I need to compute the two-dimensional Fourier transform of a function with circular symmetry: $$ \int dx dy\, \frac{e^{i (k_x x+k_y y)}}{((z'-z)^2-t^2+x^2+y^2)((z'+z)^2-t^2+x^2+y^2)} $$ For simplicity, we can assume that $k_x, k_y >0$.
One can perform the $x$-integral and obtains $$ \frac{\pi}{4}\int dy\,e^{ik_y y}\left[ \frac{e^{-k_x\sqrt{-t^2+y^2+(z'-z)^2}}}{\sqrt{-t^2+y^2+(z'-z)^2}}-\frac{e^{-k_x\sqrt{-t^2+y^2+(z'+z)^2}}}{\sqrt{-t^2+y^2+(z'+z)^2}}\right] $$ We now have to perform an integral of the form $$ \int dy\, e^{ik_y y}\frac{e^{-k_x\sqrt{y^2+a}}}{\sqrt{y^2+a}} $$ One ansatz I tried was to write this as $$ -\frac{2}{k_x}\frac{\partial}{\partial a}\int dy\, e^{ik_y y}e^{-k_x\sqrt{y^2+a}} $$ but I still don't know how to Fourier transform this simpler function. Another ansatz would be to expand the integrand in a series $$ \sum_{n=0}^\infty \int dy\, e^{ik_y y}\frac{(-k_x)^n}{n!}(y^2+a)^{\frac{n}{2}}=\sum_{n=0}^\infty \frac{(-k_x)^n}{\Gamma(n+1)\Gamma\left(-\frac{n}{2}\right)}2^{\frac{3+n}{4}}a^{\frac{1+n}{4}}k_y^{-\frac{n+1}{2}}K_{\frac{1+n}{2}}\left(\sqrt{a} k\right) $$ I do not know how to sum this series.
Does anybody know a useful resource or an ansatz (maybe using the circular symmetry of the function I want to transform from the beginning)?
Edit: Another attempt is the following: write the integral in polar coordinates $$ \int dr \int_0^{2\pi} d\phi\, \frac{r\,e^{i (k_x r \cos(\phi)+k_y r \sin(\phi))}}{((z'-z)^2-t^2+r^2)((z'+z)^2-t^2+r^2)} $$ Then the $\phi$-integral can be done to give $$ 2\pi\int dr\, \frac{r J_0\left(\sqrt{k_x^2+k_y^2}\,r\right)}{((z'-z)^2-t^2+r^2)((z'+z)^2-t^2+r^2)} $$ Unfortunately, this integral is difficult to solve as well.
After performing the $\phi$ integral the resulting radial integral is of the form $$ 2\pi\int dr\, \frac{r J_0\left(c\,r\right)}{(a+r^2)(b+r^2)}. $$ One can write the "rational function" part in the integrand as a Laplace transform $$ \frac{r}{(r^2+a)(r^2+b)}=\int_0^\infty ds\,e^{-rs}\left(\frac{\cos(\sqrt{b}s)-\cos(\sqrt{a}s)}{a-b}\right) $$ Then use the fact that the Bessel function $J_0$ has a simple Laplace transform $$ \int_0^\infty dr\,e^{-rs}J_0(cr)=\frac{1}{\sqrt{c^2+s^2}} $$ The resulting integral is $$ \int_0^\infty \frac{ds}{\sqrt{c^2+s^2}}\,\left(\frac{\cos(\sqrt{b}s)-\cos(\sqrt{a}s)}{a-b}\right)=\frac{1}{a-b}\left(K_0(c\sqrt{b})-K_0(c\sqrt{a})\right) $$ So the resulting Fourier transform is $$ \frac{\pi}{2z z'} \left[K_0\left(\sqrt{k_x^2+k_y^2}\sqrt{(z'-z)^2-t^2}\right)-K_0\left(\sqrt{k_x^2+k_y^2}\sqrt{(z'+z)^2-t^2}\right)\right]. $$