\documentclass{article}[12pt]
%\documentstyle[a4,11pt,axodraw]{article}
\usepackage{ amsmath, amsthm, amssymb }
\usepackage{axodraw}
\usepackage{color}
\usepackage{cite}
%
\topmargin -1cm
\textwidth 16.3cm
\textheight 22.5cm
\oddsidemargin 0cm
\evensidemargin 0cm
\newcommand{\be} { \begin{equation} }
\newcommand{\ee} { \end{equation} }
%
%\newcommand{\labbel}[1] { \ \{#1\} \label{#1} }
\newcommand{\labbel}[1] { \label{#1} }
%
\newcommand{\intq} { \int {\mathfrak D}^d q_1 \,{\mathfrak D}^d q_2 }
\newcommand{\intQ} { \int {\mathfrak D}^d k }
\newcommand{\sqrtR} {\sqrt{R_4(b;\,b_1,b_2,b_3,b_4)}}
\newcommand{\Cla}{{\rm Cl}_2\left( \frac{\pi}{3}\right)}
\newcommand{\Lst}{{\rm Ls}_3\left( \frac{2\,\pi}{3}\right)}
\newcommand{\bb}{ (\sqrt{u}-1)^2 }
\newcommand{\BB}{ (\sqrt{u}+1)^2 }
\newcommand{\noo}{{(9,\infty)}}
\newcommand{\zu}{{(0,1)}}
\newcommand{\un}{{(1,9)}}
\newcommand{\ooz}{{(-\infty,0)}}
\newcommand{\Tad}{{\rm Tad}}
\newcommand{\Bub}{{\rm Bub}}
\newcommand{\Tri}{{\rm Tri}}
\newcommand{\D}{\mathfrak{D}}
\newcommand{\I}{\mathcal{I}}
\newcommand{\J}{\mathcal{J}}
\newcommand{\K}{\mathcal{K}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Ima}{{\rm Im}}
\newcommand{\Rea}{{\rm Re}}
\newcommand{\Li}{{\rm Li}}
\newcommand{\intk} { \int {\mathfrak D}^d k }
\newcommand{\intl} { \int {\mathfrak D}^d l }
%
\newcommand{\bubble}[1]{
\mbox{\parbox{2.5cm}{\hspace{0.25cm}
\begin{picture}(2,1)
\thicklines
\put(0.3,0.5){\vector(1,0){0.1}}
%\put(0.5,0.5){\line(1,0){1}}
\put(-0.03,0.5){\line(1,0){0.5}}
\put(1.53,0.5){\line(1,0){0.5}}
\put(1,0.5){\circle{2}}
\put(0.85,1.12){$m_1$}
% \put(0.85,0.60){$m_2$}
\put(0.85,0.08){$m_2$}
\put(0.25,0.7){\makebox(0,0)[b]{$#1$}}
\end{picture}
}}
\hfill}
%
\newcommand{\triangleo}[3]{
\mbox{\parbox{3cm}{\hspace{-0.1cm}
\begin{picture}(2.5,1.4)
\thicklines
%\linethickness{0.30mm}
\put(0,0.7){\line(1,0){0.5}}
\put(1.0,1.2){\line(1,0){0.7}}
\put(1.0,0.2){\line(1,0){0.7}}
\put(1.0,0.2){\line(0,1){1.0}}
\put(0.3,0.7){\vector(1,0){0.1}}
\put(1.5,0.2){\vector(1,0){0.1}}
\put(1.5,1.2){\vector(1,0){0.1}}
\put(0.5,0.7){\line(1,1){0.5}}
\put(0.5,0.7){\line(1,-1){0.5}}
\put(1,1.2){\line(1,0){0.5}}
\put(1,0.2){\line(1,0){0.5}}
\thinlines
%\qbezier(1.0, 1.2)(0.5, 1.2)(0.5, 0.7)
\put(0.25,0.9){\makebox(0,0)[b]{$#1$}}
\put(1.85,1.2){\makebox(0,0)[l]{$#2$}}
\put(1.85,0.2){\makebox(0,0)[l]{$#3$}}
\end{picture}
}}
\hfill}
%
\newcommand{\sunrisetwo}[1]{
\mbox{\parbox{2.5cm}{\hspace{0.25cm}
\begin{picture}(2,1)
\thicklines
\put(0.3,0.5){\vector(1,0){0.1}}
\put(0.5,0.5){\line(1,0){1}}
\put(0,0.5){\line(1,0){0.5}}
\put(1.5,0.5){\line(1,0){0.5}}
\put(1,0.5){\circle{2}}
\put(0.85,1.12){$m_1$}
\put(0.85,0.60){$m_2$}
\put(0.85,0.08){$m_3$}
\put(0.25,0.7){\makebox(0,0)[b]{$#1$}}
\end{picture}
}}
\hfill}
\newcommand{\kite}[1]{
\mbox{\parbox{2.5cm}{\hspace{0.25cm}
\begin{picture}(2,1)
\thicklines
\put(0.3,0.5){\vector(1,0){0.1}}
\put(0,0.5){\line(1,0){0.5}}
%\Photon(0,0)(2,0){5}{2}
\DashLine(18,18)(36,0){4}
\DashLine(37,37)(55,18){4}
\put(1.0,0.0){\line(0,1){1}}
%\put(0.5,0.5){\line(1,-1){0.5}}
%\put(1.5,0.5){\line(-1,1){0.5}}
\put(1.5,0.5){\line(1,0){0.5}}
\put(0.5,0.5){\line(1,1){0.5}}
\put(1.0,0.0){\line(1,1){0.5}}
\put(0.25,0.7){\makebox(0,0)[b]{$#1$}}
\end{picture}
}}
\hfill}
%
\begin{document}
\setlength{\unitlength}{1.3cm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{titlepage}
\vspace*{-1cm}
\begin{flushright}
TTP16-005
\end{flushright}
\vskip 3.5cm
\begin{center}
\boldmath
{\Large\bf Differential equations and dispersion relations for Feynman \\
amplitudes.
The two-loop massive sunrise and the kite integral\\[3mm] }
\unboldmath
\vskip 1.cm
{\large Ettore Remiddi}$^{a,}$
\footnote{{\tt e-mail: ettore.remiddi@bo.infn.it}}
{\large and Lorenzo Tancredi}$^{b,}$
\footnote{{\tt e-mail: lorenzo.tancredi@kit.edu}}
\vskip .7cm
{\it $^a$ DIFA, Universit\`a di Bologna and INFN, Sezione di Bologna,
I-40126 Bologna, Italy } \\
{\it $^b$ Institute for Theoretical Particle Physics, KIT, 76128 Karlsruhe,
Germany }
\end{center}
\vskip 2.6cm
\begin{abstract}
It is shown that the study of the imaginary part and of the corresponding
dispersion relations of Feynman graph amplitudes
within the differential equations method can provide a powerful tool for the
solution of the equations, especially in the massive case.
\par
The main features of the approach are illustrated by discussing the
simple cases of the 1-loop self-mass and of a particular vertex amplitude,
and then used for the evaluation of the two-loop massive sunrise and the
QED kite graph (the problem studied by Sabry in 1962), up to first order
in the $(d-4)$ expansion.
\vskip .7cm
{\it Key words}: Sunrise, Kite, Master integrals, Differential equations,
Dispersion relations
\end{abstract}
\vfill
\end{titlepage}
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction} \labbel{sec:intro} \setcounter{equation}{0}
\numberwithin{equation}{section}
In the last years we have assisted to an impressive increase in our
knowledge of the mathematical
structures that appear in multiloop Feynman integrals, thanks to the
combined use of various computational techniques, such as
to the method of differential
equations~\cite{Kotikov:1990kg,Remiddi:1997ny,Gehrmann:1999as},
the introduction of a class of special functions, (dubbed originally harmonic
polylogarithms, HPLs~\cite{Remiddi:1999ew,Gehrmann:2000zt},
they came out to be a subset of the much larger class of
multiple polylogarithms, MPLs, see~\cite{Goncharov:1998kja,Goncharov:2001iea,Panzer:2014caa}
and references therein), the definition of a so-called canonical
basis~\cite{Henn:2013pwa} for dealing with increasingly larger systems
of differential equations and the use of the Magnus exponentiation~\cite{Argeri:2014qva}.\par
However, most of the above results have been obtained in the massless
limit; indeed, the situation for massive amplitudes is different,
as the two-loop massive sunrise (which has three propagators only)
is still the object of thorough
investigation~\cite{Caffo:1998du,Laporta:2004rb,Bloch:2013tra,Remiddi:2013joa,
Adams:2013nia,Adams:2014vja,Adams:2015gva,Adams:2015ydq,Bloch:2016izu}.
\par
In this paper we will show that the study of the imaginary parts
and related dispersion relations satisfied by the Feynman amplitudes,
within the differential equation frame, can provide another useful
practical tool for their evaluation in the massive case as well.
\par
The imaginary parts of Feynman graphs can be obtained in various ways.
To start with, one can use Cutkosky-Veltman
rule~\cite{Cutkosky:1960sp,Veltman:1963th,Remiddi:1981hn}
for integrating directly the loop momenta in the very definition of the
graphs. When the $d$-continuous dimensional regularization is
used, nevertheless, that is practical only in the simplest cases. Another possibility is
the extraction of the imaginary part from the solution of the differential
equations, which of course requires the knowledge of the solution itself.
More interestingly, one can observe that often the differential equations become
substantially simpler when restricted to the imaginary part only, so that their
solution can become easier. \par
In any case, once the imaginary part of some
amplitude $ A(d;u) $, say $ {\Ima}A(d;u) $, is obtained, one has at disposal the dispersive
representation for $ A(d;u) $, namely an expression of the form
$$ A(d;u) = \frac{1}{\pi}\int dt\ {\Ima}A(d;t)\ \frac{1}{t-u} $$
(where the limit of integration have been skipped for ease of typing).
Such a representation turns out to be very useful when the amplitude
$ A(d;u) $ appears within the inhomogeneous terms of some other differential
equation, regardless of the actual analytical expression of $ A(d;u) $.
