into quarks are $$\begin{aligned}
\label{fzerof}
 f^{(0)}_F(x) = (4\pi\alpha e_q / p_F^{(0)})
d \ln (e^{x^2/4} D_{p_F^{(0)}}(x))/d x ~, \end{aligned}$$ with $ p_B^{(0)} = 2e_q/(1 - e_q)^2$ . Let us stress that the forward amplitudes for the annihilation into leptons are given by Eq. (\[aleptf\]). The amplitude $f^{(0)}_{F,B}$ was obtained first in Ref. [@ggfl] for the backward scattering in QED. Obviously, the only difference between the formulae for $f_j(x)$ and $f^{(0)}_j(x)$ is in the different factors $a_j$, $p_j $ and $\lambda_j$. We can specify $C(\sigma)$, using the matching $$\label{matchj}
F_j(\omega) = f_j(\omega) - f^{(0)}_j(\omega) ~,$$ when $\eta = 0$. The invariant amplitudes $f_j$ were calculated in Ref. [@egt]: $$\label{fj}
f_j (x) = \frac{a_j}{ p_j}
\frac{d \ln (e^{x^2/4} D_{p_j}(x)) }{d x} 
= a_j \frac{ D_{p_j - 1}(x))}{D_{p_j}(x))} ~.$$

Using Eq. (\[matchj\]) we are led to $$\label{Fj}
F_j = \frac{\big( f_j(x + y) - f^{(0)}_j (x + y)\big) P(\sigma, \tau)}
{P(\sigma, \sigma) - (f_j(x + y) - f^{(0)}_j (x + y)) 
\big( Q(\sigma, \sigma) - Q(\sigma, \tau) \big)} ~$$ and finally to $$\begin{aligned}
\label{solutionaj}
A_j(s/M^2, \eta) &=& 
\int_{-\imath \infty}^{\imath \infty} \frac{d \omega}{2\pi\imath}\,
\Big( \frac {s}{\mu^2}\Big)^{\omega} f_j^{(0)} (\omega) + 
\\ \nonumber
& &\int_{-\imath \infty}^{\imath \infty} \frac{d \omega}{2\pi\imath}\,
\Big( \frac {s}{M^2}\Big)^{\omega}
\frac{\big( f_j(x + y) - 