! 
          -  \! \frac{69}{(1-y)^4} \! 
          +  \! \frac{293}{4(1-y)^3} \! 
          -  \! \frac{91}{4(1-y)^2} \! 
          -  \! \frac{77}{16(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{7}{8(1+y)^2}
          + \frac{51}{16(1+y)}
         \Bigg] H(0,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[ 
            \frac{96}{(1-y)^4} \! 
          -  \! \frac{192}{(1-y)^3} \! 
          +  \! \frac{197}{2(1-y)^2} \! 
          -  \! \frac{5}{2(1-y)} \! 
          +  \! \frac{1}{2(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{2(1+y)}
         \Bigg] H(0,1;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            \frac{63}{(1-y)^4} \! 
          -  \! \frac{126}{(1-y)^3} \! 
          +  \! \frac{257}{4(1-y)^2} \! 
          -  \! \frac{5}{4(1-y)} \! 
          +  \! \frac{1}{4(1+y)^2} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{4(1+y)}
         \Bigg] H(1,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
             \frac{36}{(1-y)^5} \! 
          -  \frac{90}{(1-y)^4} \! 
          +  \frac{75}{(1-y)^3} \! 
          - \frac{45}{2(1-y)^2} \! 
          + \frac{3}{4(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{3}{4(1+y)}
         \Bigg] H(-1,0,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[ 
             \frac{90}{(1-y)^5} \! 
          -   \! \frac{225}{(1-y)^4} \! 
          +  \! \frac{335}{2(1-y)^3} \! 
          -  \! \frac{105}{4(1-y)^2} \! 
          -  \! \frac{25}{8(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{25}{8(1+y)}
         \Bigg] H(0,-1,0;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[ 
             \frac{24}{(1-y)^5} \! 
          -  \frac{60}{(1-y)^4} \! 
          +   \! \frac{42}{(1-y)^3} \! 
          -  \frac{3}{(1-y)^2} \! 
          - \frac{3}{2(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{3}{2(1+y)}
         \Bigg]  H(0,0,-1;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[ 
            \frac{3}{2(1-y)^5} \! 
          -  \! \frac{15}{4(1-y)^4} \! 
          +  \! \frac{2}{(1-y)^3} \! 
          +  \! \frac{3}{4(1-y)^2} \! 
          -  \! \frac{11}{32(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{3}{8(1+y)^3} \! 
          +  \! \frac{9}{16(1+y)^2} \! 
          - \frac{11}{32(1+y)} \! 
         \Bigg] H(0,0,0;y) \nn \\
%
\hspace*{-5mm} & &
       + \Bigg[
            \frac{96}{(1-y)^5} \! 
          -  \! \frac{240}{(1-y)^4} \! 
          +  \! \frac{357}{2(1-y)^3} \! 
          -  \! \frac{111}{4(1-y)^2} \! 
          -  \! \frac{7}{2(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          - \frac{1}{2(1+y)^3}
          + \frac{3}{4(1+y)^2}
          - \frac{7}{2(1+y)}
         \Bigg] H(0,0,1;y) \nn \\
%
\hspace*{-5mm} & &
       - \Bigg[
            \frac{63}{(1-y)^5} \! 
          -  \! \frac{315}{2(1-y)^4} \! 
          +  \! \frac{124}{(1-y)^3}  \! 
          -  \! \frac{57}{2(1-y)^2} \! 
          -  \! \frac{7}{16(1-y)} \nn \\
\hspace*{-5mm} & & \hspace*{5mm}
          + \frac{1}{4(1+y)^3}
          - \frac{3}{8(1+y)^2}
          - \frac{7}{16(1+y)}
         \Bigg] H(0,1,0;y) \nn \\
%
\hspace*{-5mm} & + & {\mathcal O}(D-4) \, .
\eea

