I want to give formal meaning to the following integral in the distributional sense:
$$ A^{\mu}(x) = \frac{1}{4\pi} \int_{\mathbb{R^3}} d^3 \vec{y} \int_{\mathbb{R}} dy^0 \frac{1}{|\vec{x} - \vec{y}|} \; \delta \, (x^0 - y^0 - |\vec{x} - \vec{y}|) \; \theta(x^0 - y^0) \; J^{\mu}(y^0,\vec{y}) = \\ = \frac{1}{4\pi} \int_{\mathbb{R^3}} d^3 \vec{y} \; J^{\mu}(x^0 - |\vec{x}-\vec{y}|,\vec{y}) $$
Without the $\theta(x^0-y^0)$ the result follows from the $\delta(x)$ distribution definition. With the Heaviside distribution I couldn't give any formal meaning/computation as in general the product of distribution is not even defined, and in this case $$ \int_{-\infty}^{+\infty} dy^0 \theta(x^0-y^0)\delta(x^0-y^0-|\vec{x}-\vec{y}|)f(y^0) = \int_{-\infty}^{x^0} dy^0 \delta(x^0-y^0-|\vec{x}-\vec{y}|)f(y^0) \neq f(x^0-|\vec{x}-\vec{y}|). $$ Can we give meaning to the product of the Dirac and Heaviside distribution? Maybe can it be stated in form of convolution of distribution?
I failed to make sense out of it in any of the following cases:
$J^{\mu} \in S$ Schwartz space
$J^{\mu} \in C^0_c$
$J^{\mu} \in S'$ Tempered distributions, a usual expression is $J^{\mu} = \int \delta^4 (x-y) \; dy^{\mu}$
where the previous are to be interpreted componentwise in the vector field $J^{\mu}$