||Message]
$\newcommand{\dydxt}[2]{\frac{d#1}{d#2}}
\newcommand{\vv}[1]{{\bf #1}}$
As an application of the first Born approximation we calculate the Rutherford scattering by a Coulomb potential. For
\begin{eqnarray}V( r) =\frac{C}{r}, \end{eqnarray}
The integral in question is singular. We ``screen" the Coulomb potential replacing $C/r$ by $C\exp(-\epsilon r)/r$ with the understanding that $\epsilon\to 0$. Then
\begin{eqnarray*} \int_0^\infty dr e^{-\epsilon r}\sin(Kr)= \lim_{\epsilon\to 0}\frac{K}{\epsilon^2+K^2}=
\frac{1}{K},\end{eqnarray*}
and the cross section formula is the familiar
\begin{eqnarray}\dydxt{\sigma}{\Omega}=\left(\frac{C}{4E}\right)^2\frac{1}{\sin^4(\theta/2)}, \end{eqnarray}
where $K^2\hbar^2=|\vv{k}-\vv{k'}|^2=4k^2\sin^2(\theta/2)$, $\theta$ is the angle between $\vv{k}$ and $\vv{k'}$, and $E$ is the energy $E=k^2/2m$.
