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Let $\varphi:\mathbb{R}^4\to\mathbb{R}$ be a twice continuously differentiable compactly supported function, i.e. $\varphi\in C_c^2(\mathbb{R}^4)$. The following identity should hold (see note):$$\nabla_x\cdot\int_\mathbb{R^3}\varphi(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)\,\frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}\,d\mu_{\boldsymbol{y}}+\frac{1}{c}\int_\mathbb{R^3}\frac{\dot\varphi(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)}{\|\boldsymbol{x}-\boldsymbol{y}\|^2}\,d\mu_{\boldsymbol{y}}$$ $$=4\pi\varphi(\boldsymbol{x},t)$$where $\dot\varphi$ is the derivative of $\varphi:(\boldsymbol{\xi},\tau)\mapsto\varphi(\boldsymbol{\xi},\tau) $ with respect to the second, linear, variable $\tau$, the integrals are Lebesgue integrals or equivalently limits of Riemann integrals, and $\nabla_x$ is the divergence calculated with respect to the components of $\boldsymbol{x}$.

As explained here, I know that $$\nabla_x\cdot\int_\mathbb{R^3}\varphi(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)\frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}\,d\mu_{\boldsymbol{y}}= \int_\mathbb{R^3}\nabla_\xi\cdot\varphi(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)\frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}\,d\mu_{\boldsymbol{y}}$$ where I use $\nabla_\xi$ for the divergence of $\varphi:(\boldsymbol{\xi},\tau)\mapsto\varphi(\boldsymbol{\xi},\tau) $ calculated according to the components of the first, tridimensional, variable $\boldsymbol{\xi}$, but I am not able to use this property to prove the desired identity. Moreover, I know that the partial derivatives with respect to the components of $\boldsymbol{x}$ cannot just be moved under the integral sign. I suspect that integrating by parts is the best path to the proof, but I am not able to apply it...

How could the desired identity be proved? I would think that we should use some sort of integration by part, but I cannot chose a form of it enabling me to s. I heartily thank any answerer.


Note: I am convinced that the identity holds because the Lorenz gauge retarded potential of electrodynamics $V(\boldsymbol{x},t) :=\frac{1}{4\pi\varepsilon_0}\int_\mathbb{R^3}\frac{\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)}{\|\boldsymbol{x}-\boldsymbol{y}\|}d\mu_{\boldsymbol{y}}$ satisfies the equality $\nabla_x^2 V(\boldsymbol{x},t)$ $-\frac{1}{c^2}\frac{\partial^2 V(\boldsymbol{x},t)}{\partial t^2}= -\frac{\rho(\boldsymbol{x},t)}{\varepsilon_0}$. By using the reasoning used here, I see that $\nabla_x V(\boldsymbol{x},t)=-\frac{1}{4\pi\varepsilon_0}\int_\mathbb{R^3}\frac{\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}\,d\mu_{\boldsymbol{y}}- \frac{1}{4\pi\varepsilon_0 c}\int_\mathbb{R^3}\frac{\dot\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^2}\,d\mu_{\boldsymbol{y}} $ and, by taking the divergence of the two addends, and applying the same reasonings to differentiate the second addend under the integral sign, I get: $\nabla_x^2 V(\boldsymbol{x},t)=-\frac{1}{4\pi\varepsilon_0}\nabla_x\cdot\int_\mathbb{R^3}\frac{\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}\,d\mu_{\boldsymbol{y}}-$ $ \frac{1}{4\pi\varepsilon_0 c}\int_\mathbb{R^3}\frac{\dot\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)}{\|\boldsymbol{x}-\boldsymbol{y}\|^2}\,d\mu_{\boldsymbol{y}} + \frac{1}{4\pi\varepsilon_0 c^2}\int_\mathbb{R^3}\frac{\ddot\rho(\boldsymbol{y},t-c^{-1}\| \boldsymbol{x}-\boldsymbol{y} \|)}{\|\boldsymbol{x}-\boldsymbol{y}\|}\,d\mu_{\boldsymbol{y}}$. Therefore $\nabla_x^2 V-\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}= -\frac{\rho}{\varepsilon_0}$ holds if and only if the identity that I want to prove holds.

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