\frac{\eta^2}{\pi}\int
K_1^2(\eta \varrho)(1-\exp(-i{\bf q}
\mbox{\boldmath$\varrho$}))d^2\varrho,
\nonumber \\
&& F_2(x)=\frac{2x^2+1}{x\sqrt{1+x^2}}\ln(x+\sqrt{1+x^2})-1,\quad
q_{min}=m^3\omega/4\varepsilon^2 \varepsilon',
\label{4.6}\end{aligned}$$ here value $q_{min}$ is defined in c.m.s. of colliding particles. The function $F_2(x)$ encounters in the radiation theory. To calculate the corresponding contribution into the radiation spectrum we have to substitute (\[4.5\]) into Eq.(\[3.11\]) and take the integrals. After substitution of variable in (\[4.5\]) $$w=\frac{q}{2q_{min}\zeta},
\label{4.7}$$ we obtain the integral over $\zeta$ in Eq.(\[3.11\]): $$\begin{aligned}
&&\int_{1}^{\infty} \left( v -\frac{4}{\zeta} +
\frac{4}{\zeta^2}\right)\exp(-s^2\zeta^2)d\zeta \equiv f(s)
\nonumber \\
&& = \frac{\sqrt{\pi}}{2s}(v-8s^2){\rm
Erfc}(s)+4\displaystyle{e^{-s^2}}+2{\rm Ei}(-s^2),
\label{4.8}\end{aligned}$$ where $$s=w r q_{min}, \quad r^2= \Sigma_z^{-2}\cos^2\varphi+
\Sigma_y^{-2}\sin^2\varphi
\label{4.8a}$$ Making use of Eq.(\[4.4\]) we find for the spectrum $$\begin{aligned}
&& d\sigma_1^{(1)}=\frac{2\alpha^3}{m^2}
\frac{\varepsilon'}{\varepsilon}
\frac{d\omega}{\omega}f^{(1)}(\omega),
\nonumber \\
&& f^{(1)}(\omega)= -\frac{1}{\pi\Sigma_z
\Sigma_y}\int_{0}^{2\pi}\frac{d\varphi}{\Sigma_z^{-2}\cos^2\varphi+
\Sigma_y^{-2}\sin^2\varphi} \int_{0}^{\infty}F_2(z)f(s)s ds,
\nonumber \\
&& z^2=\frac{s^2}{q_{min}^2}\frac{1}{\Sigma_z^{-2}\cos^2\varphi+
\Sigma_y^{-2}\sin^2\varphi}.
\label{4.9}\end{aligned}$$ This formula is quite convenient for the numerical calculations.

In the case ${\bf x}_0 \neq 0$ we will use straightforwardly Eqs.(\[3.12\]) and (\[4.4\]). Taking into account (\[4.2\]) we have for the difference $$\begin{aligned}
\hspace{-15mm}&& \Delta^{(1)}({\bf x}_0) \equiv
\frac{1}{2\pi}(\Phi^{-1}({\bf x}_0)F^{(1)}({\bf x}_0)-\Phi^{-1}(0)
F^{(1)}(0))
\nonumber \\
\hspace{-15mm}&&=\frac{\eta^2}{\pi\zeta^2}\int
K_1^2(\eta\varrho)\exp\left[-\varrho_z^2\Sigma_z^2-
\varrho_y^2\Sigma_y^2\right] \{\exp\left[-2\varrho_zz_0\Sigma_z^2-
2\varrho_yy_0\Sigma_y^2\right]-1\}d^2\varrho,
 \label{4.10}\end{aligned}$$ where the function $F^{(1)}({\bf x}_0)$ is defined in Eqs.(\[3.12\]), (\[3.15a\]).

Using the Macdonald’s formula (see e.g.[@BE], p.53) $$2K_1^2(\eta\varrho)=\int_{0}^{\infty} \exp\left[ -\varrho^2
t-\frac{\eta^2}{2t} \right]K_1\left(
\frac{\eta^2}{2t}\right)\frac{dt}{t} \label{4.11}$$ and taking the Gaussian integrals over 