Indeed, as the whole dependence on $ u $ is in the denominator $(t-u)$
one can work out its contribution by considering only that denominator,
freezing, so to say,
the $ t $-integration and the
weight $ {\Ima}A(d;t) $ until the dependence on the variable $ u $
(the variable of the differential equation)
has been properly processed. Let us emphasize, again, that such a processing
is, obviously, fully independent of the actual form of $ {\Ima}A(d;t) $.
\par
In the following, we will illustrate the above remarks in a couple of
elementary applications and then use them in the case of the two-loop
QED-kite, {\it i.e.} the two-loop electron self-mass in QED, already
studied by Sabry~\cite{Sabry} long ago. The study of the kite amplitudes requires
in turn the knowledge of the two-loop massive sunrise, which appears as inhomogeneous
terms in their differential equations.
\par
The paper is organized as follows.
We begin in section~\ref{sec:1lB} studying the imaginary part of the one-loop self mass
and its dispersion relation for generic values of the dimensions $d$. We elaborate
on its calculation both from Cutkosky-Veltman rule and from the differential equations.
In section~\ref{sec:1lT} we study a particular vertex amplitude
through the differential equations method. The one-loop self-mass appears as inhomogenous
term in the equations and we show that their evaluation can be simplified,
once the one-loop self-mass is inserted as dispersive relation.
In section~\ref{sec:iter} we move our attention to the two-loop sunrise graph, which
we write as iteration of two one-loop bubbles. This allows us to derive an extremely
compact representation valid for generic $d$, from which one can show that,
at every order in $(d-2)$ (and therefore also in $(d-4)$),
the sunrise can be written as a one-dimensional integral over a square root of a quartic polynomial,
times a combination of multiple polylogarithms only.
The simplicity of this result motivates us to look more systematically for a similarly simple
structure using differential equations
from the very beginning for the whole integral family of the kite. In section~\ref{sec:deq}
we discuss the notation and describe the master integrals which have to be computed.
In section~\ref{sec:simple} we provide the solution of the simple topologies, which can be written
in terms of HPLs only. Then in section~\ref{sec:basis} we start a systematic study of the differential equations
of the sunrise graph. As is it well known, the solution for the sunrise graph is simplified if we
consider its Laurent series in $(d-2)$ instead of in $(d-4)$. Nevertheless, in order to consistently
study all
master integrals in the integral family, it is more convenient to expand everything in $(d-4)$ from
the beginning. We start therefore describing how to find a convenient basis of master integrals
for the sunrise whose
expansion in $(d-4)$ is as simple as the expansion of the original masters in $(d-2)$.
Once we have a basis and the corresponding differential
equations, we show how to solve them iteratively in section~\ref{sec:sun}.
We conclude the section
providing explicit analytical results for both master integrals for the first two non-zero orders
and showing
how to extract their imaginary parts and write dispersion relations for them.
We move then to the kite integral in section~\ref{sec:kite}, where we show how
the representation of the sunrise as a dispersion relation is particularly convenient, as it allows
to write a compact solution
for the first two orders of the kite integral.
Finally we conclude in section~\ref{sec:concl}. We enclose different appendices
where we provide further mathematical details and explicit derivations.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The 1-loop self-mass: imaginary part and dispersion relation }
\labbel{sec:1lB} \setcounter{equation}{0}
\numberwithin{equation}{section}
We define the integration over a loop momentum $k$ in $d$
continuous dimensions as
\be \intk = \frac{1}{C(d)} \int \frac{d^d k}{(2 \pi)^{d-2}}\,,
\labbel{defDk} \ee
with
\be C(d) = (4 \pi)^{(4-d)/2} \Gamma\left( 3 - \frac{d}{2}\right) \ ,
\labbel{defCd} \ee
so that the tadpole amplitude $ \Tad(d;m) $ reads
\be \Tad(d;m) = \intk \ \frac{1}{k^2+m^2}
= \frac{m^{d-2}}{(d-2)(d-4)} \ . \labbel{defTad} \ee
We then consider the 1-loop ``bubble"
\begin{align}
\Bub(d;-q^2,m_1,m_2) &= \bubble{q} \nonumber\\
&= \intk \ \frac{1}{(k^2+m_1^2)((q-k)^2+m_2^2)} \,.
\labbel{defB} \end{align}
We work in the Euclidean metric such that $q^2$ is positive when $q$ is spacelike.
At $ q = 0 $ one has at once
\be \Bub(d;0,m_1,m_2) = \frac{1}{m_1^2-m_2^2}
\left( \Tad(d;m_2) - \Tad(d;m_1) \right) \ . \labbel{B(d,0)} \ee
\par
Cutkosky-Veltman rule gives for the imaginary part of the bubble
amplitude in $d$-continuous dimensions and for $ s = -q^2 > (m_1+m_2)^2 $
the expression
\begin{align}
{\Ima}\Bub(d;s,m_1,m_2) = \pi \ \frac{1}{2}
\frac{B_d}{\sqrt{R_2(s,m_1^2,m_2^2)}}
\,\left( \frac{R_2(s,m_1^2,m_2^2)}{s} \right)^{(d-2)/2} \ ,
\labbel{ImBub}
\end{align}
where we introduced the usual K\"{a}llen function
\begin{align}
R_2(s,m_1^2,m_2^2) &= s^2+m_1^4+m_2^4-2m_1^2s-2m_2^2s-2m_1^2m_2^2
\nonumber\\ &= (s-(m_1+m_2)^2)(s-(m_1-m_2)^2) \ ,
\labbel{defR2} \end{align}
and the $d$ dependent coefficient
\begin{align}
B_d = \frac{4\sqrt\pi }{2^d\ \Gamma\left( 3 - \frac{d}{2}\right)
\Gamma \left( \frac{d-1}{2}\right)}\,, \labbel{defBd}
\end{align}
whose expansion for $d \approx 2$ reads
\begin{align}
B_d = 1 + \frac{1}{2} (d-2) - \frac{1}{12}
(\pi^2-3)(d-2)^2 + \mathcal{O}((d-2)^3)\,. \labbel{Bdat2}
\end{align}
As a further remark, Eq.(\ref{ImBub}) can also be written as
\be \frac{1}{\pi}\, {\Ima}B(d;s,m_1^2,m_2^2) =
\frac{B_d}{2} \left[s-(m_1+m_2)^2\right]^{\frac{d-3}{2}}
\left[s-(m_1-m_2)^2\right]^{\frac{d-3}{2}}
s^{-\frac{d-2}{2}} \ . \labbel{eq:ImB1} \ee
Once the imaginary part is given, we can write a dispersion relation for the
one-loop bubble
\begin{align}
\Bub(d;-q^2,m_1,m_2) &= \intk
\ \frac{1}{(k^2+m_1^2)((q-k)^2+m_2^2)} \nonumber\\
&= \int_{(m_1+m_2)^2}^\infty \frac{dt}{t+q^2}
\frac{1}{\pi}\, {\Ima}\Bub(d;s,m_1,m_2) \ .
\labbel{eq:dispB}
\end{align}
Note that Eq.(\ref{ImBub}) is valid for arbitrary values
of $ d $, but Eq.(\ref{ImBub}) is written in the form most convenient
for the expansion in $ (d-2) $; the same holds also for
Eq.(\ref{eq:dispB}), which is however convergent only for $ d<4 $.
To obtain a formula valid also in the $ d\approx4 $ region,
one can write a subtracted dispersion relation
\begin{align}
\Bub(d;-q^2,m_1,m_2) &= \Bub(d;0,m_1,m_2)
- q^2\int_{(m_1+m_2)^2}^\infty \frac{dt}{t(t+q^2)}
\frac{1}{\pi}\ {\Ima}\Bub(d;t,m_1^2,m_2^2) \ ,
\labbel{eq:subdispB}
\end{align}
where $ \Bub(d;0,m_1,m_2) $, which is given in Eq,(\ref{B(d,0)}), contains
a pole at $ d=4 $, while the integral is convergent for $ d < 6 $.
\par
The 1-loop self-mass amplitude Eq.(\ref{defB}), which in the following
will be written as $ \Bub(d;s) $ for ease of typing, is known to satisfy
the following differential equation in $ s $
\begin{align} \frac{d}{ds} \Bub(d;s) =
&- \frac{1}{2}\left( \frac{1}{s-(m_1+m_2)^2} + \frac{1}{s-(m_1-m_2)^2}
\right) \Bub(d;s) \nonumber\\
&- \frac{1}{2}\left( \frac{1}{s} - \frac{1}{s-(m_1+m_2)^2}
- \frac{1}{s-(m_1-m_2)^2} \right) (d-2)\Bub(d;s) \nonumber\\
&+ N(d;s) \ , \labbel{EqBd} \end{align}
where the inhomogeneous term, $ N(d;s) $ is given by
\begin{align} N(d;s) &= \frac{d-2}{4m_1(m_1^2-m_2^2)} \left(
- 2\frac{m_1}{s} + \frac{m_1-m_2}{s-(m_1+m_2)^2}
+ \frac{m_1+m_2}{s-(m_1-m_2)^2}
\right) \Tad(d;m_1) \nonumber\\
&+ \frac{d-2}{4m_2(m_1^2-m_2^2)} \left(
+ 2\frac{m_2}{s} + \frac{m_1-m_2}{s-(m_1+m_2)^2}
- \frac{m_1+m_2}{s-(m_1-m_2)^2}
\right) \Tad(d;m_2) \ . \labbel{defN}
\end{align}
The homogeneous equation associated to Eq.(\ref{EqBd}) is (obviously)
\begin{align} \frac{d}{ds} b(d;s) =
&- \frac{1}{2}\left( \frac{1}{s-(m_1+m_2)^2} + \frac{1}{s-(m_1-m_2)^2}
\right) b(d;s) \nonumber\\
&- \frac{1}{2}\left( \frac{1}{s} - \frac{1}{s-(m_1+m_2)^2}
- \frac{1}{s-(m_1-m_2)^2} \right) (d-2)b(d;s) \ .
\labbel{Eqbd} \end{align}
One sees immediately that $ {\Ima}\Bub(d;s) $, Eq.(\ref{eq:ImB1}) satisfies
the homogeneous equation, for any value of $ d $. That fact is hardly
surprising, yet it deserves some comments. \par
When looking for a solution of an equation like Eq.(\ref{EqBd}), it can be
convenient, in order to fix the boundary conditions, to start by considering
values of the variable $ s $ for which the solution is expected to be
real (typically, $ s=0 $ or $s $ negative, {\it i.e.} in the spacelike
region).
