= Poisson potential of an exponential-over-radius source
{title2=$\rho(r)=Ae^{-\beta r}/r$}
For $\beta>0$, solve $\nabla^2\phi=-\rho$ in three dimensions with $\rho(r)=Ae^{-\beta r}/r$, finite potential at the origin and decay at infinity. The <Green function> is $1/(4\pi|\mathbf r-\mathbf r'|)$ in this sign convention. The angular integral supplied by the <spherical shell theorem> gives
$$
\phi(r)=\frac1r\int_0^r s^2\rho(s)\,ds+\int_r^\infty s\rho(s)\,ds
=\boxed{\frac{A(1-e^{-\beta r})}{\beta^2r}}.
$$
The value at the origin is the limit $A/\beta$. A second derivation sets $u=r\phi$: the radial <Poisson equation> becomes $u''=-Ae^{-\beta r}$. Decay excludes the constant-potential term and boundedness at the origin cancels the remaining $1/r$ singularity. Although the source is locally <integrable>, it is singular at the origin; the resulting potential solves the equation there in the sense of <distributions>, without an extra <Dirac delta>.
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