I would like to verify that $\frac{1}{4\pi |x|}$ is the fundamental solution of the 3 dimensional laplace operator so that
$$ \triangle \frac{1}{4 \pi |x|} = \delta(x) $$
What I have tried:
I think my best approach is to apply the distribution to a function
$$ \begin{aligned} \triangle \frac{1}{4 \pi |x|}[\varphi] &= \delta[\varphi] \\ - \frac{1}{4 \pi |x|} [\triangle \varphi] &= \delta[\varphi] \\ - \int_\mathbb{R} \frac{1}{4 \pi |x|} \triangle \varphi dx^3 &= \int_\mathbb{R} \delta(x) \varphi(x) dx^3 = \varphi(0) \end{aligned} $$ However the integral on the left has to be reinterpreted, because $\frac{1}{|x|}$ is not a schwarz function and it is undefined at 0. So we have to use some combination of a convergece factor and a modefication of the region, maybe similar to this: $$ - \int_\mathbb{R} \frac{1}{4 \pi |x|} \triangle \varphi dx^3 = - \lim_{\varepsilon, \eta \to 0} \int_{\mathbb{R} \setminus B_\varepsilon(0)} \frac{e^{- \eta |x|}}{4 \pi |x|} \triangle \varphi dx^3 $$
I do not really care that much about the answer beeing super rigoros, if it works, it works. I only experimented with the convergence factor because I was stuck and thought introducing these things might help me.
I was hopeing that using integration by parts and a change into spherical coordinates I would get something like this (with volume element $4\pi r²$):
$$ - 4 \pi \int_\varepsilon^\infty \frac{1}{r} \triangle \varphi \: r² dx^3 = 4 \pi \int_\varepsilon^\infty \left( \triangle r \right) \varphi(r) dr - \left[ r \varphi \right]_\varepsilon^\infty \qquad (\varepsilon \to 0) $$ The $\left[ r \varphi \right]_\varepsilon^\infty$ part is $\varepsilon \varphi(\varepsilon)$ and the integral should somehow evaluate to $\varphi(0)$, when using the spherical laplace operator $\triangle r = \frac{2}{r}$.
( I have also looked at other questions and answers from the math stackexchance, in particular this one, but I do not understand the answer. Also this is more about verifying that it is a solution than finding the solution. There are also many questions regarding the 2 Dimensional one.)
