# An integral giving back a function which is a factor in the integrand

I suspect that, under opportune assumptions on $\varphi:\mathbb{R}^3\times\mathbb{R}\to\mathbb{R}$, $(\boldsymbol{\xi},\tau)\mapsto \varphi(\boldsymbol{\xi},\tau)$, such as $\varphi\in C_c^2(\mathbb{R}^4)$, the following identity holds, for any $\alpha\in\mathbb{R}$: $$\int_{\mathbb{R}^3}\frac{\nabla_{\boldsymbol{\xi}}^2\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} -\frac{\alpha^2\ddot\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} d\mu_{\boldsymbol{y}}=-4\pi\varphi(\boldsymbol{x},t)\quad\quad(1)$$where $\nabla_{\boldsymbol{\xi}}^2\varphi$ is the Laplacian calculated with respect to the first tridimensional variable of $\varphi$ (called $\boldsymbol{\xi}$ at the beginning of the post) and $\ddot\varphi$ is the second order derivative with respect to the second variable of $\varphi$. I am convinced that this equality holds because, if it did, it could be used to rigourously prove that the Lorenz gauge retarded potential $\boldsymbol{A}$ satisfies the equality $\nabla^2\boldsymbol{A}-\varepsilon_0\mu_0\frac{\partial^2 }{\partial t^2}\boldsymbol{A}=-\mu_0\boldsymbol{J}$ in the same way the same equality with $\alpha=0$, which holds as proved here, can be used to rigourously prove that the the magnetostatic potential is such that $\nabla^2\boldsymbol{A}=-\mu_0\boldsymbol{J}$.

Can anybody help me to prove the equality $(1)$?

As pointed out in the comments, whose author Daniel Fischer I thank again, the integral might be calculated by integrating by parts and taking the limit of the (Riemann) integral $$\int_{\mathbb{R}^3\setminus{B(\boldsymbol{x},\delta)}}\frac{\nabla_{\boldsymbol{\xi}}^2\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} -\frac{\alpha^2\ddot\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} dy_1dy_2dy_3$$as $\delta\to 0$, but, in the formula of integration by parts $\int_\Omega \frac{\partial f}{\partial x_j} g\ d^3x = \int_{\partial \Omega} fgn_j\, d\sigma -\int_{\Omega} f\frac{\partial g}{\partial x_j}\, d^3x$ (where $n_j$ is the $j$-th component of the external normal vector to $\partial\Omega$), I do not know what to chose as $f$ and $g$ in our integrand. I heartily thank any answerer.

• I'd expect that integrating this against a test function, changing the order of integration (Fubini), and then integrating the inner integral (over $x$ and $t$) by parts to move the derivatives over to the test function (and the Newton potentials) would work. But I'm not particularly inclined to try that out myself, looks too much like physics for my taste. – Daniel Fischer Aug 18 '17 at 19:54
• I meant showing that $\iint I(x,t)u(x,t)\,d\mu(x)\,d\mu(t) = -4\pi\iint \varphi(x,t)u(x,t)\,d\mu(x)\,d\mu(t)$ for every test function $u$. But on second thoughts, it shouldn't be necessary to take a test function and another integral. If you look at the integral over $\lVert x-y\rVert > \varepsilon$, then integrate by parts, and finally let $\varepsilon \to 0$, that should work. – Daniel Fischer Aug 20 '17 at 13:54
• Oh, disregard my previous comments. I missed that there were two time derivatives in the second term. Sorry about that. – Nate Eldredge Sep 11 '17 at 13:43