\section{Expansion for $Q^2 \gg m^2$ \label{Q2>>a}}

We present in this section the asymptotic limit $Q^2 \gg m^2$ ($x \to
0$) of the space-like UV-renormalized form factors, given in section 
\ref{fullyren}, putting for brevity $L = \log (Q^2/m^2) = \log (q^2)$.
The analytic continuation for large time-like $S=-Q^2$ can be carried
out with the replacement $Q^2=-(S+i \epsilon)$, generating an
imaginary parts in the logarithm $L$ (which becomes 
$L = \log(-s-i\epsilon) = \log{(s)} - i \pi\ ,$). 
\bea
F^{(2l)}_{1}(D,q^2) & = &  
%
         \frac{1}{(D-4)^2} \Bigg\{
%
            \frac{1}{2}
          - L
          + \frac{1}{2} L^2 
%
       + \frac{1}{q^2}   \Bigg(
          - 2
          + 2 L
          \Bigg)
       \nn \\
       & & \qquad \qquad \quad
       +  \frac{1}{q^4}   \Bigg(
            5
          - 5 L
          + 2 L^2
          \Bigg)
       \nn \\
       & & \qquad \qquad \quad
       +  \frac{1}{q^6}   \Bigg(
          - \frac{50}{3}
          + \frac{68}{3} L
          - 8 L^2
          \Bigg)
%
       \Bigg\} \nn \\
%
 & &
       + \frac{1}{(D-4)} \Bigg\{ 
%
          - 1
          + \frac{1}{2} \zeta(2)
          + \frac{7}{4} L
          - \frac{1}{2} \zeta(2) L 
          - L^2
          + \frac{1}{4} L^3
       \nn \\
       & & \qquad \qquad \quad
       +  \frac{1}{q^2}   \Bigg(
            \frac{5}{2}
          - \zeta(2)
          - \frac{7}{2} L
          + 2 L^2
          \Bigg)
       \nn \\
       & & \qquad \qquad \quad
       +  \frac{1}{q^4}   \Bigg(
          - 6
          + \frac{5}{2} \zeta(2)
          + \frac{55}{4} L
          - 2 \zeta(2) L 
       \nn \\
       & & \qquad \qquad \quad
          - \frac{29}{4} L^2
          + L^3
          \Bigg)
%
       +  \frac{1}{q^6}   \Bigg(
            \frac{212}{9}
          - \frac{34}{3} \zeta(2)
       \nn \\
       & & \qquad \qquad \quad
          - \frac{500}{9} L
          + 8 \zeta(2) L 
          + \frac{97}{3} L^2
          - 4 L^3
          \Bigg)
%
       \Bigg\} \nn \\
%
 & &
          + \frac{1387}{216}
          + \frac{33}{16} \zeta(2)
          - 3  \zeta(2) \ln2
          - \frac{11}{4} \zeta(3)
          - \frac{59}{40} \zeta^2(2)
          \nn \\
          & &
          - \frac{3355}{864} L
          + \frac{1}{12} \zeta(2) L 
          + 2 \zeta(3) L 
          + \frac{571}{288} L^2
          - \frac{3}{8} \zeta(2) L^2 
           \nn \\
          & &
         - \frac{4}{9} L^3
          + \frac{7}{96} L^4
          \nn \\
%
 & &
       + \frac{1}{q^2}   \Bigg(
          - \frac{6301}{432}
          - \frac{173}{24} \zeta(2)
          + 18  \zeta(2) \ln2
          - \frac{3}{2} \zeta(3)
          + \frac{8}{5} \zeta^2(2)
           \nn \\
          & & \qquad \quad
          + \frac{487}{48} L
          - \frac{15}{4} \zeta(2) L 
          + 3 \zeta(3) L 
          - \frac{97}{16} L^2
          + \frac{1}{2} \zeta(2) L^2 
           \nn \\
          & & \qquad \quad
          + \frac{7}{6} L^3
          - \frac{1}{48} L^4
