Integral of an unbounded function as a solution of $\nabla^2\boldsymbol{A}=-\mu_0\boldsymbol{J}$

Question

While studying the equivalence between the Biot-Savart and Ampère's laws I have only found proofs of the fact that$$\boldsymbol{A}(\boldsymbol{x})=\frac{\mu_0}{4\pi}\int_V \frac{\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d^3 x'$$ is a solution of $\nabla^2\boldsymbol{A}=-\mu_0\boldsymbol{J}$, where $\mu_0$ is a constant (magnetic permeability in that physical case), using Dirac's delta, which use in the tridimensional case I have not studied yet.

I have calculated - please correct me if I am wrong - that, if $V\subset\mathbb{R}^3$ is compact (as usually assumed in physics, I think), which allows us to differentiate under the integral sign, and $\boldsymbol{x}\notin V$, which allows the integral to exist finite if we are considering a Riemann integral, $\nabla^2\boldsymbol{A}=\mathbf{0}$.

Although my first tought was that it is not possible to prove that $\frac{\mu_0}{4\pi}\int_V \frac{\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d^3 x'$ is a solution of $\nabla^2\boldsymbol{A}=\mathbf{0}$ only by using the tools of multivariate calculus, without using Dirac's $\delta$, while considering the integral as a "Riemann integral", in the sense of a limit (since if $\boldsymbol{x}\in V$ it cannot obviously be a Riemann integral proper), I have been told in PSE that it may well be possible and suggested to ask here how to prove it. I heartily thank any answerer!

Answer 1

I hope I have a possible solution. Le us assume that, for all the components $J_i$ of $\boldsymbol{J}$, there is a function $f_i\in C^4(\mathring{A})$ such that $\nabla^2 f_i=J_i$, with $\bar{V}\subset \mathring{A}$ and $\boldsymbol{x}\in \mathring{V}$ and there exists a $\delta$ such that, for all $\epsilon\le \delta$, $\epsilon >0$, the region $V\setminus B_\epsilon(r)$ satisfies the condition of the divergence theorem. Then this result, applied to the components of $\boldsymbol{J}$, guarantess that

$$f_i(\boldsymbol{x})=\frac{1}{4\pi}\int_{\partial V}\left(\frac{\nabla' f_i(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}-f(\boldsymbol{x}')\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\right)\cdot d\boldsymbol{S}'-\frac{1}{4\pi}\int_V\frac{\nabla^2f_i(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu_{\boldsymbol{x}'}$$

where $\mu_{\boldsymbol{x}'}$ is the usual Lebesgue measure on $\mathbb{R}^3$ and the integral is to be intended as a Lebesgue integral, which is the same as the limit, as $\epsilon\to0$, of the Riemann integral $\int_{V\setminus B_{\epsilon}(r)}\frac{\nabla^2f_i(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dV'$.

By differentiating under the surface integral sign, which is possible under our assumptions of continuity and with $\boldsymbol{x}\notin\partial V$, and by taking into account that $$\forall\boldsymbol{x}\ne\boldsymbol{x}'\quad\left(\nabla^2\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]=0\quad\land\quad\nabla^2\left[\frac{\boldsymbol{x}'-\boldsymbol{x}}{\|\boldsymbol{x}-\boldsymbol{x}'\|^3}\right]=\mathbf{0}\right)$$we see that, for all $i=1,2,3$, $$-\nabla^2\left[\frac{1}{4\pi}\int_V\frac{\nabla^2f_i(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu_{\boldsymbol{x}'}\right]=\nabla^2 f_i(\boldsymbol{x}).$$Therefore, by letting $\boldsymbol{f}:=(f_1,f_2,f_3)$ and $$\boldsymbol{A}(\boldsymbol{x}):=\frac{\mu_0}{4\pi}\int_V \frac{\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu_{\boldsymbol{x}'}=\frac{\mu_0}{4\pi}\int_V \frac{\nabla^2\boldsymbol{f}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu_{\boldsymbol{x}'}$$ we see that $$\nabla^2\boldsymbol{A}(\boldsymbol{x})=-\mu_0\nabla^2\boldsymbol{f}(\boldsymbol{x})=-\mu_0\boldsymbol{J}(\boldsymbol{x}).$$

I do not accept my own answer because I am not sure of its correctness nor of the correctness of its premise, reserving that act for future answers adding details, possibly dispensing with the restrictive assumption that there is a $\boldsymbol{f}$ such that $\nabla^2\boldsymbol{f}=\boldsymbol{J}$, and correcting mistakes if mine is not correct.

Answer 2

By applying this result, for which I thank Daniel Fischer again, with $f(x)=\|x\|^{-1}$ and $g$ in place of the components of $\frac{\mu_0}{4\pi}\boldsymbol{J}$, we can see that, if $\boldsymbol{J}\in C^2(\mathbb{R}^3)$ is supported within the compact set $V\subset\mathbb{R}^3$, then $$\nabla_x^2\left[\frac{\mu_0}{4\pi}\int_{V} \frac{\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d \mu_{ \boldsymbol{x}'}\right]=\nabla_x^2\left[\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3} \frac{\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d \mu_{ \boldsymbol{x}'}\right]=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3} \frac{\nabla'^2\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d \mu_{ \boldsymbol{x}'}$$where the integrals are Lebesgue integrals with respect to the $3$-dimensional Lebesgue measure, and this last integral can be evaluated, by using this argument, due to our kind and brilliant Daniel again, as $$\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3} \frac{\nabla'^2\boldsymbol{J}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d \mu_{ \boldsymbol{x}'}=-\mu_0\boldsymbol{J}(\boldsymbol{x}).$$