(${\bf x}_0=0,~J=0$), $$\begin{aligned}
&& J_{-}=(1+\delta)\int_{0}^{\infty}
\frac{ds}{[s(1+\delta)+2+\delta][s\delta +1+\delta]}
\nonumber \\
&& =(1+\delta) \ln \frac{(1+\delta)^2}{\delta(2+\delta)}.
\label{5.10}\end{aligned}$$ In the limiting cases the function $J_{-}$ has the form $$J_{-}(\delta \gg 1) \simeq \frac{1}{\delta},\quad
 J_{-}(\delta=1) = 2\ln\frac{4}{3}, \quad J_{-}(\delta \ll 1)
 \simeq \ln \frac{1}{2\delta}.
 \label{5.11}$$ In the first case the subtraction term $J_{-}$ is small. For the beams of the same size the subtraction term $J_{-}$ contributes to the constant entering into the expression for the cross section. The subtraction term $J_{-}$ modifies essentially the cross section in the case when the radius of target beam is much smaller than the radius of radiating beam. In this case the cross section (\[5.7\]) contains the combination $$\ln \frac{m^2}{4\Sigma^2}-J_{-} \simeq \ln \frac{m^2\Delta^2}{2}
 -\ln \frac{\Delta^2}{2\sigma^2}=\ln (m\sigma)^2.
\label{5.12}$$ So in the all cases considered above the cross section defines the transverse dimension of target beam.

When the axes of round beams are displaced with respect each other in the transverse plane the integral in (\[5.4\]) is $$\begin{aligned}
&&J=\int_{0}^{\infty}
\left[\exp\left(\frac{d}{x+1}\right)-1\right]\frac{dx}{x+1}={\rm
Ei}(d)-C-\ln d ,
\nonumber \\
&& d={\bf x}_0^2\Sigma^2 =
\frac{x_0^2+y_0^2}{2(\Delta^2+\sigma^2)}.
\label{5.13}\end{aligned}$$

It is convenient in this case to calculate the function $J_{-}$ using straightforwardly Eq.(\[4.18\]) where we omit the term with $\eta^2$ in the exponent of integrand $${\bf I}_{cr}= \mbox{\boldmath$\varrho$}\int_{0}^{\infty}
