= Solution
For a short-range potential, an incident <plane wave> of unit amplitude and wave vector $k\mathbf e_z$ has the large-distance <scattering wavefunction>
$$
\boxed{\psi(\mathbf r)=e^{ikz}+f(\theta,\phi)\frac{e^{ikr}}r+O(r^{-2})\qquad(r\to\infty)}.
$$
The coefficient $f$ is the <scattering amplitude>. For a central potential it depends only on the scattering angle $\theta$. The incident <probability current> is $\hbar k/m$, while the scattered radial current is $(\hbar k/m)|f|^2/r^2$ to leading order. Their ratio gives
$$
\boxed{\frac{d\sigma}{d\Omega}=|f(\theta,\phi)|^2,\qquad
\sigma_{\mathrm{tot}}=\int |f(\theta,\phi)|^2\,d\Omega}.
$$
These are the <differential scattering cross-section> and total <cross section> for elastic scattering with the same incident and outgoing wave number.
The expression is \i[asymptotic]: it specifies the outgoing wave's leading $1/r$ term, not its finite-distance radial corrections or its form inside the interaction region. In particular, for angle-dependent $f$, the spherical term by itself has an angular-Laplacian contribution of order $r^{-3}$ to the free wave equation, which is cancelled by subleading terms in the exact solution.
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