$$\begin{aligned}
&& J^{(2)}(\omega, {\bf x}_0)= \frac{\sqrt{a b}}{\Sigma_z
\Sigma_y} \exp(z_0^2 \Sigma_z^2+y_0^2 \Sigma_y^2)\int_{0}^{\infty}
ds_1 \int_{0}^{\infty} ds_2  g\left( \frac{q_{min}\sqrt{s}}{2}
\right)G(s_1, s_2, {\bf x}_0 ),
\nonumber \\
&& G(s_1, s_2, {\bf x}_0 )= \left( \frac{a_1a_2 b_1 b_2}{A
B}\right)^{1/2}\left[\frac{a_1a_2}{A}\left( \frac{1}{2}+
\frac{z_0^2a^2}{A}\right)+\frac{b_1b_2}{B}\left( \frac{1}{2}+
\frac{y_0^2b^2}{B}\right)\right]
\nonumber \\
&& \times \exp\left[-\frac{z_0^2a}{A}(a_1+a_2)
-\frac{y_0^2b}{B}(b_1+b_2) \right].
\label{4.20}\end{aligned}$$ Here the function $g$ appears as a result of integration over $\zeta$: $$\begin{aligned}
\hspace{-15mm}&&g(q)=\int_{1}^{\infty}\left( v-\frac{4}{\zeta}+
\frac{4}{\zeta^2}\right)\exp(-q^2\zeta^2)\frac{d\zeta}{\zeta^2}=\left(
v -\frac{2}{3}\right)\exp(-q^2)
\nonumber \\
\hspace{-15mm}&&-2q^2 \int_{1}^{\infty}\left( v-\frac{2}{\zeta}+
\frac{4}{3\zeta^2}\right)\exp(-q^2\zeta^2)d\zeta
\nonumber \\
\hspace{-15mm}&&= \left( v -\frac{2}{3}\right)\exp(-q^2)
-2q^2\left[ \frac{\sqrt{\pi}}{2q}\left(v-\frac{8}{3}q^2\right){\rm
Erfc}(q)+\frac{4}{3}\displaystyle{e^{-q^2}}+{\rm Ei}(-q^2)\right].
\label{4.21}\end{aligned}$$ In (\[4.20\]) we introduced the following notations $$\begin{aligned}
&& a=\frac{1}{2\Delta_z^2}, \quad b=\frac{1}{2\Delta_y^2}, \quad
a_{1,2}=\frac{1}{s_{1,2}+2\sigma_z^2},\quad
b_{1,2}=\frac{1}{s_{1,2}+2\sigma_y^2},
\nonumber \\
&& A=a_1+a_2+a, \quad B=b_1+b_2+b,\quad s=s_1+s_2.
\label{4.22}\end{aligned}$$

Narrow beams
============

This is the case when the ratio $q_{min}/(\Sigma_z+\Sigma_y) \ll
1$, so that the main contribution to the integral (\[4.9\]) gives the region $q \sim \zeta \sim 1,~z \gg 1$. Using the asymptotics of function $F_2(z)$ at $z \gg 1$ $$F_2(z) \simeq \ln (2z)^2-1
\label{5.1}$$ and the following integrals $$\begin{aligned}
&&\hspace{-10mm}
\frac{1}{2\pi\Sigma_z\Sigma_y}\int_{0}^{2\pi}\frac{d\varphi}
{\Sigma_z^{-2}\cos^2\varphi+\Sigma_y^{-2}\sin^2\varphi}=1,
\nonumber \\
&&\hspace{-10mm}
\frac{1}{2\pi\Sigma_z\Sigma_y}\int_{0}^{2\pi}\frac{d\varphi}
{\Sigma_z^{-2}\cos^2\varphi+\Sigma_y^{-2}\sin^2\varphi}\ln
\frac{4}{\Sigma_z^{-2}\cos^2\varphi+\Sigma_y^{-2}\sin^2\varphi}=\ln
(\Sigma_z+\Sigma_y)^2,
\nonumber \\
&&\hspace{-10mm}\int_{1}^{\infty}dq^2 \left(\alpha-\beta \ln
q^2\right)\int_{1}^{\infty}\left( v-\frac{4}{\zeta}+
\frac{4}{\zeta^2}\right)\exp(-q^2\zeta^2)d\zeta
\nonumber \\
&&
\hspace{-10mm}=\left(v-\frac{2}{3}\right)[\alpha+\beta(2+C)]+\frac{2}{9}\beta,
\label{5.2}\end{aligned}$$ where C is Euler’s constant $C = 0.577...$, we get for the function $f^{(1)}(\omega)$ (\[4.13\]) the following expression $$f^{(1)}(\omega) \simeq \left(v-\frac{2}{3}\right)\left(2\ln
