 Solution

Write $r=|\mathbf{r}|$, $s=|{\mathbf{r}}^{\prime }|$. For a spherical source, the angular integral needed for the Green function of the Laplacian isThis is the angular form of the spherical shell theorem. Substituting the source into the Green function integral therefore yieldsThe elementary integrals areThe terms proportional to $e^{-\beta r}/\beta$ cancel, leavingThe value at the origin is the continuous limit. At large radius $\varphi \sim -1/({\beta }^{2}r)$, consistent with the integrated source $\int \rho dV=-4\pi /{\beta }^2$ and the factor $1/(4\pi r)$ in the far-field potential.

For a direct check, set $u=r\varphi$. Then $u^{\prime \prime }=e^{-\beta r}$, so ${\nabla }^2\varphi =u^{\prime \prime }/r=e^{-\beta r}/r=-\rho$ for $r>0$. Boundedness of $\varphi$ at zero cancels the singular $1/r$ terms and excludes an extra point source. The original density is itself singular but locally integrable; the equation at the origin is understood in the sense of distributions. This is the Poisson potential of an exponential-over-radius source with amplitude $A=-1$.