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For scattering of a spinless particle from a central potential $V=V(r)$, the first Born amplitude can be written in the basis $\kk{\vv{k}}$, the momentum states (also the energy eigenstates).
\begin{eqnarray*} T_1(\vv{k},\vv{k'})=\bb{\vv{k}}|V\kk{\vv{k}'}
=\int d^3\vv{r}\bb{\vv{k}}\kk{\vv{r}}
\bb{\vv{r}}|V\kk{\vv{k}'}\end{eqnarray*}
This can be simplified, using
$\bb{\vv{r}}\kk{\vv{k}}=\exp(i\vv{k}.\vv{r}/\hbar)/(2\pi\hbar)^{3/2}$,
\begin{eqnarray*} \bb{\vv{k}}|V\kk{\vv{k}'}=-\frac{2}{(2\pi\hbar)^2\hbar}
\int_0^\infty r^2drV(r)\frac{\sin(Kr)}{Kr} \end{eqnarray*}
where $K=|\vv{k}-\vv{k}'|/\hbar$. With this the differential cross section can
be written
\begin{eqnarray}d\sigma=\left|-\frac{2m}{\hbar^2}\int_0^\infty
r^2drV(r)\frac{\sin(Kr)}{Kr}\right|^2 d\Omega_{\vv{k}'} \end{eqnarray}
where we have purposely written the cross sectrion in a form that is traditional for Borm approximation (see, for example, L. I. Schiff, Quantum Mechanics, Third Edition, p.325).
