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I have found a derivation of the fact that, under the assumptions made in physics on ##\rho## (to which we can give the physical interpretation of charge density) the function defined by

$$V(\mathbf{x},t):=\frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|} d^3y,$$

where I think the integral to be a Lebesgue integral or limit of a Riemann integral (since the singularity at ##\mathbf{y}=\mathbf{x}## does not allow, as far as I know, to define a Riemann integral), which corresponds to the retarted electric potential of electrodynamics if the constant ##\varepsilon_0## is interpretated as vacuum permittivity, is such that

$$\nabla_x^2V(\mathbf{x},t)=-\frac{1}{\varepsilon_0} \rho(\mathbf{x},t)+ \frac{1}{c^2} \frac{\partial^2 V(\mathbf{x},t) }{\partial t^2} .$$

But I am not able to understand the mathematical meaning of the steps I find in that derivation. Could anybody help me in understanding it, with a rigourous justification of its steps by known theorems, or giving me an alternative mathematically rigourous proof?

To simplify the notation I will omit the arguments of the functions ##V## and ##\rho##. The derivation that I have found is as follows (blue text):

$$\nabla_x V= \frac{1}{4\pi\varepsilon_0} \int \frac{1}{\|\mathbf{x}-\mathbf{y}\|} \nabla\rho+\rho\nabla\left(\frac{1}{\|\mathbf{x}-\mathbf{y}\|}\right) d^3y$$

I notice that differentiation under the integral sign is used, but I have no idea why it is legitimate. Given the physical nature of the problem, I think that ##\rho(-,t):\mathbf{y}\mapsto \rho(\mathbf{y},t)## can be assumed to be compacly supported as a function of ##\mathbf{y}## for all ##t## and such that ##\rho\in C^2(\mathbb{R}^4)##, but the discontinuity of the integrand at ##\mathbf{x}## does not allow me to use a theorem I know*.

##\nabla_x\rho=-c^{-1}\dot\rho\nabla_x \|\mathbf{x}-\mathbf{y}\|## (where the dot above denotes the partial derivative with respect to the second variable), ##\nabla_x \|\mathbf{x}-\mathbf{y}\|=\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|}## and ##\nabla_x\|\mathbf{x}-\mathbf{y}\|^{-1}=-\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} ##, so $$\nabla_x V= \frac{1}{4\pi\varepsilon_0} \int -\frac{\dot\rho}{c}\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} -\rho\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} d^3y.$$Taking the divergence $$\nabla_x^2 V= \frac{1}{4\pi\varepsilon_0} \int -\frac{1}{c}\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2}\cdot\nabla_x\dot\rho -\frac{\dot\rho}{c}\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right) $$$$-\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\cdot\nabla_x \rho - \rho \nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)\,d^3y $$

Where differentiation under the integral sign is assumed to be legitimate, but it is not clear to me why it is. The derivation follows:

But $$\nabla_x \dot\rho=-\frac{\ddot\rho}{c} \nabla_x \|\mathbf{x}-\mathbf{y}\|=-\frac{\ddot\rho}{c} \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|},$$ $$\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right)=\frac{1}{\|\mathbf{x}-\mathbf{y}\|^2} $$and $$\nabla\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)=4\pi\delta^3(\mathbf{y}-\mathbf{x}).$$

Where we see that here, unlike the other derivatives, the "derivatives" are not meant to be the ordinary derivatives (which would give ##\nabla\cdot( (\mathbf{x}-\mathbf{y}) / \|\mathbf{x}-\mathbf{y}\|^3 )=0## everywhere except for ##\mathbf{x}=\mathbf{y}##, where it does not exist, but one point would not change the integral), if it were a Lebesgue integral. The derivation follows:

So $$\nabla_x^2 V=\frac{1}{4\pi\varepsilon_0} \int\frac{\ddot\rho}{c^2\|\mathbf{x}-\mathbf{y}\|}-4\pi\delta^3(\mathbf{y}-\mathbf{x})\,d^3y=\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}-\frac{\rho}{\varepsilon_0}.$$

It is not the first time I find unexplained derivations under the integral sign, which suddenly becomes something different from a Lebesgue integral at some step, and introductions of ##\delta## without any explanation of why such introductions are legimitate. I know, as proved here, that, provided that we give the special sense used in functional analysis to the derivatives of the Laplacian, ##\nabla^2\left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right)= -4\pi\delta^3(\mathbf{y}-\mathbf{x})##, **but** that is in the sense that $$\forall\varphi\in C_c^2(\mathbb{R}^3)\quad \int_{\mathbb{R}^3}\frac{\nabla_y^2\varphi(\mathbf{y})}{\| \mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=-4\pi \varphi(\mathbf{x}), $$ while, if we read ##\nabla_x^2## as a Laplacian where the derivatives are what they are in calculus, $$\int_{\mathbb{R}^3} \varphi(\mathbf{y}) \nabla_x^2 \left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right) d\mu_{\mathbf{y}}=\int_{\mathbb{R}^3} \varphi(\mathbf{y})\cdot 0 \,d\mu_{\mathbf{y}}=0.$$

I also notice that the singularity at ##\mathbf{y}=\mathbf{x}## does not prevent to consider ##-\frac{\dot\rho}{c}\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right) -\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\cdot\nabla_x \rho=0##, while ##\nabla\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)## is not considered constantly ##0##. (Honestly speaking, from the answers I get when talking about these issues with other students - I study by myself and I am not enrolled at a university yet -, I am beginning to think that many see that sort of differentiations under the integral sign and introductions of ##\delta## as tricks to pretend to do mathematics.)

Could anybody help me with a rigourous proof of ##\nabla_x^2V=-\frac{1}{\varepsilon_0} \rho+ \frac{1}{c^2} \frac{\partial^2 V }{\partial t^2} ## including the mention of the theorems (I think that this and this may help, but I am not able to produce a complete proof) used to justify the not-so-trivial steps?

I ##\infty##-ly thank any answerer!

*The Leibniz rule in the form: if ##V\subset \mathbb{R}^n## is compact and ##f:V\times[a,b]\to\mathbb{R}##, ##(\mathbf{x},s)\mapsto f(\mathbf{x},s)## is such that ##\frac{\partial f}{\partial s} \in C(V\times[a,b])## (and therefore,by the Heine-Cantor theorem, uniformly continuous on ##V\times[a,b]##, which is compact), then

$$\forall s\in[a,b]\quad \frac{d}{ds}\int_V f(\mathbf{x},s)d^n x=\int_V \frac{\partial f(\mathbf{x},s)}{\partial s} d^n x$$

where the integrals are (proper) Riemann integrals.

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# I Laplacian of retarded potential

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