          \Bigg) \nn \\
%
 & &
       + \frac{1}{q^4}   \Bigg(
            \frac{12871}{144}
          - \frac{109}{12} \zeta(2)
          - 72  \zeta(2) \ln2
          + \frac{123}{2} \zeta(3)
           \nn \\
          & & \qquad \quad
          - \frac{503}{20} \zeta^2(2)
          - \frac{7487}{216} L
          + \frac{401}{12} \zeta(2) L 
          - 37 \zeta(3) L 
           \nn \\
          & & \qquad \quad
          + \frac{1261}{36} L^2
          - \frac{27}{4} \zeta(2) L^2 
          - \frac{295}{72} L^3
          + \frac{7}{12} L^4
          \Bigg) \nn \\
%
 & &
       + \frac{1}{q^6}   \Bigg(
          - \frac{94525}{144}
          + \frac{5317}{18} \zeta(2)
          + 288  \zeta(2) \ln2
          - \frac{1447}{3} \zeta(3)
           \nn \\
          & & \qquad \quad
          + 227 \zeta^2(2)
          + \frac{11401}{216} L
          - \frac{1175}{6} \zeta(2) L 
          + 296 \zeta(3) L 
           \nn \\
          & & \qquad \quad
          - \frac{4451}{24} L^2
          + 55 \zeta(2) L^2 
          + \frac{403}{18} L^3
          - \frac{9}{4} L^4
          \Bigg) \nn \\
 %
 & + & {\mathcal O} \Bigg(\frac{1}{q^8} \Bigg) \ , \\
F^{(2l)}_{2}(D,q^2) & = & 
         \frac{1}{(D-4)} \Bigg\{
%
         \frac{1}{q^2}   \Bigg(
            L
          - L^2
          \Bigg)
%
       + \frac{1}{q^4}   \Bigg(
            2
          - 6 L
          + 2 L^2
          \Bigg)
       \nn \\
       & & \qquad \qquad \quad
%
       + \frac{1}{q^6}   \Bigg(
          - 11
          + 20 L
          - 8 L^2
          \Bigg)
%
       \Bigg\} \nn \\
%
 & & 
       + \frac{1}{q^2}   \Bigg(
            \frac{17}{3}
          + \frac{11}{4} \zeta(2)
          - 12  \zeta(2) \ln2
           \nn \\
          & & \qquad \quad
          + \zeta(3)
          - \frac{379}{72} L
          + 3 \zeta(2) L 
          + \frac{55}{24} L^2
          - \frac{1}{2} L^3
          \Bigg) \nn \\
%
 & &
       + \frac{1}{q^4}   \Bigg(
          - \frac{1717}{36}
          + \frac{31}{2} \zeta(2)
          + 48  \zeta(2) \ln2
          - 56 \zeta(3)
           \nn \\
          & & \qquad \quad
          + \zeta(3)
          + \frac{66}{5} \zeta^2(2)
          + \frac{931}{36} L
          - 19 \zeta(2) L 
          + 28 \zeta(3) L 
           \nn \\
          & & \qquad \quad
          - \frac{289}{12} L^2
          + 4 \zeta(2) L^2 
          + \frac{7}{3} L^3
          - \frac{1}{12} L^4
          \Bigg) \nn \\
%
 & &
       + \frac{1}{q^6}   \Bigg(
          + \frac{34471}{72}
          - 254 \zeta(2)
          - 192 \zeta(2) \ln2 
          + 448 \zeta(3)
           \nn \\
          & & \qquad \quad
          - \frac{858}{5} \zeta^2(2)
          - \frac{295}{6} L
          + 129 \zeta(2) L 
          - 252 \zeta(3) L 
           \nn \\
          & & \qquad \quad
          + 147 L^2
          - 44 \zeta(2) L^2 
          - \frac{83}{6} L^3
          + \frac{1}{4} L^4
          \Bigg) \nn \\
%
& + & {\mathcal O}\Bigg(\frac{1}{q^8} \Bigg) \, .
\eea



