{s}{M^2}\Big)^{\omega}
\frac{2\phi_1^{(0)}(\phi_1 -\phi _1^{(0)})e^{2c\phi_1^{(0)}\eta}}
{\phi_1^{(0)} + \phi_1 - (\phi_1 -\phi_1^{(0)})e^{2c\phi_1^{(0)}\eta}} ~.$$ Obviously, when $\mu \to M$, Eqs. (\[aleptf\]) converges to the same amplitude obtained with using only one cut-off. Indeed, substituting $\mu = M$ and $\eta = 0$ leads to $$\label{aleptf1}
L_1 = 
\int_{-\imath \infty}^{\imath \infty} \frac{d \omega}{2\pi\imath}\,
\Big( \frac {s}{M^2}\Big)^{\omega} \phi_1(\omega)~.$$

According to Eqs. (\[qedlept\], \[mellinaqed\]), the QED amplitude $\phi_1^{(0)}$ is easily expressed in terms of Mellin amplitude $\phi_F^{(0)}$ for the forward $e^+e^- \to \mu^+\mu^-$ annihilation: $$\label{phi0f}
\phi_1^{(0)} = 2 \phi_F^{(0)}~.$$ The expression for $\phi_F^{(0)}$ can be taken from Refs. [@ggfl],[@flmm] and [@egt]: $$\begin{aligned}
\label{phi0}
\phi_F^{(0)} = 
4\pi^2 \big( \omega - \sqrt{\omega^2 - \chi^2_0} \big), \end{aligned}$$ with $$\label{chi0}
\chi^2_0 = 2\alpha/\pi.$$ On the other hand, the amplitudes $\phi_j$ were calculated in Ref. [@egt]. In particular, $$\label{phi1}
\phi_1 = 4 \pi^2\big( \omega - \sqrt{\omega^2 - \chi^2} \big)~,$$ where $\chi^2$ is expressed through the electroweak couplings $g$ and $g'$: $$\label{chi}
\chi^2 = [3 g^2 + {g'}^2]/(8 \pi^2) ~.$$

Next, let us solve Eq. (\[master\]) for the general case of non-zero factor $b_j^{(\gamma)}$. Then, this equation describes the backward $e^+ e^-$ annihilation into a lepton pair (e.g. $\mu^+ \mu^-$) and also the forward and backward annihilation into quarks. Eq. (\[master\]) looks simpler 