I have the following function
\begin{equation} f(\textbf{r}-\textbf{r}')=\frac{e^{-a|\textbf{r}-\textbf{r}'|}}{|\textbf{r}-\textbf{r}'|}, \end{equation}
with $\textbf{r}=(x,y,z)$ and $a>0$. Now I am trying to take the spatial 2D-Fourier transform with respect to $\textbf{r}_{\parallel}=(x,y)$, namely
\begin{equation} \int d\textbf{r}_{\parallel}\frac{e^{-a\left|\textbf{r}_{\parallel}-\textbf{r}_{\parallel}'+\left(z-z'\right)\hat{z}\right|}}{\left|\textbf{r}_{\parallel}-\textbf{r}_{\parallel}'+\left(z-z'\right)\hat{z}\right|}e^{-i\textbf{k}_{\parallel}\cdot\textbf{r}_{\parallel}}. \end{equation}
I have tried to use polar coordinate to solve the above integral but I did not manage to.
We might have luck if we work in polar coordinate, $\textbf{r}_{\parallel}=\left(r,\phi\right)$. Now, recall that if $g\left(\theta-\phi\right)=g\left(\theta-\phi+2\pi n\right)$, we have $\int_{0}^{2\pi}d\theta g\left(\theta-\phi\right)=\int_{0}^{2\pi}d\phi g\left(\phi\right)$ and hence we can write \begin{align*} {\cal F}_{2D}^{\textbf{k}_{\parallel},\omega}\left[\left\langle \delta\rho\left(\textbf{r},t\right)\delta\rho\left(0,0\right)\right\rangle \right] & =\int_{0}^{\infty}drr\int_{0}^{2\pi}d\phi\frac{e^{-a\sqrt{r^{2}+\left(z'-z\right)^{2}}}}{\sqrt{r^{2}+\left(z'-z\right)^{2}}}e^{-ikr\cos\theta}\\ & =\int_{0}^{\infty}dr\thinspace r\frac{e^{-a\sqrt{r^{2}+\left(z'-z\right)^{2}}}}{\sqrt{r^{2}+\left(z'-z\right)^{2}}}\int_{0}^{2\pi}d\theta e^{-ikr\cos\theta}\\ & =\int_{0}^{\infty}dr\thinspace r\frac{e^{-\left(a\sqrt{r^{2}+\left(z'-z\right)^{2}}\right)}}{\sqrt{r^{2}+\left(z'-z\right)^{2}}}\times2\pi J_{0}\left(kr\right)\\ & = ? \end{align*}
Any thoughts?