\section{Expansion for $Q^2 \ll m^2$ \label{Q2<<a}}

We present in this section the expansion in $Q^2=m^2q^2$ around $Q^2=0$, 
valid for $Q^2 \ll m^2$ ($x \to 1$), of the form factors. Since we are 
in the analyticity region, the expansions are valid for space-like as 
well as time-like values of $Q^2$. The expansion reads 
\bea
F^{(2l)}_{1}(D,q^2) & = & 
            \frac{q^4}{(D-4)^{2}} \Biggl\{ \frac{1}{18} 
%
                          - q^2   \biggl( \frac{1}{60} \biggr)
                                  \Biggr\} \nn\\
& - & \frac{q^4}{(D-4)} \Biggl\{ \frac{1}{24} 
%
                          - q^2   \biggl( \frac{31}{1440} \biggr) 
                        \Biggr\} \nn\\
&  & +  \,  q^2  \, \biggl(
                             \frac{4819}{5184}
                           + \frac{49}{72} \zeta(2)
                           - 3 \zeta(2) \ln{2}
                           + \frac{3}{4} \zeta(3)
                    \biggr) \nn\\
%
& &  - \, q^4   \,  \biggl(
            \frac{1349}{6480}
          + \frac{8731}{28800} \zeta(2)
          - \frac{11}{10} \zeta(2) \ln{2}
          + \frac{11}{40} \zeta(3)
          \biggr) \nn\\
%
& &        + \, q^6   \,  \biggl(
            \frac{109069}{1814400} \! 
          +  \! \frac{723901}{5644800} \zeta(2) \! 
          -  \! \frac{113}{280} \zeta(2) \ln{2} \! 
          +  \! \frac{113}{1120} \zeta(3) \! 
          \biggr) \nn\\
& + & {\mathcal O} \bigl( q^8 
\bigr) \, , \\
F^{(2l)}_{2}(D,q^2) & = & - \frac{q^2}{(D-4)} \Biggl\{ \frac{1}{6} 
%
       -  \, q^2  \,  \biggl(
            \frac{19}{360}
          \biggr)
%
       +  \, q^4   \,  \biggl(
            \frac{73}{5040}
          \biggr)
\Biggr\} \nn\\
& &       +   \frac{197}{144}  
          +   \frac{1}{2} \zeta(2)  
          -   3 \zeta(2) \ln{2}  
          +   \frac{3}{4} \zeta(3)  \nn\\
%
& &        -  \, q^2   \,  \biggl(
            \frac{1031}{2160}  
          + \frac{13}{20} \zeta(2)
          -   \frac{23}{10} \zeta(2) \ln{2}
          + \frac{23}{40} \zeta(3)
          \biggr) \nn\\
%
& &        +   \,  q^4   \,  \biggl(
            \frac{2551}{15120}  
          + \frac{187}{525} \zeta(2)
          - \frac{15}{14} \zeta(2) \ln{2}  
          + \frac{15}{56} \zeta(3)
          \biggr) \nn\\
%
& &        -  \, q^6   \,  \biggl(
            \frac{69019}{1209600}
          + \frac{140951}{940800} \zeta(2)
          - \frac{29}{70} \zeta(2) \ln{2}
          + \frac{29}{280} \zeta(3)
          \biggr) \nn\\
& + & {\mathcal O} \bigl( q^8 \bigr) \, .
\eea

It is to be noted that the absence of the first order term in $q^2$ 
from the (infrared) singular part of $F_1^{(2l)}(D,q^2)$ and the absence
of the zeroth order term from the (infrared) singular part of
$F_2^{(2l)}(D,q^2)$ guarantee the finiteness of the 2-loop value for the 
charge slope of the electron ($-F_{1}^{(2l)}$$'(0)$ in the current 
notation) and of the magnetic anomaly $F_{2}^{(2l)}(0)$. The corresponding 
values are of course in agreement with the well known results in the 
literature (see for example \cite{2loop1}). 


\section{Summary}

We carried out the analytic calculation of the renormalized on shell 
form factors of the electron vertex at two loops in perturbative QED. 
We gave the full results, for arbitrary momentum 
transfer $S$ and on-shell fermionic external lines with finite mass 
$m$, in the space-like region, $S < 0$, and the imaginary parts in the 
time-like region above the threshold, $S > 4m^2$. 
The results are expressed in term of 1-dimensional HPLs 
of maximum weight 4.

We gave also the expansions of the form factors in the two kinematical 
regions of large momentum transfer, $S \to \infty$, and small momentum 
transfer, $S \to 0$, recovering in particular the known results for the 
charge slope and the anomalous magnetic moment of the electron. 

For the calculation we used the MIs calculated in a previous paper,
regularized in the dimensional regularization scheme. Both UV and soft
IR divergences are regularized with the parameter $D$, dimension of the
space-time. In the final results, after the renormalization of the UV
divergences, the form factors still possess polar singularities in $(D-4)$, 
because of the soft IR divergences (the poles in $1/(D-4)$ cancel out 
from any physical observable when soft real photons are also accounted 
for). 


\section{Acknowledgment}
We are grateful to J. Vermaseren for his kind assistance in the use
of the algebra manipulating program {\tt FORM}~\cite{FORM}, by which
all our calculations were carried out.

We wish to thank A. Ferroglia for useful discussions. 

