+2
\nonumber \\
&& +J -J_{-}\Bigg]+\frac{2}{9}\Bigg\},\quad
v=\frac{\varepsilon}{\varepsilon'}+\frac{\varepsilon'}{\varepsilon},\quad
\varepsilon'=\varepsilon - \omega,
 \label{5.7}\end{aligned}$$ where $J$ is given in (\[5.4\]), $$J_{-}=\frac{\sqrt{a b}}{\Sigma_z
\Sigma_y} \exp(z_0^2 \Sigma_z^2+y_0^2 \Sigma_y^2)\int_{0}^{\infty}
ds_1 \int_{0}^{\infty} ds_2 G(s_1, s_2, {\bf x}_0 ),
\label{5.8}$$ where entering functions defined in Eqs.(\[4.20\]) and (\[4.22\]).

In the case of coaxial beams ${\bf x}_0=0,~J=0$ one can take the integral in (\[5.8\]) over one of variables (for definiteness over $s_2$) using the formula $$\int_{0}^{\infty}\frac{dx}{(a_z+b_z x)^{3/2}(a_y+b_y x)^{1/2}}=
\frac{2}{a_z\sqrt{b_z b_y}+b_z\sqrt{a_z a_y}}.
\label{5.8a}$$ After this we have the simple integral over $s \equiv s_1$ $$\begin{aligned}
&& J_{-}(0)=\sqrt{1+\delta_z}\sqrt{1+\delta_y}(J_z+J_y),
\nonumber \\
&& J_{z,y}=\int_{0}^{\infty}D_{z,y}(s)ds,\quad
D_{z,y}=\frac{1}{a_{z,y}\sqrt{b_z b_y}+b_{z,y}\sqrt{a_z a_y}},
 \label{5.8b}\end{aligned}$$ where $$a_{z,y}=s(1+\delta_{z,y})+2\sigma_{z,y}^2 (2+\delta_{z,y}),\quad
b_{z,y}=\frac{s}{2\Delta_{z,y}^2}+1+\delta_{z,y},\quad
\delta_{z,y}= \frac{\sigma_{z,y}^2}{\Delta_{z,y}^2},
\label{5.8c}$$

The cross section (\[5.7\]) differs from Eq.(24) of [@BKS1] because the subtraction term $J_{-}$ is included. Without this term, generally speaking, the bremsstrahlung cross section would be qualitatively erroneous. In particular an appearance of the term $J_{-}$ violates, generally speaking, the symmetry of radiation cross section in opposite directions in $e^{-}e^{-}~(e^{-}e^{+})$ collisions.

To elucidate the qualitative features of narrow beams bremsstrahlung process we consider the case of round beams where the calculation becomes more simple: $$\begin{aligned}
&& \sigma_z=\sigma_y=\sigma,\quad \Delta_z=\Delta_y=\Delta, \quad
\Sigma_z^2=\Sigma_y^2=\Sigma^2=\frac{1}{2(\sigma^2+\Delta^2)},
\nonumber \\
&& b=a,\quad b_{1,2}=a_{1,2},\quad B=A, \quad
\delta=\frac{\sigma^2}{\Delta^2}.
\label{5.9}\end{aligned}$$ We consider first the case of coaxial beams 