I am studying Dirac-delta function and not in a rigorous way. How to $\nabla^2 \frac{e^{-\alpha r}}{r}$ , $\alpha$ is a constant, in a delta function of spherical coordinate system?
I know that we can reduce it to $\nabla^2 (\frac{1}{r})$, which is solved. But is there another method to solve this kind of problem?
In general, is it possible to write any function that blows up to infinity at a point in a delta function?
For $\alpha>0$ what you seek doesn't happen, because point charges aren't involved in the generation of such a potential. Integrating$$\nabla^2\frac{e^{-\alpha r}}{r}=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\frac{e^{-\alpha r}}{r}\right)=\frac{\alpha^2}{r}e^{-\alpha r}$$over $\Bbb R^3$ gives $4\pi\alpha^2\int_0^\infty re^{-\alpha r}dr=4\pi$ if $\alpha>0$. For $\alpha=0$, the potential's $r=0$ divergence can be interpreted as a unit charge with a Dirac delta density, so the total charge multiplied by $4\pi$ is the above integral. But we cannot write the $\alpha=0$ Laplacian as $f(r)\delta^{(3)}(x)$, because it doesn't vanish outside the origin.
