$\varrho_z$ and $\varrho_y$ we get $$\Delta^{(1)}({\bf x}_0) =\frac{1}{\zeta^2}\int_{0}^{\infty}
\frac{\exp\left( -\frac{\eta^2}{2t}\right)K_1\left(
\frac{\eta^2}{2t}\right)}{\sqrt{t+\Sigma_z^2}\sqrt{t+\Sigma_y^2}}
\left\{ \exp\left[ \frac{z_0^2\Sigma_z^4}{t+\Sigma_z^2}+
\frac{y_0^2\Sigma_y^4}{t+\Sigma_y^2}\right]-1
\right\}\frac{\eta^2 dt}{2t}. \label{4.12}$$ For the correction to the cross section (see Eqs.(\[4.4\]) and (\[4.9\])) we have correspondingly $$d\sigma_1^{(1)}=\frac{2\alpha^3}{m^2}
\frac{\varepsilon'}{\varepsilon}
\frac{d\omega}{\omega}[f^{(1)}(\omega)+J^{(1)}(\omega, {\bf x}_0
)],
\label{4.13}$$ where $$J^{(1)}(\omega, {\bf x}_0)= \int_{1}^{\infty} U(\zeta)
 \Delta^{(1)}({\bf x}_0)d\zeta.
\label{4.14}$$

Now we pass over to the calculation of the second (subtraction) term $F^{(2)}(\omega, \zeta)$ in Eq.(\[3.12\]). Using Eq.(\[4.1\]) we get $${\bf I}= \eta \int K_1(\eta
\varrho)\frac{\mbox{\boldmath$\varrho$}}{\varrho}w_c({\bf
x}-\mbox{\boldmath$\varrho$})d^2\varrho =
\frac{\eta}{(2\pi)^2}\int S(q) \frac{{\bf q}}{q}\exp(-i{\bf q
x})w_c({\bf q})d^2q, \label{4.15}$$ where $$\begin{aligned}
&& S(q)=  \int K_1(\eta\varrho)\frac{{\bf q
}\mbox{\boldmath$\varrho$}}{q\varrho}\exp(i{\bf q
}\mbox{\boldmath$\varrho$})d^2\varrho
\nonumber \\
&& = 2\pi i \int K_1(\eta\varrho) J_1(q\varrho) \varrho d\varrho
=  2\pi i~ \frac{q}{\eta} \frac{1}{q^2+\eta^2}.
\label{4.16}\end{aligned}$$ Using the exponential parametrization $$\frac{1}{q^2+\eta^2}=\frac{1}{4}\int_{0}^{\infty}
\exp\left[-\frac{s}{4}(q^2+\eta^2) \right]ds
\label{4.17}$$ and taking the Gaussian integrals over $q_z$ and $q_y$ we obtain $$\begin{aligned}
&& {\bf I}= \int_{0}^{\infty} \exp\left[-\frac{\eta^2 s}{4} -
\frac{z^2}{s+2\sigma_z^2} - \frac{y^2}{s+2\sigma_y^2} \right]
\nonumber \\
&& \times \left[ \frac{z {\bf e}_z}{s+2\sigma_z^2}+ \frac{y {\bf
e}_y}{s+2\sigma_y^2}\right]
\frac{ds}{\sqrt{s+2\sigma_z^2}\sqrt{s+2\sigma_y^2}}, \label{4.18}\end{aligned}$$ where ${\bf e}_z$ and ${\bf e}_y$ are the unit vectors along axes $z$ and $y$. Substituting (\[4.18\]) into Eq.(\[3.12\]), taking the Gaussian integrals over $z$ and $y$ and using Eq.(\[4.4\]) we get the correction to the cross section $$d\sigma_1^{(2)}=-\frac{2\alpha^3}{m^2}
\frac{\varepsilon'}{\varepsilon}
\frac{d\omega}{\omega}J^{(2)}(\omega, {\bf x}_0), \label{4.19}$$ where 