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 $\alpha\in\mathbb{R}$: $$ \int_{\mathbb{R}^3}\frac{\nabla_{\boldsymbol{\xi}}^2\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} -\frac{\alpha^2\ddot\varphi(\boldsymbol{y},t-\alpha\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} d\mu_{\boldsymbol{y}}=-4\pi\varphi(\boldsymbol{x},t)\label{1}\tag{1}$$ 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 \eqref{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.