= Solution
For <elastic scattering> by a real short-range scalar potential, write $C=m/(2\pi\hbar^2)$, $\mathbf k_i=\mathbf k$ and $|\mathbf k_f|=k$. In the far field, $|\mathbf r-\mathbf r'|=r-\widehat{\mathbf r}\cdot\mathbf r'+O(r^{-1})$ and $\mathbf k_f=k\widehat{\mathbf r}$. The outgoing spherical-wave coefficient is
$$
f(\mathbf k_f,\mathbf k_i)=-C\int d^3r'\,
e^{-i\mathbf k_f\cdot\mathbf r'}V(\mathbf r')\psi_{\mathbf k_i}(\mathbf r').
$$
Substitution of the incident <plane wave> inside this integral gives
$$
\boxed{f^{(1)}=-C\widetilde V(\mathbf q),\qquad
\frac{d\sigma^{(1)}}{d\Omega}=C^2|\widetilde V(\mathbf q)|^2},\qquad
\widetilde V(\mathbf q)=\int e^{-i\mathbf q\cdot\mathbf r}V(\mathbf r)\,d^3r.
$$
The incoming and outgoing <flux> factors cancel because their speeds agree. Here $\mathbf q=\mathbf k_f-\mathbf k_i$ is a <wavevector> transfer; the actual <momentum transfer> is $\hbar\mathbf q$. This distinction preserves dimensions. The approximation requires the scattered correction to the interior wave to be small, and does not in general apply to an unscreened long-range potential.
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