But as one is also interested in the value of the solution for timelike,
physical values of $s $, one is naturally lead to consider the
solution as a complex analitical function of the argument $s$, to
be evaluated along the whole line $ s+i\epsilon $, with
$ s $ real and varying in the range $ -\infty < s < +\infty $
and $ \epsilon $ small and positive (the Feynman prescription).
As the singular points of the equation correspond to real values of
$ s $, such as for instance $ s =(m_1\pm m_2)^2 $, the function has
no singularities along the $ s+i\epsilon $ line, so that its
value is fully determined by the analytic continuation in terms of the
initial boundary conditions. \par
Moreover, one might be interested in considering separately the real part
$ {\Rea}\Bub(d;s) $ and the imaginary part $ {\Ima}\Bub(d;s) $ of the solution
$ \Bub(d;s) = {\Rea}\Bub(d;s) + i{\Ima}\Bub(d;s) $. In so doing, as the
inhomogeneous term is real, Eq.(\ref{EqBd}) splits into the two equations
\begin{align} \frac{d}{ds} {\Rea}\Bub(d;s) =
&- \frac{1}{2}\left( \frac{1}{s-(m_1+m_2)^2} + \frac{1}{s-(m_1-m_2)^2}
\right) {\Rea}\Bub(d;s) \nonumber\\
&- \frac{1}{2}\left( \frac{1}{s} - \frac{1}{s-(m_1+m_2)^2}
- \frac{1}{s-(m_1-m_2)^2} \right) (d-2){\Rea}\Bub(d;s) \nonumber\\
&+ N(d;s) \labbel{EqReBd} \\
\frac{d}{ds} {\Ima}\Bub(d;s) =
&- \frac{1}{2}\left( \frac{1}{s-(m_1+m_2)^2} + \frac{1}{s-(m_1-m_2)^2}
\right) {\Ima}\Bub(d;s) \nonumber\\
&- \frac{1}{2}\left( \frac{1}{s} - \frac{1}{s-(m_1+m_2)^2}
- \frac{1}{s-(m_1-m_2)^2} \right) (d-2){\Ima}\Bub(d;s) \ ,
\labbel{EqImBd} \end{align}
where Eq.(\ref{EqImBd}) is of course identical to Eq.(\ref{Eqbd}).
One can now look at the {\it real} solution of Eq.(\ref{EqImBd}) for
{\it real} values of $ s $. One finds easily: \\
if $ 0 < s < (m_1-m_2)^2 $ the solution is
\be {\Ima}B(d,s) = c_1 \left[(m_1+m_2)^2-s\right]^{\frac{d-3}{2}}
\left[(m_1-m_2)^2-s\right]^{\frac{d-3}{2}}
s^{-\frac{d-2}{2}} \ ; \nonumber \ee
if $ (m_1-m_2)^2 < s < (m_1+m_2)^2 $ the solution is
\be {\Ima}B(d,s) = c_2 \left[(m_1+m_2)^2-s\right]^{\frac{d-3}{2}}
\left[s-(m_1-m_2)^2\right]^{\frac{d-3}{2}}
s^{-\frac{d-2}{2}} \ ; \nonumber \ee
if $ (m_1+m_2)^2 < s < +\infty $ the solution is
\be {\Ima}B(d,s) = c_3 \left[s-(m_1+m_2)^2\right]^{\frac{d-3}{2}}
\left[s-(m_1-m_2)^2\right]^{\frac{d-3}{2}}
s^{-\frac{d-2}{2}} \ . \nonumber \ee
The evaluation of the solutions in the various regions is almost trivial,
but one needs the knowledge of three constants, $ c_1, c_2, c_3 $ to
actually recover the imaginary part Eq.(\ref{eq:ImB1})
(in that case the constants are,
obviously, $ c_1 = c_2 = 0 $, $ c_3 = \pi B_d/2 $). \par
Summarising, the evaluation of the imaginary parts alone within the
differential equation approach is much simpler than the evaluation of
the complete solution (real and imaginary parts), but requires
some additional external information (such as the knowledge of the
regions in which the imaginary part vanishes and
its normalization when not vanishing).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The 1-loop self-mass and the 1-loop equal mass triangle }
\labbel{sec:1lT} \setcounter{equation}{0}
\numberwithin{equation}{section}
In this section we show how to use Eq.(\ref{eq:dispB}) in the solution
of the differential equation for a particular massive triangle amplitude,
namely
\begin{align}
\Tri(d;s) &= \triangleo{q}{p_1}{p_2}\nonumber \\
&= \int \D^d k \frac{1}{(k^2+m^2)((k-p_1)^2+m^2)((k-p_1-p_2)^2+m^2)}
\labbel{defTri} \end{align}
with $ q=p_1+p_2$, $\ p_1^2 = p_2^2 =0 $ and $ -q^2 = s$ ($q^2$ is positive
when $ q $ is spacelike).
The differential equation in the variable $ s $ for the amplitude
$ \Tri(d;s) $ reads
\begin{align}
\frac{d}{ds} \Tri(d;s) &= -\frac{1}{s} \Tri(d;s) +
\frac{(d-2)}{8 m^4}\left( \frac{1}{s-4 m^2} - \frac{1}{s}\right)
\Tad(d;m)\nonumber \\
&+\frac{(d-3)}{4\,m^2} \left( \frac{1}{s-4 m^2} - \frac{1}{s}\right)
\Bub(d;s) \ ,
\end{align}
where $ \Tad(d;m) $ is the tadpole defined in Eq.(\ref{defTad}),
and $ \Bub(d;s) $ is the equal mass limit of the 1-loop self-mass of the previous
section, {\it i.e.} $ \Bub(d;s) = \Bub(d;s,m,m) $, see Eq.(\ref{defB}).
\par
Now we notice that the homogeneous part of the equation is independent of
$d$ and reads
$$\frac{d}{ds} h(s) = -\frac{1}{s} h(s) \ ; $$
its solution (apart from a multiplicative constant) is
\begin{equation}
h(s) = \frac{1}{s}\,. \labbel{defh}
\end{equation}
We can then use Euler's method to write the general solution for the triangle
as follows
\begin{align}
\Tri(d;s) = c(d;m) \frac{1}{s}\,
&+ \frac{(d-2)}{2\,s} \int_0^s\, \frac{du}{u-4 m^2} \Tad(d;m) \nonumber \\
&+ \frac{(d-3)}{s} \int_0^s\, \frac{du}{u-4 m^2} \, \Bub(d;u) \,,
\labbel{EulTri} \end{align}
where $c(d,m)$ is an integration constant, depending in general
on $d$ and $m$.
We can fix the integration constant requiring that for $s \to 0 $ the
amplitude is not divergent, which implies
\be c(d;m) = 0\ . \ee
The tadpole, Eq.(\ref{defTad}), is of course independent of $u$
and for the bubble we use its dispersive representation Eq.(\ref{eq:dispB})
\begin{align}
\Bub(d;u) = \frac{1}{\pi} \int_{4 m^2}^\infty \frac{dt}{t-u - i\epsilon}
{\Ima}\Bub(d;t) \,. \labbel{dispBub}
\end{align}
Let us recall that the above integral is convergent for $ d\approx 2$ and that
if one is interested in $ d\approx 4 $, one can use its subtracted version
Eq.(\ref{eq:subdispB}).
Assuming for definiteness $ s < 4m^2 $, and therefore ignoring for the
moment the $+i\epsilon$ prescription in Eq.(\ref{dispBub}),
the triangle amplitude becomes
\begin{align}
\Tri(d;s) &= \frac{m^{d-2}}{2\,s\,(d-4)}
\int_0^s\, \frac{du}{u-4 m^2} \nonumber \\
&+ \frac{(d-3)}{s} \frac{1}{\pi} \int_{4 m^2}^\infty \,\frac{dt}{t-4 m^2}\,
{\Ima}\Bub(d;t)
\int_0^s\, du\, \left( \frac{1}{u-4 m^2} + \frac{1}{t-u} \right)\,.
\labbel{eq:Tri0} \end{align}
The integration in $u$ is trivial and we get
\begin{align}
\Tri(d;s) &= \frac{m^{d-2}}{2\,s\,(d-4)}
\ln{\left( 1- \frac{s}{4 m^2}\right)} \nonumber \\
&+ \frac{(d-3)}{s} \frac{1}{\pi}
\int_{4 m^2}^\infty \, \frac{dt}{t-4 m^2}\, {\Ima}\Bub(d;t)
\left[ \ln{\left( 1- \frac{s}{4 m^2}\right)}
-\ln{\left( 1- \frac{s}{t}\right)} \right]
\,. \labbel{eq:Tri1l}
\end{align}
Note that the above result holds for any $d$ (within the considered range)
independently of the actual
explicit form of the inserted amplitude $ \Bub(d;u) $.
If $0~~4 m^2$ it develops an imaginary part.
In order to properly extract it, it is enough to notice that, for $s>4 m^2$,
the $s\to s+ i\epsilon$ prescription gives
\begin{align}
\ln{\left( 1- \frac{s+i\epsilon}{4 m^2} \right)} =
\ln{\left( \frac{s}{4 m^2} -1 \right)} - i\, \pi\,,
\end{align}
and
\begin{align}
\int_{4 m^2}^\infty \, \frac{dt}{t-4 m^2}\, {\Ima}\Bub(d;t)
\ln{\left( 1- \frac{s+i\epsilon}{t}\right)} &=
\int_{4 m^2}^s \, \frac{dt}{t-4 m^2}\, {\Ima}\Bub(d;t) \,\left[
\ln{\left( \frac{s}{t} - 1 \right)} - i\, \pi\right]\nonumber \\
&+\int_s^\infty \, \frac{dt}{t-4 m^2}\, {\Ima}\Bub(d;t)
\ \ln{\left( 1- \frac{s}{t}\right)} \ .