\appendix

\section{Propagators \label{app1}}

We list here the denominators of the integral expressions appeared in
the paper.
\bea
{\mathcal D}_{1} & = & k_{1}^{2} \, , \\
{\mathcal D}_{2} & = & k_{2}^{2} \, , \\
{\mathcal D}_{3} & = & (k_{1}-k_{2})^{2} \, , \\
{\mathcal D}_{4} & = & (p_{1}-k_{1})^{2} \, , \\
{\mathcal D}_{5} & = & (p_{2}-k_{2})^{2} \, , \\
{\mathcal D}_{6} & = & [k_{1}^{2}+m^2] \, , \\
{\mathcal D}_{7} & = & [k_{2}^{2}+m^2] \, , \\
{\mathcal D}_{8} & = & [(k_{1}+k_{2})^{2}+m^2] \, , \\
{\mathcal D}_{9} & = & [(p_{1}-k_{1})^{2}+m^2] \, , \\
{\mathcal D}_{10} & = & [(p_{2}+k_{1})^{2}+m^2] \, , \\
{\mathcal D}_{11} & = & [(p_{2}-k_{2})^{2}+m^2] \, , \\
{\mathcal D}_{12} & = & [(p_{1}-k_{1}+k_{2})^{2}+m^2] \, , \\
{\mathcal D}_{13} & = & [(p_{2}+k_{1}-k_{2})^{2}+m^2] \, .
\eea

\section{One-loop results \label{app2}}

For completeness we give in this appendix the 1-loop form factors, 
corresponding to the first order in $\alpha / \pi$ in 
Eqs. (\ref{b00020}-\ref{b00022}), expanded up to the first order 
in $(D-4)$. 

The Feynman diagrams involved are those in Fig. \ref{fig3}, where the
order $\alpha / \pi$ Feynman diagram and the relative subtraction for
the renormalization of the charge form factor are shown.


%%%%%%%%%%%%%%%%%%%% 1-loop Vertex %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bfig
\bc
%
\subfigure[]{
\begin{fmfgraph*}(30,30)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmf{fermion}{i1,v1}
\fmf{fermion}{i2,v2}
\fmf{photon}{v5,o}
\fmflabel{$p_2$}{i1}
\fmflabel{$p_1$}{i2}
\fmf{fermion,tension=.3}{v2,v5}
\fmf{fermion,tension=.3}{v1,v5}
\fmf{photon,tension=0}{v2,v1}
\end{fmfgraph*}}
%
%%%%%%%%%%%%%%%%%%%%%%%
%
\hspace{10mm}
\subfigure[]{
\begin{fmfgraph*}(30,30)
\fmfleft{i1,i2}
\fmfright{o}
\fmfforce{0.2w,0.93h}{v2}
\fmfforce{0.2w,0.07h}{v1}
\fmfforce{0.8w,0.5h}{v5}
\fmfforce{0.66w,0.5h}{v55}
\fmf{plain}{i1,v1}
\fmf{plain}{i2,v2}
\fmf{photon}{v5,o}
\fmflabel{$p_2$}{i1}
\fmflabel{$p_1$}{i2}
%\fmfv{d.s=hexacross}{v5}
\fmfv{label=$\otimes$}{v55}
\fmfv{l=$Z_{1}^{(1l)}$,l.a=70,l.d=.12w}{v5}
\fmf{fermion,tension=.3}{v2,v5}
\fmf{fermion,tension=.3}{v1,v5}
\end{fmfgraph*}}
%
%%%%%%%%%%%%%%%%%%%%%%%
%
\vspace*{8mm}
\caption{\label{fig3} 1-loop vertex diagrams for the QED form factor.}
\ec
\efig
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The unrenormalized form factors ${\mathcal F}_1^{(1l)}(D,q^2)$, 
${\mathcal F}_2^{(1l)}(D,q^2)$ and ${\mathcal F}_3^{(1l)}(D,q^2)$ for the 
diagram {\tt(a)} in Fig. \ref{fig3} were already given in 
Eqs. (\ref{1loopF1}--\ref{c7}), while 
$Z^{(1l)}_1(D)$ was given in Eq. (\ref{c001}). The counter-term in Fig. 
\ref{fig3} {\tt(b)} contributes only to the renormalization of the {\it
charge} form factor, being the product of the renormalization constant
$Z_{1}^{(1l)}(D)$ at 1-loop level, Eq. (\ref{c001}), and the 
tree-level vertex.

