I am looking for the best coordinate system to help simplify this problem I am looking at. Suppose I have a beam of light rays entering normally into a focusing device (blue rectangle in figure). Upon entering the focusing element the light rays will undergo circular motion with a radius of $\rho=L/ax$ where $a$ is some constant which is inversely proportional to the light ray's horizontal displacement; light rays at $x=0$ will not undergo circular motion and go straight through the focusing device. Upon exiting the focusing element, the light rays will undergo linear motion and will focus to a point along $\hat{z}$.
My goal is to solve integrals of the form:
$$A(\vec{r},t) = \int_{Blue \ Rectangle} dV' \frac{\vec{j}(\vec{r'},t')}{|\vec{r}-\vec{r'}|}, $$
where $\vec{r'}$ is the displacement vector within the focusing device only (blue rectangle), $\vec{r}$ is the displacement vector anywhere outside and to the right of the focusing device, $\vec{j}(\vec{r'},t')$ is the "current density" of the light ray within the device at some "past" time $t'$. The distinction between $t$ (the present) and $t'$ (the past) is similar to that of retardation effects in radiation studies i.e. the information of the light rays velocity in the focusing device will take time to propagate from $\vec{r'}$ to $\vec{r}$. The time it will take is $|\vec{r}-\vec{r'}|/c$ where $c$ is the speed of light, so $t'=t-|\vec{r}-\vec{r'}|/c$.
The current density $\textit{before}$ entering the focusing device is a bi-Gaussian traveling along $+\hat{z}$ with speed of $v_o$:
$$ \vec{j}(\vec{r'},t') = \frac{Q_{tot}}{2\pi\sigma_x\sigma_z}\exp{\left({-\frac{x'^2}{2\sigma_x^2}}\right)}\exp{\left({-\frac{\left({z'-v_ot'}\right)^2}{2\sigma_z^2}}\right)}\left({v_o \hat{z}}\right)$$
Now, the problem with computing $\vec{A(\vec{r},t)}$ is the complexity that putting $t'=t-|\vec{r}-\vec{r'}|/c$ in the expression gives and the rotation of the current density inside the focusing device. In addition, the integration over the volume, $dV'$, in cartesian coordinates will have a complicated $x-$dependence due to the circular trajectories of the light rays within the focusing device.
My attempt to a solution:
Assume the density of the light rays along $\hat{z}$ is constant and the light rays are infinitely long: $$ \vec{j}(\vec{r'},t') = \frac{Q_{tot}}{2\pi\sigma_x\sigma_z}\exp{\left({-\frac{x'^2}{2\sigma_x^2}}\right)}\lambda_o\left({v_o \hat{z}}\right).$$
Do the integration over cylindrical coordinates ($dV'=\rho'd\rho d\theta' dz'$) to simplify the current density inside the focusing device i.e. for a given $\rho'$ value the transverse density will be constant and will only have a $j_\theta$ component:
$$ \begin{align} A(\vec{r},t) &= \int\int\int \left({\rho' d\rho' d\theta' dz'}\right) \frac{\frac{Q_{tot}}{2\pi\sigma_x\sigma_z}\exp{\left({-\frac{\left(\frac{lq}{a\rho'}\right)^2}{2\sigma_x^2}}\right)}\lambda_o\left({v_o \hat{\theta}}\right)}{\sqrt{\rho^2+\rho'^2-2\rho\rho'\cos\left({\theta-\theta'}\right)}}\\ \end{align} $$
As you integrate over $d\rho'$ the origin will shift along $\hat{x}$ an amount of $x_{origin} = -(\rho'-x')=-(|\frac{L}{ax'}|-x')$.
The description of $\vec{r}$ is become more complicated in this coordinate system: $$\rho^2 = (L+d)^2+((\rho'-x')+x)^2 $$ $$ \theta = \tan\left({ \frac{(\rho'-x')+x}{L+d} }\right)$$
So, even with my simplifications and the transformation into another coordinate system the integration does not give an analytical result. I was wondering if anyone had any ideas if theres a formalism to solve such an integration. I was going to try Frenet-Serret coordinate system next but I dont have high hopes for it.