\exp\left(-\frac{\mbox{\boldmath$\varrho$}^2}{s+2\sigma^2}\right)
\frac{ds}{(s+2\sigma^2)^2}=
\frac{\mbox{\boldmath$\varrho$}}{\mbox{\boldmath$\varrho$}^2}\left[1-
\exp\left(-\frac{\mbox{\boldmath$\varrho$}^2}{2\sigma^2}\right)\right]
\label{5.14}$$ Substituting this expression ([**I**]{} is defined in Eq.(\[4.15\])) into the subtraction term Eq.(\[3.12\]) and using the exponential parametrization $$\frac{1}{\mbox{\boldmath$\varrho$}^2}=\int_{0}^{\infty}
\exp(-\mbox{\boldmath$\varrho$}^2s)ds$$ we obtain $$\begin{aligned}
&& J_{-}=\frac{a e^d}{\pi \Sigma^2} \int_{0}^{\infty}ds \int
d^2\varrho \exp(-\mbox{\boldmath$\varrho$}^2s) \left[1-
\exp\left(-\frac{\mbox{\boldmath$\varrho$}^2}{2\sigma^2}\right)\right]
\exp(-a(\mbox{\boldmath$\varrho$}+{\bf x}_0)^2)
\nonumber \\
&& = \frac{a e^{d-d_1}}{\Sigma^2}
\int_{0}^{\infty}\Bigg[\frac{1}{s+a}\exp\left( \frac{d_1  a
}{s+a}\right)-2\frac{1}{s+a+\sigma^{-2}/2}\exp\left(
d_1\frac{a}{s+a+\sigma^{-2}/2} \right)
\nonumber \\
&& + \frac{1}{s+a+\sigma^{-2}}\exp\left(
d_1\frac{a}{s+a+\sigma^{-2}} \right)\Bigg]ds= \frac{a
e^{d-d_1}}{\Sigma^2}\Bigg[ {\rm Ei}(d_1)-2{\rm Ei}\left(d_1
\frac{\sigma^2}{\sigma^2+\Delta^2} \right)
\nonumber \\
&& +{\rm Ei}\left(d_1 \frac{\sigma^2}{\sigma^2+2\Delta^2}
\right)\Bigg],\quad d_1=a{\bf
x}_0^2=\frac{z_0^2+y_0^2}{2\Delta^2}.
\label{5.15}\end{aligned}$$ In the limit $d_1 \rightarrow 0$ the last expression goes over to Eq.(\[5.10\]).

When the displacement of the axes of colliding beams is large enough $({\bf x}_0^2 \gg \sigma^2+\Delta^2)$ one use the asymptotic expansion of the function Ei($z$) in (\[5.15\]): $${\rm Ei}(z) \simeq \frac{e^z}{z}\left( 1+\frac{1}{z}\right),\quad
z \gg 1.
\label{5.16}$$ In this case the main terms in the difference $J-J_{-}$ in Eq.(\[5.7\]) are canceled: $$J-J_{-} \simeq \frac{e^{d}}{d}\left(
\frac{1}{d}-\frac{1}{d_1}\right) = \frac{2 e^{d}}{d}
\frac{\sigma^2}{{\bf x}_0^2}.
\label{5.16a}$$ The compensation of the main terms in (\[5.16\]) is due to the fact that the incoherent scattering originates on the fluctuations of the potential of the target (scattering) beam. Correspondingly we have for the mean square of the momentum transfer dispersion at the large distance from the target beam $$\begin{aligned}
&& 