The UV-renormalized form factors at 1-loop level, in the space-like
region $-S=Q^2=m^2q^2 > 0$, up to the first order in $(D-4)$, are the 
following: 
\bea
F^{(1l)}_{1}(D,q^2) & = &   \frac{1}{(D-4)} \Biggl\{ 
1 - \biggl[ 1 - \frac{1}{(1-x)} - \frac{1}{(1+x)} \biggr] H(0;x)
\Biggr\} 
\nn\\
& &       - 1 
          + \frac{1}{2} \biggl[ 
	    \frac{1}{2} \! 
	  - \! \frac{1}{(1\! +\! x)}
          \biggr] H(0;x) \nn\\
& & - \! \frac{1}{2} \biggl[ 1 \! - \! 
   \frac{1}{(1\! -\! x)} \! - \! \frac{1}{(1\! +\! x)} \biggr] \bigl[ 4 
   \zeta(2)  - 2 H(0;x) - H(0,0;x)  \nn\\
& &   - 2 H(1,0;x) \bigr] \nn\\
& + & (D-4) \Biggl\{ 1 
       - \frac{1}{8} \biggl[ 
	    1 \! 
	  -  \! \frac{2}{(1 \! + \! x)} \biggr]
         \bigl[ \zeta(2)  \! 
	  -  \! H(0,0;x)  \! 
	  -  \! 2 H( \! -1,0;x) \bigr] \nn\\
& & \hspace*{18mm}
       + \frac{1}{4} \biggl[ 
            1 
	  - \frac{1}{(1-x)} 
	  - \frac{1}{(1+x)} \biggr] 
	 \bigl[
            2( \zeta(2) + \zeta(3) )  \nn\\
& & \hspace*{23mm}
	  + ( \zeta(2)  \! -  \! 4 ) H(0;x)  \! 
	  -  \! 2 \zeta(2) H( \! 1;x) \! 
          -  \! 2 H(0, \! 0;x)   \nn\\
& & \hspace*{23mm}
	  +  \! 2 H( \! -1,0;x)  \! 
	  -  \! H(0,0,0;x)  \! 
	  -  \! 4 H( \! -1, \! -1,0;x)   \nn\\
& & \hspace*{23mm}
	  +  \! 2 H(-1,0,0;x) \! 
          +  \! 2 H(0,-1,0;x) \bigr] \Biggr\}  \nn\\
& + &  {\mathcal O} \left( (D-4)^2 \right)
\label{appb1}
\, , \\
F^{(1l)}_{2}(D,q^2) & = &  -  \frac{1}{2} \left[ \frac{1}{(1-x)} -
\frac{1}{(1+x)} \right] H(0;x) \nn\\
& - &   (D-4) \Biggl\{ \frac{1}{4} \left[ \frac{1}{(1-x)} -
\frac{1}{(1+x)} \right] \bigl[ \zeta(2) 
           - 4 H(0;x)   \nn\\
& & \hspace*{23mm}
           - 4 H(0,0;x) 
	   + 2 H(-1,0;x)
\bigr] \Biggr\}   \nn\\
& + &  {\mathcal O} \left( (D-4)^2 \right)
\label{appb2} \, , \\
F^{(1l)}_{3}(D,q^2) & = &  0 \, .
\eea 
where $F_{2}^{(2l)}(D,q^2)$ is exactly equal to 
${\mathcal{F}}_{2}(D,q^2)$, given in Eq. (\ref{1loopF2}) and 
$F^{(1l)}_{1}(D,q^2)$ is obtained by subtracting $Z^{(1l)}_1$ of
Eq. (\ref{c001}) from ${\mathcal{F}}_{1}(D,q^2)$ given in 
Eq. (\ref{1loopF1}). 

