\frac{q_{min}}{\Sigma_z+\Sigma_y}+3+C \right) + \frac{2}{9},~\quad
q_{min} \ll (\Sigma_z+\Sigma_y). \label{5.3}$$ This expression agrees with Eq.(24) of [@BKS1].

Under the assumption used in (\[5.3\]) and the additional condition $q_{min}(z_0+y_0)\ll 1$ the main contribution to the integral in (\[4.12\]) gives the region $t \gg \eta^2$. In this case one can use the asymptotic expansion $K_1(z) \simeq 1/z (z
\ll 1)$. Then we have for the function $J^{(1)}(\omega, {\bf
x}_0)$ in Eq.(\[4.13\]) the following expression $$\begin{aligned}
&& J^{(1)}(\omega, {\bf x}_0) \simeq \left(v-\frac{2}{3}\right)J
\nonumber \\
&&
J=\int_{0}^{\infty}\left[\exp\left(\frac{z_0^2\Sigma_z^4}{t+\Sigma_z^2}
+\frac{y_0^2\Sigma_y^4}{t+\Sigma_y^2}\right)-1\right]
\frac{dt}{\sqrt{t+\Sigma_z^2}\sqrt{t+\Sigma_y^2}}. \label{5.4}\end{aligned}$$ The expression (\[5.4\]) is consistent with Eq.(26) of [@BKS1].

In the case $({\bf x}_0^2+\sigma_z^2+\sigma_y^2)q_{min}^2 \ll 1$ the main contribution to the integral in (\[4.20\]) gives the interval $s q_{min}^2 \sim ({\bf
x}_0^2+\sigma_z^2+\sigma_y^2)q_{min}^2 \ll 1$. Keeping the main term of expansion over $q^2$ in Eq.(\[4.21\]) we get $$g\left(\frac{q_{min}\sqrt{s}}{2}\right) \simeq v-\frac{2}{3}.
\label{5.5}$$ The same result can be obtained if one neglects the term containing $\eta^2$ in the exponent of integrand in Eq.(\[4.18\]).

Summing the cross section $d\sigma =
d\sigma_1^{(1)}+d\sigma_1^{(2)}$ with the standard QED bremsstrahlung cross section $$d\sigma_0= \frac{2\alpha^3}{m^2} \frac{\varepsilon'}{\varepsilon}
\frac{d\omega}{\omega}\left(v-\frac{2}{3}\right)\left(\ln
\frac{m^2}{q_{min}^2}-1 \right),
\label{5.6}$$ we get the cross section for the case of interaction of narrow beams $$\begin{aligned}
&& d\sigma_{\gamma}= d\sigma_0 + d\sigma_1 =
\frac{2\alpha^3}{m^2} \frac{\varepsilon'}{\varepsilon}
\frac{d\omega}{\omega}\Bigg\{\left(v-\frac{2}{3}\right)\Bigg[2\ln
\frac{m}{\Sigma_z+\Sigma_y} +C 