\end{align}
Collecting results and combining the various terms, the imaginary part
of $ \Tri(d;s) $ for $ s>4m^2$ becomes
\begin{align}
\frac{1}{\pi}{\Ima}\Tri(d;s) = - \frac{m^{d-2}}{2\,s\,(d-4)}
- \frac{(d-3)}{s} \frac{1}{\pi} \int_s^\infty \,
\frac{dt}{t-4m^2} \, {\Ima} \Bub(d;t) \,. \labbel{ImTri}
\end{align}
It is to be noted, again, that the above result has been obtained
from Eq.(\ref{eq:Tri1l}) independently of the explicit analytic
expression of $ {\Ima}\Bub(d;u) $.
As a check, we can write the dispersion relation for the triangle
amplitude in terms of its imaginary part (we take $ s<4m^2 $ for
simplicity)
\begin{align}
\Tri(d;s) = \frac{1}{\pi} \int_{4 m^2}^\infty \frac{du}{u-s}
{\Ima} \Tri(d;u) \,. \labbel{dispTri}
\end{align}
By exchanging the order of integrations according to
$$ \int_{4 m^2}^\infty \,du \, \int_u^\infty \, dt\,
= \int_{4 m^2}^\infty \,dt \, \int_{4 m^2}^t \, du\,,$$
Eq.(\ref{eq:Tri1l}) is easily recovered.
\par
%\noindent
Summarising, the use of the dispersive representation of the inserted
amplitude $ \Bub(d;u) $ in the Euler form of the solution of the
differential equation for the triangle amplitude gives, almost at once,
the explicit form of $ \Tri(d;s) $, Eq.(\ref{eq:Tri1l}) in terms of
$ {\Ima}\Bub(d;u) $. The imaginary part $ {\Ima}\Tri(d;s) $ of the triangle amplitude
Eq.(\ref{ImTri}) can also be written in terms of $ {\Ima}\Bub(d;u) $,
without explicit reference to the analytic form of the latter.
The resulting dispersion relation Eq.(\ref{dispTri}) can be useful
if $ \Tri(d;u)$ appears within the inhomogeneous terms of the equations
for the amplitudes of some other process (such as for instance the
QED light-light graphs).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The sunrise as iteration of the bubble graph}
\labbel{sec:iter} \setcounter{equation}{0}
\numberwithin{equation}{section}
Let us consider the sunrise scalar amplitude defined as
\vspace{0.3cm}
\begin{align}
S(d;-p^2,m_1,m_2,m_3) &= \sunrisetwo{p} \nonumber \\& = \intk \intl
\ \frac{1}{(k^2+m_1^2)(l^2+m_2^2)((p-k-l)^2+m_3^2)}\,, \labbel{eq:sunscal}
\end{align}
where the integration measure is defined in Eq.(\ref{defDk}) and we work in
the Euclidean metric for simplicity.
It is well known that the sunrise graph with different masses possesses
four master integrals,
which reduce to two in the case of equal masses~\cite{Caffo:1998du}.
In this section we will not try to give a full solution for all the masters
integrals, but instead we will limit ourselves to considering the scalar
integral~\eqref{eq:sunscal} only and try to study its iterative structure
in $(d-2)$, or equivalently in $(d-4)$.
One possible way to do this is by noting that the sunrise integral can be
written as
\begin{align}
S(d;-p^2,m_1,m_2,m_3) = \intk \frac{1}{k^2+m_1^2}
\intl \frac{1}{(l^2+m_2^2)((p-k-l)^2+m_3^2)}\,, \labbel{sun1}
\end{align}
and according to Eq.(\ref{defB}) the integral in the momentum $l$ is simply
a one-loop bubble with masses $m_2$ and $m_3$ and momentum $q = (p-k) $,
\begin{equation}
\Bub(d;-q^2,m_2,m_3) = \intl \frac{1}{(l^2+m_2^2)((q-l)^2+m_3^2)}\,.
\end{equation}
The dispersive representation Eq.(\ref{eq:dispB}) then gives
\be \intl \frac{1}{(l^2+m_2^2)((q-l)^2+m_3^2)} =
\int_{(m_2+m_3)^2}^\infty \frac{dt}{t+q^2}
\frac{1}{\pi}\ {\Ima}\Bub(d;t,m_2,m_3) \ , \ee
with $ {\Ima}\Bub(d;t,m_2,m_3) $ given by Eq.(\ref{ImBub}).
As $ q=p-k$, Eq.(\ref{sun1}) becomes
\begin{align}
S(d;-p^2,m_1,m_2,m_3) &= \int_{(m_2+m_3)^2}^\infty dt
\ \frac{1}{\pi}\ {\Ima}\Bub(d;t,m_2,m_3) \nonumber\\
&\times \intk \frac{1}{(k^2+m_1^2)((p-k)^2+t)} \,.
\end{align}
Now clearly the integral in $k$ can be seen again as a one-loop
bubble amplitudes, this time with (squared) masses $m_1^2$ and $t$.
Using again the formula~\eqref{eq:dispB} we get
\begin{align}
S(d;-p^2,m_1,m_2,m_3) &= \int_{(m_2+m_3)^2}^\infty dt
\ \frac{1}{\pi}\ {\Ima}\Bub(d;t,m_2,m_3) \nonumber\\
&\times \int_{(\sqrt{t}+m_1)^2}^\infty \frac{dv}{v+p^2}
\ \frac{1}{\pi}\ {\Ima}\Bub(d;v,\sqrt{t},m_1)
\,. \labbel{eq:dispsun1}
\end{align}
We can obtain an equivalent representation by further exchanging the
integrations in the variables $ t $ and $ v $
$$\int_{(m_2+m_3)^2}^\infty dt \,\int_{(\sqrt{t}+m_1)^2}^\infty \, dv =
\int_{(m_1+m_2+m_3)^2}^\infty \, dv
\int_{(m_2+m_3)^2}^{(\sqrt{v}-m_1)^2} \, dt $$
such that, by recalling Eq.(\ref{ImBub}), we are left with
\begin{align}
S(d;-p^2,m_1,&m_2,m_3) =
\frac{B_d^2}{4} \int_{(m_1+m_2+m_3)^2}^\infty \frac{dv}{v+p^2}\nonumber \\
&\times \int_{(m_2+m_3)^2}^{(\sqrt{v}-m_1)^2}
\frac{dt}{\sqrt{R_2(t,m_2^2,m_3^2) R_2(v,t,m_1^2)}}
\left( \frac{R_2(t,m_2^2,m_3^2)}{t} \frac{R_2(v,t,m_1^2)}{v}
\right)^{(d-2)/2}\,. \labbel{eq:dispsun2}
\end{align}
Eq.(\ref{eq:dispsun2}) is the main result of this section. From it we
obtain at once, when $ -p^2 = s > (m_1+m_2+m_3)^2, $
\begin{align}
\frac{1}{\pi}{\Ima}&S(d;s,m_1,m_2,m_3) \nonumber \\ &=
\frac{B_d^2}{4} \int_{(m_2+m_3)^2}^{(\sqrt{s}-m_1)^2}
\frac{dt}{\sqrt{R_2(t,m_2^2,m_3^2) R_2(s,t,m_1^2)}}
\left( \frac{R_2(t,m_2^2,m_3^2)}{t}
\frac{R_2(s,t,m_1^2)}{s}\right)^{(d-2)/2}\,.
\labbel{eq:impartsun}
\end{align}
Note that
Eq.\eqref{eq:impartsun} is nothing but the
$d$-dimensional three-body massive phase space and Eq.(\ref{eq:dispsun2})
could indeed have been obtained also by computing first
the imaginary part of the sunrise graph using Cutkosky-Veltman rule,
and then writing a dispersion relation for it.
Remarkably, the complexity of the result in the general mass case
is practically the same as in the equal mass case $m_1=m_2=m_3=m$.
\par
Let us further emphasize that Eqs.~\eqref{eq:dispsun1},~\eqref{eq:dispsun2}
and~\eqref{eq:impartsun} are all true for generic, continuous values of
$d$. Furthermore, their expansion in $(d-n)$, where $n$ is virtually any
positive integer (and in particular in $(d-2)$), is completely
straightforward and generates only products of logarithms\footnote{Note that
in odd numbers of dimensions, $d=2\,n + 1$, the imaginary part becomes particularly simple
since the square root in Eq.\eqref{eq:impartsun} cancels.}.
This implies in turn that, at every order in $(d-2)$, the integral
in $ v $ in Eq.\eqref{eq:dispsun1} can
\textsl{always be performed} in terms of multiple polylogarithms only.
This shows that, at every order in $(d-2)$, the sunrise integral can be written
as a one-fold integral over the root of a quartic polynomial, times combinations
of multiple polylogaritms. The result is interesting and it resembles
similar results found for the finite term of a completely unrelated
massless double box in $\mathcal{N}=4$~\cite{Paulos:2012nu,
CaronHuot:2012ab}\footnote{One should compare in particular our Eq.\eqref{eq:dispsun1}
with Eq.(3.23) in~\cite{CaronHuot:2012ab}.}.
Finally, the relation of this representation of the imaginary part of the sunrise
Eq.\eqref{eq:impartsun} with the results
obtained by the explicit solution of the system of differential equations
for the two amplitudes of the sunrise problem (which involves two
pairs of solutions, {\it i.e.} four functions altogether,
see for instance section~\ref{sec:sun} of this paper) is also intriguing, but will not
be further investigated here. Starting from the next section we will instead focus
on the more general problem of computing the full set of master integrals
of the kite graph using the differential equations method.
\section{The differential equations for the kite master integrals}
\labbel{sec:deq} \setcounter{equation}{0}
\numberwithin{equation}{section}
Let us consider the family of the integrals of the QED kite graph
with three massive propagators and two massless ones, defined as
\begin{align}
\I(n_1,n_2,n_3,n_4,n_5) &= \kite{p} \nonumber \\ &=
\int \D^d k \,\D^d l\, \frac{1}{D_1^{n_1} D_2^{n_2} D_3^{n_3}
D_4^{n_4} D_5^{n_5}}
\end{align}
where dashed lines represent massless propagators. The five denominators are chosen as
\begin{align}
&D_1 = k^2 + m^2\,, \qquad D_2 = l^2 \,,
\qquad D_3 = (k-l)^2 + m^2\,,\nonumber \\
&D_4 = (k-p)^2\,, \qquad D_5 = (l-p)^2 + m^2\,, \label{topo}
\end{align}
with $ -p^2 = s $ and $ p^2>0 $ when $ p $ is spacelike.