Eqs. (\ref{appb1},\ref{appb2}) can be analytically continued in the
time-like region $S=-Q^2>0$ and in particular above the physical
threshold $S>4m^2$, where an imaginary part appears. Using the
substitution of Eq. (\ref{analyticY}) and writing ($s=S/m^2$) 
\bea
F^{(1l)}_{1}(D,-s-i\epsilon) & = & \Re \, F^{(1l)}_{1}(D,-s) 
                         + i \pi \, \Im \, F^{(1l)}_{1}(D,-s) \, , \\ 
F^{(1l)}_{2}(D,-s-i\epsilon) & = & \Re \, F^{(1l)}_{2}(D,-s) 
                         + i \pi \, \Im \, F^{(1l)}_{2}(D,-s) \, ,
\eea
the imaginary parts have the following expressions:
\bea
\Im \, F^{(1l)}_{1}(D,-s) & = & - \frac{1}{(D-4)} \Biggl\{ 
1 - \frac{1}{(1-y)} - \frac{1}{(1+y)} 
\Biggr\}  \nn\\
& &  +\frac{1}{4} - \frac{1}{2(1+y)} + \frac{1}{2} \biggl[ 1 - 
\frac{1}{(1-y)} - \frac{1}{(1+y)} \biggr] \bigl[ 1 + H(0;y) \nn\\
& & + 2 H(1;y)
\bigr] \nn\\
& - & (D-4) \Biggl\{ \frac{1}{8} \biggl[ 
          1 
	- \frac{2}{(1+y)} \biggr] \bigl[
	  H(0;y)
	+ 2 H(1;y) \bigr] \nn\\
& & \hspace*{18mm}
      + \frac{1}{4} \biggl[ 
          1 \! 
	-  \! \frac{1}{(1-y)}  \! 
	-  \! \frac{1}{(1+y)} \biggr] \bigl[ 
          4 \! 
	-  \! 2 \zeta(2) \! 
	+  \! H(0;y) \nn\\
& & \hspace*{23mm}
	+ 2 H(1;y)
	+ H(0,0;y)
	+ 2 H(0,1;y) \nn\\
& & \hspace*{23mm}
	+ 2 H(1,0;y)
	+ 4 H(1,1;y) \bigr] \Biggr\}   \nn\\
& + &  {\mathcal O} \left( (D-4)^2 \right) \, , 
\label{appb3} \\
\Im \, F^{(1l)}_{2}(D,-s) & = & \frac{1}{2} \left[ \frac{1}{(1-y)} -
\frac{1}{(1+y)} \right] \nn\\
& - & (D-4) \Biggl\{  \frac{1}{4} \left[ \frac{1}{(1-y)} -
\frac{1}{(1+y)} \right] \bigl[ 
          \zeta(2) 
	- 4 H(0;y) \nn\\
& & \hspace*{23mm}
        - H(0,0;y) 
	+ 2 H(-1,0;y) \bigr]
\Biggr\}  \nn\\
& + &  {\mathcal O} \left( (D-4)^2 \right) \, .
\label{appb4}
\eea
The presence of $1/(D-4)$ singularities even in the on-shell renormalized 
form factors is due to the fact that soft IR divergences are still present. 
As already recalled, in any physical quantity they
will cancel against similar divergences due to soft real photons. 

\begin{thebibliography}{99} 
\def    \np     #1#2#3{{\it Nucl. Phys.} {\bf #1} (19#2) #3}
\def    \nptwoth     #1#2#3{{\it Nucl. Phys.} {\bf #1} (20#2) #3}
\def    \prep   #1#2#3{{\it Phys. Rep.} {\bf #1}  (19#2) #3}
\def    \pl     #1#2#3{{\it Phys. Lett.} {\bf #1} (19#2) #3}
\def    \pltwoth     #1#2#3{{\it Phys. Lett.} {\bf #1} (20#2) #3}
\def    \plold  #1#2#3{{\it Phys. Lett.} {\bf #1B} (19#2) #3}
\def    \prl    #1#2#3{{\it Phys. Rev. Lett.} {\bf #1}  (19#2) #3} 
\def    \pr     #1#2#3{{\it Phys. Rev.} {\bf #1}  (19#2) #3}
\def    \prd    #1#2#3{{\it Phys. Rev.} {\bf D#1}  (19#2) #3} 
\def    \zeit   #1#2#3{{\it Z. Phys.} {\bf C#1}  (19#2) #3}
\def    \cmp    #1#2#3{{\it Comm. Math. Phys.} {\bf #1}  (19#2) #3}
\def    \ibid   #1#2#3{{\it ibid.} {\bf #1} (19#2) #3}
\def    \nc     #1#2#3{{\it Nuovo Cim.} {\bf #1} (19#2) #3}
\def    \acta   #1#2#3{{\it Acta Phys. Polon.} {\bf #1} (19#2) #3}
\def    \tmp    #1#2#3{{\it Theor. Math. Phys.} {\bf #1} (19#2) #3}
\def    \comp    #1#2#3{{\it Comput. Phys. Commun.} {\bf #1} (20#2) #3}
\def    \hepph  #1 {{\tt hep-ph/#1}}
\def    \hepex  #1 {{\tt hep-ex/#1}}
\parskip 0pt
\itemsep=0pt


%%%%%%%%%%%%%% 2-loop %%%%%%%%%%%%%%%%%%%
%
%%%% Up to 3-point functions
%
\bibitem{2loop1}
  R. Barbieri, J. A. Mignaco and E. Remiddi,  \nc{11A}{72}{824}; 
  \nc{11A}{72}{865}. 
%%CITATION = NUCIA,A11,824;%%

%%%%%%%%%%%%%%% Polylog %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{Nielsen}
  N. Nielsen, {\it Nova Acta Leopoldiana} (Halle) {\bf 90} (1909) 123.

