Calculation of an integral with retarded time

[FunWarrior]

Good morning.

I would like to prove that the integral

h^{\mu \nu} (\vec{r},t) = \int d \zeta \int d^3 \vec{y} \frac{F^{\mu \nu} (\zeta,\tilde{\tau}) \delta^{(3)} (\vec{r} - \vec{x}(\zeta,\tilde{\tau}))}{|\vec{r}-\vec{y}|}

where \tilde{\tau} = t - |\vec{r}-\vec{y}|, is equal to

\int d \zeta \frac{F^{\mu \nu} (\zeta,\tau)}{|\vec{r}-\vec{x}(\zeta,\tau)| (1-\hat{n} \cdot \dot{\vec{x}}(\zeta,\tau))}

where \displaystyle \hat{n}= \frac{\vec{r}-\vec{x}(\zeta,\tau)}{|\vec{r}-\vec{x}(\zeta,\tau)|}, \tau = t - |\vec{r}-\vec{x}(\zeta,\tau)| and \dot{\vec{x}}(\zeta,\tau) is the derivative of \vec{x} with respect to his second variable.

I would like to integrate with respect to y using the Dirac function \delta^{(3)} but I don't manadge to find the value of \vec{y} such that \vec{r} - \vec{x}(\zeta,\tilde{\tau}) vanishes. I also tried to use

\delta^{(3)} ( \vec{x} - \vec{a}) = \frac{\delta^{(3)} ( \vec{\xi} - \vec{\alpha})}{|J|}

where \vec{\xi} = \vec{\xi} (\vec{x}), \vec{\alpha} = \vec{\xi} (\vec{a}) and J is the Jacobian of the transformation of \vec{x} into \vec{\xi} but without success.

Thank you in advance for your help.

 

[lalbatros]

This looks very similar to the question of deriving the Liénard-Wiechert potentials from the general Retarded potential formula. This is all about how to handle the dirac distribution.
By pure chance, I found this paper which probably will help you, specially in page 4:

http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap14/chap14.pdf