The integration measure is defined as in Eq.(\ref{defDk})
such that according to Eq.(\ref{defTad}) the one-loop tadpole reads
\begin{align}
\int \frac{\D^d k}{k^2 + m^2} = \frac{m^{d-2}}{(d-2)(d-4)}\,.
\end{align}
The integral family~\eqref{topo} can be very easily reduced to master integrals
using, for example, Reduze 2~\cite{Studerus:2009ye,vonManteuffel:2012np}.
In order to simplify the notation
we put $m=1$ and define $u = s/m^2 $.
We find $8$ independent master integrals which we choose as follows
\begin{align}
&M_1(d;u) = \I(2,0,2,0,0)\,, \quad M_2(d;u) = \I(2,0,2,1,0)\,, \nonumber \\
&M_3(d;u) = \I(0,2,2,1,0)\,, \quad M_4(d;u) = \I(0,2,1,2,0)\,, \nonumber \\
&M_5(d;u) = \I(2,1,0,1,2)\,, \quad M_6(d;u) = \I(1,0,1,0,1)\,, \nonumber \\
&M_7(d;u) = \I(2,0,1,0,1)\,, \quad M_8(d;u) = \I(1,1,1,1,1)\,.
\end{align}
Most of the master integrals are very simple and have been already studied
thoroughly
in the literature. In particular $M_1$,...,$M_5$ are known and
can be written in terms of HPLs only.
The remaining three integrals, $M_6$, $M_7$ and $M_8$, cannot be expressed
in terms of MPLs and will be the main topic of this paper. Note that
$M_6$ and $M_7$ are the two master integrals of the two-loop massive
sunrise with equal masses, see Eq.\eqref{eq:sunscal}. As we will see,
$ M_6 $ and $ M_7 $ satisfy a system of two coupled differential
equations, with $ M_6 $ appearing further within the
inhomogeneous terms of the differential equation for $ M_8 $.
\par
As usual, we are interested in the Laurent expansion of the master
integrals for $d \approx 4$. The computation of the first five integrals
in terms of HPLs is straightforward.
In particular, it can be simplified by the choice of a canonical basis
in $d\approx4$, which can be found following the methods described
in~\cite{Gehrmann:2014bfa,Henn:2014qga}.
For the last three integrals, instead, a canonical basis in the usual sense
cannot be found and we will have to resort to different arguments in order
to put the system of differential equations in a
form that is suitable for their integration.
We choose the following canonical basis for the simple topologies
\begin{align}
&f_1(d;u) = 4\,(d-4)^2 \,M_1(d;u)\,, \quad
f_2(d;u) = (d-4)^2\, u\, M_2(d;u)\,,
\nonumber \\
&f_3(d;u) = (d-4)^2\, u\, M_3(d;u)\,, \quad f_4(d;u) = (d-4)^2\,
\left[ 2\,M_3(d;u) + (1-u) M_4(d;u)\right]\,, \nonumber \\
&f_5(d;u) = (d-4)^2\, u^2\, M_5(d;u)\,, \labbel{eq:basisca}
\end{align}
while for the non-trivial topologies we introduce
\begin{align}
&f_6(d;u) = (d-4)^2\, M_6(d;u)\,, \nonumber\\
&f_7(d;u) = (d-4)^2\, M_7(d;u)\,, \nonumber\\
&f_8(d;u) = (d-4)^3\, (d-3)\,u\, M_8(d;u)\,. \labbel{eq:basis1}
\end{align}
The system of differential equations for the first five masters integrals
can then be written as
\begin{align}
\frac{d}{du} f_i(d;u) = (d-4)\, \sum_{j=1}^5\, A_{ij}(u)\, f_j(d;u)\,,
\quad \forall i=1,...,5 \labbel{eq:deqca}
\end{align}
where the matrix $A(u)$ reads
\begin{align}
A(u) = \frac{1}{u}\; \left( \begin{array}{ccccc}
0 & 0 & 0 & 0 & 0 \\
0 & -1/2 & 0 & 0 & 0 \\
0 & 0 & -1/2 & 0 & 0 \\
0 & 0 & 3/2 & 0 & 0 \\
0 & 0 & 0 & 0 & -1
\end{array} \right)\,
+ \frac{1}{u-1}\; \left( \begin{array}{ccccc}
0 & 0 & 0 & 0 & 0 \\
1/8 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2 & 0 \\
0 & 1 & 0 & 0 & 2
\end{array} \right)\,. \labbel{eq:matrix}
\end{align}
The differential equations for the last three integrals cannot be put in
a similarly simple form
and we will write them explicitly later on, once we come to study them.
\section{The simple kite master integrals}
\labbel{sec:simple} \setcounter{equation}{0}
\numberwithin{equation}{section}
Let us focus on the first five integrals. If we start from the differential
equations~\eqref{eq:deqca}, carrying out the integration
in terms of harmonic polylogarithms is straightforward. As it is well known, harmonic polylogarithms are
a special case of multiple polylogarithms and
for convenience of the reader, we recall here their iterative definition.
We start at weight one defining
\begin{align}
G(0,x) = \ln{(x)}\,, \qquad G(a,x) = \int_0^x \frac{dt}{t-a} = \ln{\left( 1 - \frac{x}{a}\right)}\,.
\end{align}
The multiple polylogarithms are then iteratively defined at weight $n$ as follows
\begin{align}
G(\underbrace{0,...,0}_{n};x) = \frac{1}{n!} \ln^n{x}\,, \qquad
G(a_1,a_2,...,a_n; x) = \int_0^x \frac{dt}{t-a_1} G(a_2,...,a_n; x)\,.
\labbel{eq:defMPLs}
\end{align}
Note that all integrals have a cut at $s=m^2$, $u=1$, {\it i.e.} they are real for $u<1$ and
develop an imaginary part for $u>1$ whose sign is fixed by Feynman's prescription $u \to u + i\, 0^+$.
We present here the solution valid for for $0~~__1$ with $u \to u + i\, 0^+$.
Thanks to the choice of a canonical basis the solution takes a particularly compact form and all
sub-topologies, up to weight 4, fit in one single page.
\begin{align}
f_1(d;u) &= 1\,, \label{f1du} \\ \nonumber \\
%%%%%%%%
f_2(d;u) &= \frac{(d-4)}{8}\, G(1,u) + \frac{(d-4)^2}{8} \left[ G(1,1,u) - \frac{1}{2} G(0,1,u) \right] \nonumber \\
&+\frac{(d-4)^3}{8} \left[ G(1,1,1,u) - \frac{1}{2}G(1,0,1,u) - \frac{1}{2} G(0,1,1,u) + \frac{1}{4}G(0,0,1,u) \right]
\nonumber \\
&+ \frac{(d-4)^4}{8} \left[ G(1,1,1,1,u) - \frac{1}{2}G(1,1,0,1,u) - \frac{1}{2}G(1,0,1,1,u)
+ \frac{1}{4}G(1,0,0,1,u)
\right. \nonumber \\
&\qquad \qquad \; \left. -\frac{1}{2}G(0,1,1,1,u) + \frac{1}{4}G(0,1,0,1,u) + \frac{1}{4} G(0,0,1,1,u) - \frac{1}{8}G(0,0,0,1,u)
\right]\,,\\ \nonumber \\
%%%%%%%%
f_3(d;u) &= - \frac{(d-4)}{8}\, G(1,u) + \frac{(d-4)^2}{4} \left[ - G(1,1,u) + \frac{1}{4} G(0,1,u) \right] \nonumber \\
&+\frac{(d-4)^3}{2} \left[ - G(1,1,1,u) + \frac{3}{8}G(1,0,1,u) - \frac{\pi^2}{48} G(1,u)
+ \frac{1}{4} G(0,1,1,u) - \frac{1}{16}G(0,0,1,u) \right]
\nonumber \\
&+ (d-4)^4 \left[ -G(1,1,1,1,u) + \frac{3}{8}G(1,1,0,1,u) - \frac{\pi^2}{48} G(1,1,u) + \frac{3}{8}G(1,0,1,1,u)
\right. \nonumber \\
&\qquad \qquad \; - \frac{3}{32}G(1,0,0,1,u)
-\frac{\zeta_3}{32} G(1,u) + \frac{1}{4}G(0,1,1,1,u) - \frac{3}{32}G(0,1,0,1,u) \nonumber \\
&\qquad \qquad \; \left.
+ \frac{\pi^2}{192} G(0,1,u)
- \frac{1}{16} G(0,0,1,1,u) + \frac{1}{64}G(0,0,0,1,u)
\right]\,, \\ \nonumber \\
%%%%%%%%
f_4(d;u) &= \frac{1}{8} + \frac{(d-4)}{4}\, G(1,u) + \frac{(d-4)^2}{2} \left[ G(1,1,u) - \frac{3}{8} G(0,1,u) + \frac{\pi^2}{48}
\right] \nonumber \\
&+(d-4)^3 \left[ G(1,1,1,u) - \frac{3}{8}G(1,0,1,u) +\frac{\pi^2}{48} G(1,u)
- \frac{3}{8} G(0,1,1,u) + \frac{3}{32}G(0,0,1,u) + \frac{\zeta_3}{32} \right]
\nonumber \\
&+ (d-4)^4 \left[ 2\,G(1,1,1,1,u) - \frac{3}{4}G(1,1,0,1,u) + \frac{\pi^2}{24} G(1,1,u) - \frac{3}{4}G(1,0,1,1,u)
\right. \nonumber \\
&\qquad \qquad \; + \frac{3}{16}G(1,0,0,1,u)
+ \frac{\zeta_3}{16} G(1,u) - \frac{3}{4}G(0,1,1,1,u) + \frac{9}{32}G(0,1,0,1,u) \nonumber \\
&\qquad \qquad \; \left.