\bibitem{Kolbig}
  K. S. K\"olbig, J. A. Mignaco and E. Remiddi, {\it BIT} {\bf 10} 
  (1970) 38.

%%%%%%%%%%%%%% 2-loop %%%%%%%%%%%%%%%%%%%

\bibitem{Pie}
  P. Mastrolia and E. Remiddi, ({\tt hep-ph/0302162}),
  to appear in {\it Nucl. Phys.} {\bf B}.
%%CITATION = HEP-PH 0302162;%%

%%%%%%%%%%%%%%% Harm-Polylog %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{Polylog}
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  {\bf A15} (2000) 725 ({\tt hep-ph/9905237}). 
%%CITATION = HEP-PH 9905237;%%

\bibitem{Polylog3}
  T. Gehrmann and E. Remiddi, \comp{141}{01}{296} 
  ({\tt hep-ph/0107173}).
%%CITATION = HEP-PH 0107173;%%

%%%%%%%%%%%%%%%%%% Dim Reg %%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{DimReg}
  G. 't Hooft and M. Veltman, \np{B44}{72}{189}.\\
%%CITATION = NUPHA,B44,189;%%
  C. G. Bollini and J. J. Giambiagi, \plold{40}{72}{566}; 
  \nc{12B}{72}{20}. \\
%%CITATION = NUCIA,B12,20;%%
  J. Ashmore, {\it Lett. Nuovo Cimento} {\bf 4} (1972) 289.\\
%%CITATION = NCLTA,4,289;%%
  G. M. Cicuta and E. Montaldi, {\it Lett. Nuovo Cimento} {\bf 4} 
  (1972) 289. \\
%%CITATION = NCLTA,4,329;%%
  R. Gastmans and R. Meuldermans, \np{B63}{73}{277}.
%%CITATION = NUPHA,B63,277;%%

%%%%%%%%%%%%%% 2-loop %%%%%%%%%%%%%%%%%%%

\bibitem{Bon1}
  R. Bonciani, P. Mastrolia and E. Remiddi, \nptwoth{B661}{03}{289}
  ({\tt hep-ph/0301170}).
%%CITATION = HEP-PH 0301170;%%

%%%%%%%%%%%%%%%%%% IBP %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

\bibitem{Chet}
  F.V. Tkachov, \pl{B100}{81}{65}.\\
%%CITATION = PHLTA,B100,65;%%
  K.G. Chetyrkin and F.V. Tkachov, \np{B192}{81}{159}.
%%CITATION = NUPHA,B192,159;%%


%%%%%%%%%%%%%%%%%% LI %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{Rem3}
  T. Gehrmann and E. Remiddi, \nptwoth{B580}{00}{485} 
  ({\tt hep-ph/9912329}). 
%%CITATION = HEP-PH 9912329;%%
  
%%%%%%%%%%%%%%%%%%% Programs %%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{FORM} J.A.M.\ Vermaseren, Symbolic Manipulation with
               {\tt FORM}, Version 2, CAN, Amsterdam, 1991; \\
               New features of {\tt FORM}, [math-ph/0010025].
%%CITATION = MATH-TH 0010025;%%

%%%%%%%%%%%%%%%%%% Diff. Eqs. Method %%%%%%%%%%%%%%

\bibitem{Kot}
  A. V. Kotikov, \pl{B254}{91}{158}.
%%CITATION = PHLTA,B254,158;%%

\bibitem{Rem1} 
  E. Remiddi, \nc{110A}{97}{1435} ({\tt hep-th/9711188}).
%%CITATION = HEP-TH 9711188;%%

\bibitem{Rem2}
  M. Caffo, H. Czy\.{z}, S. Laporta and E. Remiddi, 
  \acta{B29}{98}{2627} ({\tt hep-ph/9807119}).\\
%%CITATION = HEP-TH 9807119;%%
  M. Caffo, H. Czy\.{z}, S. Laporta and E. Remiddi, 
  \nc{A111}{98}{365} ({\tt hep-ph/9805118}).
%%CITATION = HEP-TH 9805118;%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bibitem{YFS}
  D. R. Yennie, S. C. Frautschi and H. Suura, {\it Ann. of Phys.} {\bf
  13} (1961) 379.
%%CITATION = APNYA,13,379;%%



\end{thebibliography}
\end{fmffile}
\end{document} 


