\begin{equation}
\frac{d{\sigma}_c}{dy}=\frac{{\pi} {\alpha}^2}{x m_e^2} \left[f_0+\pee\pge f_1
   +\pee\pgse f_2 + \pge\pgse f_3 \right] 
\label{cross_section}
\end{equation}
\begin{displaymath}
 f_0=\frac{1}{1-y}+1-y-4r(1-r)
\qquad\qquad
 f_1=xr(1-2r)(2-y)
\end{displaymath}
\begin{displaymath}
 f_2=xr \left(1+(1-y)(1-2r)^2  \right)  
\qquad\qquad
 f_3= (1-2r) \left( \frac{1}{1-y} +1-y \right) 
\end{displaymath}
\begin{displaymath}
 x=\frac{4E{\omega}_0}{m_e^2}
\qquad\qquad
 y=1-\frac{pk}{pk_0}\approx\frac{\omega}{E}
 \qquad\qquad
r=\frac{y}{x(1-y)}
\end{displaymath}
\begin{displaymath}
 0 \leq y \leq y_{max} = \frac{x}{x+1}
\end{displaymath}