- \frac{\pi^2}{64} G(0,1,u)
+ \frac{3}{16} G(0,0,1,1,u) - \frac{3}{64}G(0,0,0,1,u) + \frac{\pi^4}{1280}
\right]\,,\\ \nonumber \\
f_5(d;u) &= \frac{(d-4)^2}{8} G(1,1,u)
+ \frac{(d-4)^3}{8} \left[ 3\,G(1,1,1,u) - \frac{1}{2}G(1,0,1,u) + G(0,1,1,u) \right] \nonumber \\
&+ \frac{(d-4)^4}{8} \left[ 7\, G(1,1,1,1,u) - \frac{3}{2} G(1,1,0,1,u) - \frac{5}{2} G(1,0,1,1,u) + \frac{1}{4} G(1,0,0,1,u) \right.
\nonumber \\
&\qquad \qquad \; \left. -3\,G(0,1,1,1,u) + \frac{1}{2} G(0,1,0,1,u) + G(0,0,1,1,u)
\right]\,. \labbel{eq:subeasy}
\end{align}
\section{The choice of the basis for the sunrise amplitudes.}
\labbel{sec:basis} \setcounter{equation}{0}
\numberwithin{equation}{section}
We move now to consider the last three integrals. First of all we need to
focus on the two master
integrals of the two-loop sunrise graph, {\it i.e.} $f_6(d;u)$ and $f_7(d;u)$.
They satisfy a system of two coupled differential equations
\begin{align}
u\,\frac{d}{du} f_6(d;u) & = -f_6(d;u) + 3f_7(d;u)
+ (d-2)f_6(d;u)\,,\nonumber \\&\nonumber \\
u(u-1)(u-9)\,\frac{d}{du} f_7(d;u) & = (u-3)f_6(d;u)
- (u^2-9)f_7(d;u) \nonumber \\
& + (d-2)\left[ -\frac{5}{2}(u-3)f_6(d;u)
+\frac{u^2+10 u -27}{2} f_7(d;u)\right]\nonumber \\
& + (d-2)^2 \frac{3\,(u-3)}{2}f_6(d;u)
- \,\frac{u}{2}\,f_1(d;u)\,. \labbel{eq:deq1}
\end{align}
Let us recall here that amplitude $ f_1(d;u) $ appearing within the
inhomogeneous term corresponds to the product of two tadpoles and is
in fact constant, according to Eq.(\ref{f1du}). \\
Using the methods described in~\cite{Tancredi:2015pta} one can show
that it is not possible to decouple the system,
in any even number of dimensions $d=2\,n$, $n \in \mathbb{N}$, by taking
simple linear combinations of the masters integrals with rational
coefficients. Indeed, it is very well known that the solution cannot be
expressed in terms of MPLs only and elliptic generalizations of the
latter must be introduced~\cite{Laporta:2004rb,Bloch:2013tra,Remiddi:2013joa,
Adams:2013nia,Adams:2014vja,Adams:2015gva}.
In order to simplify the integration of these two integrals
we will proceed as follows. We will start considering the integrals for
$d \approx 2$. The reason for this is two-fold. First, when $d=2$ the two
master integrals $f_6(2;u)$ and $f_7(2;u)$ are finite.
Second, as we will see explicitly, their imaginary parts when $d=2$ are
particularly simple. That is important because the imaginary parts are
related to the solutions of the corresponding homogeneous system, which
in turn are the building blocks for the iterative solution of the
$2 \times 2$ differential system~\eqref{eq:deq1} through Euler's
method. Those considerations will allow us
to determine a basis of master integrals for which we can easily solve
the differential equations as a Laurent series in $(d-2)$.
At this point, we could solve the system as Laurent series in $(d-2)$ and then
use Tarasov shifting identities~\cite{Tarasov:1996br}
in order to obtain the corresponding coefficients of their Laurent series
in $(d-4)$, which are the physically relevant results.
Instead of proceeding in this way, though, we will use the technique
described in~\cite{Tancredi:2015pta} (see appendix B therein) in order to
build up a \textsl{new basis} of master integrals which fulfills the very
same differential equations, but this time with $d \to d-2$.
This implies that the series expansion in $(d-4)$ of the new basis will be
\textsl{formally identical} to that of the former basis in $(d-2)$.
This will allow us to treat more consistently everything in $d\approx 4$
from the very beginning.
\subsection{Simplifying the differential equations in $d=2$}
The system of differential equations~\eqref{eq:deq1} has four regular
singular points, {\it i.e.} $u=0$, $u=1$, $u=9$ and $u = \pm \infty$, we will
therefore need to consider the solution in the four different regions
$$-\infty____9$.
Now, since the tadpole does not have any cut in $u$, the imaginary parts
of the master integrals $f_6(d;u)$ and $f_7(d;u)$
must satisfy the associated homogeneous system.
We have already computed the imaginary part of the first master integral in
section~\ref{sec:iter} for generic $d$.
For $d=2$, a straightforward application of Cutkosky-Veltman's rule to
$ f_7(2;u) $ as well gives
\begin{align}
& \frac{1}{\pi} \Ima f_6(2;u) = I(0,u) \nonumber \\
& \frac{1}{\pi} \Ima f_7(2;u) = \frac{1}{(u-1)(u-9)}
\left[ \frac{u^2-6\,u + 21}{6}\,I(0,u)
- \frac{1}{2} I(2,u) \right]\,,\labbel{eq:impartseq} \\&\nonumber
\end{align}
where the functions $I(n,u)$ are defined as
\begin{align}
&I(n,u) = \int_4^{(\sqrt{u}-1)^2} db\, \frac{b^n}{\sqrt{R_4(b,u)}}\,,
\labbel{defInu} \end{align}
and $R_4(d,u)$ is the fourth-order polynomial
\begin{align}
R_4(b,u) = b(b-4)((\sqrt{u}-1)^2-b)((\sqrt{u}+1)^2-b)\,. \labbel{defR4}
\end{align}
Some of the properties of these functions are discussed
in appendices~\ref{App:Ell} and~\ref{App:CEllInt}.
In particular, there it is shown that all functions $I(n,u)$ can be
expressed in terms of two independent functions only, say $I(0,u)$
and $I(2,u)$, which can be in turn expressed in terms of the complete
elliptic integrals of first and second kind, and can be therefore considered
as known analytically. Eqs.~\eqref{eq:impartseq} suggest to perform
the following change of basis
\begin{align}
g_6(d;u) &= f_6(d;u) \nonumber \\
g_7(d;u) &= - 2 (u-1) (u-9) f_7(d;u) + \frac{1}{3}( u^2 - 6 \,u + 21 )
f_6(d;u)\,,\labbel{eq:transf1}
\end{align}
so that obviously the imaginary parts of the functions
$g_6(d;u)$ and $g_7(d;u)$ in $d=2$ become
\begin{align}
\frac{1}{\pi} \Ima \,g_6(2;u) &= I(0,u)\,,\nonumber\\
\frac{1}{\pi} \Ima \,g_7(2;u) &= I(2,u)\,. \labbel{Img67}
\end{align}
Note that these relations are true in $d=2$ and do not change if we
modify~\eqref{eq:transf1} by a term proportional to $(d-2)$.
This freedom can be used to get rid of the term
proportional to $(d-2)^2$ in the second of Eqs.~\eqref{eq:deq1}.
While this is not strictly required, it indeed helps in simplifying
the structure of the solution.
We modify therefore Eq.\eqref{eq:transf1} as follows
\begin{align}
g_6(d;u) &= f_6(d;u) \nonumber \\
g_7(d;u) &= - 2 (u-1) (u-9) f_7(d;u) + \frac{1}{3}( u^2 - 6 \,u + 21 )
f_6(d;u) + (d-2) C(u) f_6(d;u)\,,\labbel{eq:transf2}
\end{align}
where $C(u)$ is a function of $u$ only, to be determined by imposing that
the term proportional to $(d-2)^2$ in Eqs~\eqref{eq:deq1} is zero.
By writing down explicitly the differential equations for $g_6(d;u)$ and
$g_7(d;u)$ one easily finds that there are two values of $C(u)$ which would
eliminate the unwanted term, namely
$$C(u) = 6(u-1)\,,\qquad C(u) = -\frac{(u-3)(u-9)}{3}\,.$$
At this level, there is no reason to prefer one choice over the other,
we choose therefore $C(u) = 6(u-1)$, since this produces the most compact
results. With this choice the differential equations become
\begin{align}
\frac{d}{du} g_6(d;u) & = \frac{1}{2 u (u-1)(u-9)} \left[
(3+14 u - u^2)g_6(d;u) - 3\,g_7(d;u) \right]
+ (d-2)\,\frac{1}{u-9}\,g_6(d;u)\,,\nonumber \\&\nonumber \\
\frac{d}{du} g_7(d;u) & = \frac{1}{6\, u (u-1)(u-9)}
\left[ (u+3)(3+75u-15u^2+u^3)g_6(d;u)
- 3(3+14u-u^2) g_7(d;u)\right]\nonumber \\
& + \frac{(d-2)}{6\, u(u-1)(u-9)}
\left[ (u+3)(9+63u-9u^2+u^3)g_6(d;u)
+ 3(u+1)(u-9)g_7(d;u) \right]
+ 1\,, \labbel{eq:deq2}
\end{align}
where we used $f_1(d;u) = 1$, Eq.(\ref{f1du}).
The system can be written in matrix form as follows
\begin{align}
\frac{d}{du} \left( \begin{array}{c} g_6\\g_7 \end{array} \right)
= B(u) \left( \begin{array}{c} g_6\\g_7 \end{array} \right)
+ (d-2) \, D(u)\, \left( \begin{array}{c} g_6\\g_7 \end{array} \right)
+ \left( \begin{array}{c} 0\\1 \end{array} \right)\,, \labbel{eq:sysd2}
\end{align}
where the two matrices $ B(u), D(u) $ are defined as
\begin{align}
B(u) = \frac{1}{6\, u (u-1)(u-9)}
\left( \begin{array}{cc} 3(3+14u-u^2) & - 9 \\ (u+3)(3+75u-15u^2+u^3)
& - 3(3+14u-u^2) \end{array} \right)\,,
\end{align}
\begin{align}
D(u) = \frac{1}{6\,u(u-9)(u-1)}
\left( \begin{array}{cc} 6\,u(u-1) & 0 \\
(u+3)(9+63u-9u^2+u^3) & 3(u+1)(u-9) \end{array} \right)\,.
