My textbook of physics, Gettys's Physics (Italian language edition), says that the Lorenz gauge choice uses the magnetic vector potential $$\mathbf{A}(\mathbf{x},t):=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3} \frac{\mathbf{J}(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}} $$ where $\mathbf{J}:\mathbb{R}^4\to\mathbb{R}^3$ is a function satisfying the assumptions given in physics, like differentiability and being zero outside $R\times\mathbb{R}$ where $R\subset\mathbb{R}^3$ is a bounded domain, $\mu_0$ is magnetic permeability and $c$ is the speed of light.

How can we prove that such an $\mathbf{A}$ satisfies Maxwell's equation $$\nabla\cdot(\nabla\times\mathbf{A})=0?$$ Problems arise since I am not sure whether and how we could differentiate under the integral sign, and I am asking here rather than in PSE because I notice that questions focussing on mathematical derivations tend to be redirected here.

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    $\begingroup$ HINT: Let $\vec r=\vec y-\vec x$. Then, we have $$\begin{align} \int_{\mathscr{R}^3}\frac{\vec J(\vec y,t-\frac1c|\vec y-\vec x|)}{|\vec y-\vec x|}d\mu_{\vec y}&=\int_{\mathscr{R}^3} \frac{\vec J(\vec r+\vec x,t-\frac1c|\vec r|)}{|\vec r|}d\mu_{\vec r}\\\\ &=\int_0^{2\pi}\int_0^\pi \int_0^\infty \vec J\left(\vec r+\vec x,t-\frac rc \right)\,r\sin(\theta)\,dr\,d\theta\,d\phi \end{align}$$ $\endgroup$ – Mark Viola Nov 19 '16 at 19:35
  • $\begingroup$ @Dr.MV I thank you so much for the hint! Nevertheless, I don't see how this facilitates differentiation (under the integral sign?)... $\endgroup$ – Self-teaching worker Nov 19 '16 at 22:33
  • $\begingroup$ There is no singularity now. $\endgroup$ – Mark Viola Nov 19 '16 at 23:20
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    $\begingroup$ That sounds correct. $\endgroup$ – Mark Viola Nov 20 '16 at 18:04
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    $\begingroup$ And in case you were unaware, $\nabla \cdot \nabla. \times \vec A=0$ is simply a vector identity. $\endgroup$ – Mark Viola Nov 20 '16 at 19:11

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