Explaining the integral of Laplacian of $r \mapsto \frac1r$ Consider the function $\Bbb R^3 \to \Bbb R$ defined by
$$f(x,y,z) = \frac1{\left( x^2 + y^2 + z^2 \right)^{\frac12}}$$
or, written in polar coordinates, $f (r) = \frac1r$.
The Laplacian $\nabla^2 f \equiv \nabla \cdot (\nabla f)$ can be computed as $0$ everywhere except the origin. Is there a way to rigorously define its value at the origin, and a concept of integration, such that
$$\int_{\Bbb R^3} \nabla^2 f = -4\pi?$$
(This is a result I saw in a physics textbook that was computed by non-rigorous means.)
If not, what does the previous integral expression actual mean, mathematically?
 A: Note that $\nabla\frac1{|r|}=-\frac{r}{|r|^3}$. From there, we can compute that away from $0$, $\Delta\frac1{|r|}=0$. However, if we use the Divergence Theorem,
$$
\begin{align}
\int_{|r|\lt\alpha}\Delta\frac1{|r|}\,\mathrm{d}V
&=\int_{|r|\lt\alpha}\nabla\cdot\left(-\frac{r}{|r|^3}\right)\mathrm{d}V\tag1\\
&=\int_{|r|=\alpha}n\cdot\left(-\frac{r}{|r|^3}\right)\mathrm{d}S\tag2\\
&=\int_{|r|=\alpha}\frac{r}{|r|}\cdot\left(-\frac{r}{|r|^3}\right)\mathrm{d}S\tag3\\
&=-\int_{|r|=\alpha}\frac1{|r|^2}\,\mathrm{d}S\tag4\\[6pt]
&=-4\pi\tag5
\end{align}
$$
Explanation:
$(1)$: $\nabla\frac1{|r|}=-\frac{r}{|r|^3}$
$(2)$: Divergence Theorem
$(3)$: $n=\frac{r}{|r|}$
$(4)$: simplify the dot product
$(5)$: the surface area of the sphere $|r|=\alpha$ is $4\pi\alpha^2$
Since $\Delta\frac1{|r|}=0$ away from $0$, we get
$$
\int_\Omega\Delta\frac1{|r|}\,\mathrm{d}V=\left\{\begin{array}{rl}
0&\text{if }0\not\in\overline\Omega\\
-4\pi&\text{if }0\in\overline\Omega
\end{array}\right.\tag6
$$
That is, $\Delta\frac1{|r|}$ is $-4\pi\delta$ where $\delta$ is the Dirac delta distribution.