\end{align}
In order to be able to solve~\eqref{eq:sysd2}
as a Laurent series in $(d-2)$, as a first step we need to solve the
homogeneous system for $d=2$, {\it i.e.} we need to find a pair of two solutions,
say $(I_1(u),I_2(u))$ and $(J_1(u), J_2(u))$, such that the matrix of the
solutions
\begin{align}
G(u) = \left( \begin{array}{cc} I_1(u) & J_1(u) \\ I_2(u) & J_2(u)
\end{array} \right) \labbel{defG}
\end{align}
fulfills
\begin{align}
\frac{d}{du} G(u) = B(u)\, G(u)\,. \labbel{eqG}
\end{align}
Note in particular that since
\begin{equation}
{\rm Tr}{(B(u))} = 0\,\,,
\end{equation}
the Wronskian of the four solutions,
$W(u) = I_1(u)\,J_2(u) - I_2(u)\,J_1(u)$, must be independent of $u$.
From its very definition,
\begin{align}
W(u) = \det{(G(u))} = I_1(u)\,J_2(u) - I_2(u)\,J_1(u)\,, \labbel{eq:Wronsk}
\end{align}
we find
\begin{align}
\frac{d}{du} W(u) = \frac{d}{du} \det{(G(u))}
= {\rm Tr}(G^{-1}(u)\,B(u)\,G(u))\, \det{(G(u))}
= {\rm Tr}(B(u))\, \det{(G(u))} = 0\,,
\end{align}
and $W(u)$ must be a constant. This property is of fundamental importance
to simplify the iterative solution of the differential equations, as we
will see later on.
\subsection{The choice of the basis in $d=4$}
In the previous section we showed how to choose a basis of master integrals
for the sunrise graph, whose differential equations assume a particularly
convenient form as far as their Laurent series in $(d-2)$ are considered.
Here we want to obtain a corresponding basis which has the very
same properties once expanded in $(d-4)$. In order to do this
we follow the method described in appendix B of~\cite{Tancredi:2015pta}. In
particular we define a new basis of master integrals by shifting the
basis~\eqref{eq:transf2} from $d \to d-2$
\begin{align}
h_6(d;u) = g_6(d-2,u)\,, \qquad h_7(d,u) = g_7(d-2,u)\,.
\end{align}
Using Tarasov's relations we find that the new basis $h_6(d;u),\, h_7(d;u)$
can be written in terms of the original master integrals $f_6(d;u)$ and
$f_7(d;u)$ (and their sub-topology $f_1(d,u)$) as follows
\begin{align}
h_6(d,u) &= \frac{12\,(d-3)\,(3\,d-8)}{(u-1)(u-9)}\,f_5(d,u)
+ \frac{24\,(d-3)(u+3)}{(u-1)(u-9)} \,f_6(d,u)
- \frac{3 \,(u-3) }{(u-1)(u-9)}\,f_1(d,u)\,,
\nonumber \\ & \nonumber \\
%%%%%%
h_7(d,u) &= \frac{4(d-3)(3\,d-8)(3 - (58-18 \, d)\,u
+ (7-2\,d)\,u^2)}{(u-1)(u-9)} \, f_5(d,u) \nonumber \\
&+ \frac{8\,(d-3) ( 9 + 9 (9 \, d - 29)\, u - 9 (2\,d-7)\, u^2
+ (d-3)\, u^3)}{(u-1)(u-9)}\, f_6(d,u) \nonumber \\
&+ \frac{ (9 - (51 - 18\, d) \, u - (61 - 16\, d) \,u^2
+ (7 - 2\, d) \,u^3)}{(u-1)(u-9)}\, f_1(d,u)\,. \labbel{eq:transf3}
\end{align}
It is straightforward by direct calculation, and using~\eqref{eq:sysd2},
to prove that the new basis~\eqref{eq:transf3} satisfies
the new system of differential equations
\begin{align}
\frac{d}{du} \left( \begin{array}{c} h_6\\h_7 \end{array} \right)
= B(u) \left( \begin{array}{c} h_6\\h_7 \end{array} \right)
+ (d-4) \, D(u)\, \left( \begin{array}{c} h_6\\h_7 \end{array} \right)
+ \left( \begin{array}{c} 0\\1 \end{array} \right)\,. \labbel{eq:sysd4}
\end{align}
As expected the system~\eqref{eq:sysd4} is identical to~\eqref{eq:sysd2},
upon the formal substitution $d \to d-2$. This also implies that all the
properties fulfilled by $g_6(d;u)$ and $g_7(d,u)$ in the limit $d \to 2$,
are also fulfilled by $h_6(d;u)$ and $h_7(d;u)$ in the limit
$d \to 4$. In particular the new master integrals are \textsl{finite}
in $d=4$ and their imaginary parts read
\begin{align}
\frac{1}{\pi} \Ima \,h_6(4;u) &= I(0,u)\,, \nonumber\\
\frac{1}{\pi} \Ima \,h_7(4;u) &= I(2,u)\,, \labbel{Imh67}
\end{align}
as Eqs.(\ref{Img67}).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The solution of the differential equations}
\labbel{sec:sun} \setcounter{equation}{0}
\numberwithin{equation}{section}
In this section we will show how to build the complete solution for
the sunrise master integrals up to any order in $(d-4)$. We will solve
the system as Laurent series in $(d-4)$.
In order to do this, we first need to find the homogeneous solution in the
limit $d \to 4$, such that we can then use Euler's method of variation of
constants in order to build up the complete non-homogeneous solution.
\subsection{The homogeneous solution}
As explained above, as a first step we need now to find two independent
pairs of solutions for the homogeneous system associated
to~\eqref{eq:sysd4}
\begin{equation}
\frac{d}{du} \left( \begin{array}{c} I_1 \\ I_2 \end{array} \right)
= B(u) \left( \begin{array}{c} I_1 \\ I_2 \end{array} \right) \,.
\labbel{eq:homd4}
\end{equation}
The discussion in previous section already suggests how to find the first
of the two pairs. Taking the imaginary part of~\eqref{eq:sysd4} at
$ d=4 $ gives at once
\begin{equation}
\frac{d}{du} \left( \begin{array}{c} \Ima\,h_6\\ \Ima\, h_7
\end{array} \right)
= B(u) \left( \begin{array}{c} \Ima\, h_6\\ \Ima\, h_7 \end{array}
\right) \,, \labbel{eq:imhomd4}
\end{equation}
so that Eqs.(\ref{Imh67}) provide obviously with a first pair of solution,
valid for $9____ 9 $, but it is straightforward to extend those results and build up
a complete set of solutions valid in the remaining regions, i.e.
$-\infty____9$ where the master integrals
develop an imaginary part. By straightforward use of the formulas
in the appendix we find, for $9____9$,
and the absence of an imaginary part for the intermediate region $1____9 $ threshold.
Having the imaginary part, we can write an alternative representation
for the solution~\eqref{eq:solsun0} as a dispersion relation
\begin{align}
h_6^{(0)}(u) &= \int_9^\infty\, \frac{dt}{t-u-i\, \epsilon}\,
I_1^\noo(t)\,,\nonumber \\
h_7^{(0)}(u) &= \frac{1}{\sqrt{3}} \,\Cla + u \left( \frac{5}{6}
+ \sqrt{3}\,\Cla \right) \nonumber \\
&+ u^2\, \int_9^\infty\, \frac{dt}{t^2(t-u-i\, \epsilon)}\, I_2^\noo(t)\,,
\labbel{eq:solsun0disp}
\end{align}
where for $h_7^{(0)}(u)$ we have used a doubly subtracted dispersion relation
and fixed the boundary terms matching~\eqref{eq:solsun0disp}
to~\eqref{eq:solsun0} for $u=0^+$ and $u=1^-$.
As showed in section~\ref{sec:1lT}, this representation is also particularly
convenient if we need to integrate once more over it, for example
whenever the sunrise appears as subtopology in the differential equations of
more complicated graphs, see section~\ref{sec:kite}.
\subsection{The two-loop massive sunrise at order one}
The order zero of the sunrise graph is special since its inhomogeneous term is
very simple. In order to understand the general structure we want to integrate
Eq.\eqref{eq:deqord1}, which implies integrating over the matrix $M(u)$
in~\eqref{Matrix}, using~\eqref{eq:solmord0} as inhomogenous term. Again
specializing the formulas for $0____9$, in order to extract their imaginary parts and use them to write an
alternative representation of the solutions as dispersion relations.
Also in this case, the analytic continuation is straightforward using the
results in appendix~\ref{App:AnCont} and for simplicity we give only the
result for the imaginary parts
\begin{align}
\frac{1}{\pi} \Ima\left( h_6^{(1)}(u)\right) &= \theta(u-9) \left[
\frac{1}{4} \, I_1^\noo(u) \, \bar{l}(u) - \frac{\pi}{2} J_1^\noo(u)\right]
\nonumber \\
\frac{1}{\pi} \Ima\left( h_7^{(1)}(u)\right) &= \theta(u-9) \left[
\frac{1}{4} \, I_2^\noo(u) \, \bar{l}(u) - \frac{\pi}{2} J_2^\noo(u)
+\frac{( u + 3 )^2}{6} \, I_1^\noo(u) \right]\,, \labbel{eq:Imaord1}
\end{align}
where $\bar{l}(u)$ is the real part of the function $l(u)$ defined above
threshold, i.e. for $u > 9$,
\be
\bar{l}(u) = 2\ln{(u-1)} + 2\ln{(u-9)} - \ln{(u)} \,. \labbel{eq:comblogsb}
\ee
Note that formulas~\eqref{eq:Imaord1} are extremely simple and do not involve any
integral over the functions $I_k(t)$ and $J_k(t)$.
