Getting stuck trying to solve electromagnetic wave equation using Green's function I've recently learned about Green's function and am trying to derive an equation similar to that of the Biot-Savart law but for the electric field around a wire of changing current using the electric component of the electromagnetic wave equation:
$$\frac{\partial^2 E}{\partial t^2} = c^2\nabla^2E -c^2\mu_0 \frac{\partial J}{\partial t}$$
However, I am getting stuck on a step involving a dirac delta function in an integral and I'm not sure how to proceed next. Here are my steps:
Rewrite the wave equation so that we can use a Green's function for the 3-D wave equation:
$$\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}-\nabla^2E=-\mu_0 \frac{\partial J(t)}{\partial t}$$
From Wikipedia, the green's function is:

Resulting in the following equation to find $E$,
$$E=\int\limits_{\mathbb{R}^3}G(x)f(x) dr^3=-\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\frac{\delta^{3}(t-r/c)}{r}\cdot\frac{\partial J(t)}{\partial t}dr^3$$
At this point I'm a bit stuck, I'm not sure how to deal with the dirac delta inside the integral. One step I have tried is rewriting the current density of a thin wire as $J=I\cdot\delta_2(r)$,
$$E=-\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\frac{\delta_3(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t}\delta^{2}(r)\cdot dr^3$$
Letting us rewrite the equation as a line integral, similar to how the Biot-Savart law is written,
$$E=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \left[ \iint \frac{\delta_3(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t}\delta^{2}(r)\cdot dr^2 \right] dl$$
$$=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \frac{\delta^{3}(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t} dl$$
But after this I'm completely lost. Based on Jefimenko's equations I need to get something like,
$$=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \frac{1}{r} \cdot \frac{\partial I(t-r/c)}{\partial t} dl$$
Which does seem to give the correct behavior, but I'm not exactly sure how multiplying $I(t)$ with a dirac-delta results in $I(t-r/c)$. I've tried finding a property that results in this, but I can't find anything.
I think I'm missing a step or doing something wrong. Because I'm a newbie in Green's functions and just this area of math in general, I would really appreciate an easy to understand answer. But really any help would be appreciated.
 A: General considerations
I think it will help to carefully write out all the expressions. The PDE should be read component-wise
$$
\partial_{tt}\psi_j(x,t)-\nabla^2\psi_j(x,t)=F_j(x,t)
$$
Where $x\in\mathbb{R}^3$. I've replaced $J$ with a generic source $\mathbf{F}$, $\mathbf{E}$ with a generic field $\boldsymbol{\psi}$, and set all constants to unity. The causal Green's function is
$$
G(r,t)=\frac{\delta(t-r)}{4\pi r}
$$
The solution for $\psi_j$ is
$$
\psi_j(x,t)=\int dt' \int d^3x' \ G(x-x',t-t')F_j(x',t') \ \ + \ \ \text{surface terms}
$$
The surface terms are to match initial conditions which we can specify to vanish (see eg. Zangwill Modern Electrodynamics chapter 20). The delta collapses the $t'$ integral and we are left with
$$
\psi_j(x,t)=\frac{1}{4 \pi}\int d^3x' \ \frac{F_j(x',t-|x-x'|)}{|x-x'|}
$$
Note the arguments of the source appearing in the integrand.
The wave equation for $E$
Because $J$ and $\rho$ form part of a continuity equation, they cannot be specified completely independently. For $\psi=E$, the source should be $F_j=-(\partial_t J_j+\partial_j \rho)$, not just $\partial_t J$. You can check this by deriving it from Maxwell's equations. The expression for $E$ is
$$
E_j(x,t)=-\frac{1}{4\pi} \int d^3x' \ |x-x'|^{-1} \bigg[\partial'_{j} \rho(x',t-|x-x'|) \ + \ \partial'_{t}J_j(x',t-|x-x'|) \bigg]
$$
Note the primes on partial derivatives within the integral: the retarded time is $t_r:=t-|x-x'|$. You can see why it's preferable to write the integral as
$$
E_j(x,t)=-\frac{1}{4\pi} \int d^3x' \ |x-x'|^{-1} \bigg[\partial'_{j} \rho(x',t') \ + \ \partial'_{t}J_j(x',t') \bigg]_{t'=t_r}
$$
With $R_j:=x_j-x_j'$, and using the multivariate chain rule we find (Jackson chapter 6.5)
$$
\big[\partial'_j \rho(x',t') \big]_{t'=t_r} = \partial'_j \big[ \rho(x',t')\big]_{t'=t_r} - \hat{R} \big[ \partial'_t \rho(x',t')\big]_{t'=t_r}
$$
On integrating the term $|x-x'|^{-1}\partial'_j \big[ \rho(x',t')\big]_{t'=t_r}$ by parts, we find Jefimenko's equation for $E_j$.
Finally, if you want to specify a thin wire of arbitrary shape, the expression is not as simple as you may think. Start by parametrizing in $\tau$ such that the wire is the set of points $(x,y,z)=(p_1(\tau),p_2(\tau),p_3(\tau))$, then (up to a proportionality constant)
$$
J_j(\mathbf{x})=  \int d\tau \ \frac{dp_j}{d\tau} \delta^{(3)}(\mathbf{x}-\mathbf{p}(\tau))
$$