They allow us to write
equally simple dispersion relations for the two master integrals
\begin{align}
h_6^{(1)}(u) &= \,\int_9^\infty \frac{dt}{t-u-i\,\epsilon}
\left( \frac{1}{4}\,I_1^\noo(t) \, \bar{l}(t)
- \frac{\pi}{2} \,J_1^\noo(t) \right) \labbel{eq:solsun1dispa}
\end{align}
and
\begin{align}
h_7^{(1)}(u) &= \sqrt{3}\left[ \frac{1}{6} \Cla \ln{(3)}
- \frac{1}{4}\Lst - \frac{\pi^3}{72} \right] \nonumber \\
&+ u\, \left[-\frac{5}{12} + \sqrt{3}\left( \frac{1}{2} \Cla\ln{(3)}
- \frac{3}{4}\Lst +\frac{14}{27}\Cla- \frac{\pi^3}{24} \right)\right]
\nonumber \\
&+ u^2\, \int_9^\infty \frac{dt}{t^2(t-u-i\,\epsilon)}
\left( \frac{1}{4}\,I_2^\noo(t) \, \bar{l}(t)
- \frac{\pi}{2} \,J_2^\noo(t) + \frac{(t+3)^2}{6} I_1^\noo(t) \right)\,,
\labbel{eq:solsun1dispb}
\end{align}
where, again, the dispersion relation for $h_7^{(1)}(u)$ is
doubly subtracted in $u=0$.
It is clear that, at least in principle, the techniques described here for
the integration of the first two orders of the two-loop massive sunrise,
can be used also for the higher orders.
The formulas are of course more cumbersome and, in general, it is not granted
that the result can always be written in terms of one-fold integrals only,
as for order zero and one, like in
Eqs.~\eqref{eq:solsun0},~\eqref{eq:solsun1a} and~\eqref{eq:solsun1b}.
Nevertheless one can show that, by using integration by parts as we did for
the order one, also the order $(d-4)^2$ can be substantially simplified.
One last comment is in order. The basis of master integrals that we have
been considering, $h_6(d;u)$ and $h_7(d;u)$, was build by the shift
$d \to d-2$ of the previous basis $g_6(d;u)$, $g_7(d;u)$, see
Eq.\eqref{eq:transf2}. That implies that if we expand the latter as Laurent
series in $(d-2)$
\begin{align}
g_6(d;u) = \sum_{a=0}^\infty\, g_6^{(a)}(u) (d-2)^a\,,\qquad
g_7(d;u) = \sum_{a=0}^\infty\, g_7^{(a)}(u) (d-2)^a\,,
\end{align}
the coefficients of this expansion can be directly related to the
coefficients of the Laurent expansion in $(d-4)$ of $h_6(d;u)$
and $h_7(d;u)$ as follows
\begin{align}
g_6^{(a)}(u) = h_6^{(a)}(u)\,,\qquad g_7^{(a)}(u) = h_7^{(a)}(u)\,,
\qquad \forall\; a\,.
\end{align}
\section{The solution for the kite integral}
\labbel{sec:kite} \setcounter{equation}{0}
\numberwithin{equation}{section}
As a last step we will use the results of the previous sections in order to
write compact expressions for the first two non-zero orders of the kite
integral. We will do this using the method sketched in section~\ref{sec:1lT},
namely we will derive the differential equations for the kite integral and
then we will insert into it the solution for the sunrise graph given as a
dispersive relation, see Eqs.~\eqref{eq:solsun0disp},~\eqref{eq:solsun1dispa}
and~\eqref{eq:solsun1dispb}.
We start by writing the differential equations for the master integral
$f_8(d;u)$, defined in~\eqref{eq:basis1}, while for the sunrise we use the
modified basis defined in~\eqref{eq:transf3}. The differential equation reads
\begin{align}
\frac{d}{du}f_8(d;u) &=
(d-4)\,\left(\frac{1}{u-1} - \frac{1}{2\,u}\right) f_8(d;u)
+ \frac{(d-4)^3}{24}\left( 1 - \frac{8}{u-1} \right)\, h_6(d;u) \nonumber \\
&+ \frac{(d-4)}{u-1}\left( - \frac{1}{8}f_1(d;u) + 2\, f_3(d;u) + f_4(d;u)
\right) + (d-4)\frac{1}{u}\, f_5(d;u)\,. \labbel{eq:deqkite1}
\end{align}
Two properties are worth noticing in Eq.\eqref{eq:deqkite1}. First, only
one of the two master integrals of the sunrise subgraph appears, namely
$h_6(d;u)$. Second, it appears multiplied by a factor $(d-4)^3$.
This fact is a consequence of the normalization adopted
in~\eqref{eq:basis1}, where, attempting to build up a basis similar to a
canonical one, we rescaled all master integrals of suitable powers
of $(d-4)$.
Note however that, even if in~\eqref{eq:basis1} also $f_6(d;u)$ and $f_7(d;u)$
are rescaled by $(d-4)^2$, one should recall that the masters integrals
that we are effectively calculating for the sunrise graph (and which enter
in the differential equation for the kite) are not $f_6(d;u)$ and $f_7(d;u)$,
but instead $h_6(d;u)$ and $h_7(d;u)$, as defined in~\eqref{eq:transf3}.
The latter are obtained shifting~\eqref{eq:basis1} from $d \to d-2$, such that
the factor $(d-4)^2$ in front of $f_6(d;u)$ and $f_7(d;u)$ becomes
effectively a $(d-6)^2$.
In order to make the equations more symmetric, we could have therefore
rescaled also $h_6(d;u)$ and $h_7(d;u)$ by $(d-4)^2$, reabsorbing in this
way the corresponding factor in~\eqref{eq:deqkite1}.
We preferred, nevertheless, not to do that in order to avoid the confusion
of one more change of basis. With the present normalization, the sunrise
integrals start at order zero in $(d-4)$, which shows that their first
contribution to the Laurent expansion of the kite integral is at order
$(d-4)^3$.
We can now move to the actual integration of the equations.
Once more we work in the region $0____u$.
In order to obtain real-valued polylogarithms it is convenient to perform
the change of variables
\be
v = \frac{u-1}{8}\,, \qquad \mbox{such that}
\qquad 1____9$, the
logarithm $G(t,u)$ develops an imaginary part whenever $t < u$. To keep
track of this, it is enough to split the corresponding integral in $t$
into two pieces
\begin{align}
\int_9^\infty\, dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1} \right)\, G(t,u)
&= \int_9^u \, dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1} \right)\,
\ln{\left( 1 - \frac{u+i\,0^+}{t} \right)} \nonumber \\
&+ \int_u^\infty dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1} \right)\,
G(t,u) \nonumber \\
&= \int_9^u \, dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1} \right)\,
\ln{\left( \frac{u}{t} -1\right)} \nonumber \\
&+ \int_u^\infty dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1} \right)\,
G(t,u) \nonumber\\
&- i\, \pi \, \int_9^u \, dt\, I_1^\noo(t) \, \left( 1- \frac{8}{t-1}
\right)\,,
\end{align}
where we used, as always, $u \to u + i 0^+$.
On the other hand, the multiple-polylogarithms of $v$ remain real since,
for $9____ 0 \qquad \mbox{if} \qquad 4__**9$.
Clearly, not all functions are linear independent.
Using integration-by-parts identities
$$\int_{\beta_1}^{\beta_2} db \frac{d}{db}\left(\,b^n\, \sqrt{R_4(b,u)} \right)= 0\,,
\qquad \forall \beta_i \in \{ 0,\, 4,\, (\sqrt{u}-1)^2,\, (\sqrt{u}+1)^2\}\,,$$
it is easy to prove that, for each family of functions, only three can be linear independent. We
choose for definiteness
\begin{align}
&J(0,u),\,\, J(1,u),\,\, J(2,u)\,, \nonumber \\
&I(0,u),\,\, I(1,u),\,\, I(2,u)\,, \nonumber \\
&K(0,u),\,\, K(1,u),\,\, K(2,u)\,.
\end{align}
Moreover one more relation can be written for each family of functions. We find
\begin{align}
\int_4^{(\sqrt{u}-1)^2} db \, \frac{d}{db}\,\ln{ \left( \frac{b(u+3-b) + \sqrt{R_4(b,u)} }{b(u+3-b) - \sqrt{R_4(b,u)} }\right) }
= \int_4^{(\sqrt{u}-1)^2} db \,\frac{(3b-u-3)}{\sqrt{R_4(b,u)}} = 0\,,
\end{align}
\begin{align}
\int_0^4 db \, \frac{d}{db}\,\ln{ \left( \frac{b(u+3-b) +i\, \sqrt{-R_4(b,u)} }{b(u+3-b) - i\,\sqrt{-R_4(b,u)} }\right) }
= \int_0^4 db \,\frac{i\,(3b-u-3)}{\sqrt{R_4(b,u)}} = -i\,\pi\,,
\end{align}
\begin{align}
\int_{\bb}^{\BB} db \, \frac{d}{db}\,\ln{ \left( \frac{b(u+3-b) +i\, \sqrt{-R_4(b,u)} }{b(u+3-b) - i\,\sqrt{-R_4(b,u)} }\right) }
= \int_{\bb}^{\BB} db \,\frac{i\,(3b-u-3)}{\sqrt{R_4(b,u)}} = 2\,i\,\pi\,,
\end{align}
which imply respectively
\begin{align}
J(1,u) = \frac{(u+3)}{3} J(0,u) - \frac{\pi}{3}\,,\nonumber
\end{align}
\begin{align}
I(1,u) = \frac{(u+3)}{3} I(0,u)\,,\nonumber
\end{align}
\begin{align}
K(1,u) = \frac{(u+3)}{3} K(0,u) + \frac{2\,\pi}{3}\,.
\end{align}
Finally, as expected from~\eqref{eq:dipfun},
one can prove that the functions $K(n,u)$ and $J(n,u)$
are not linearly independent from each other, in particular it holds
\begin{align}
&K(0,u) = J(0,u)\,,\nonumber \\
&K(1,u) = J(1,u) + \pi \,,\nonumber \\
&K(2,u) = J(2,u) + \pi(u+3)\,.
\end{align}
All together these relations imply that only 4 functions are linearly independent.
We choose our basis as follows
$$I(0,u),\,\, I(2,u),\,\,J(0,u),\,\, J(2,u).\,\,$$
\section{The analytic continuation of the homogeneous solutions} \labbel{App:AnCont} \setcounter{equation}{0}
\numberwithin{equation}{section}
In the main text we showed how to find the solution of the homogeneous system for the sunrise graph,
the matrix $G^{(a,b)}(u)$, in the region $9**