2D Fourier transform of $1/(x^2-y^2+q)$

Question

How can I calculate the following 2D Fourier integral: $$ \iint \frac{{\rm e}^{{\rm i}(ax+by)}}{x^2-y^2+q} {\rm d}x\,{\rm d}y, $$ where $q$ is a complex number?

If there was a "+" sign in the denominator: $(x^2+y^2+q)^{-1}$, I'd use polar coordinates to get to the Hankel transform of a simple function $(r^2+q)^{-1}$, that can be calculated in terms of modified Bessel function. But I've no idea how to deal with the $(x^2-y^2+q)^{-1}$ function.

Update:

I am also interested in calculating even more general integral $$ \iint \frac{{\rm e}^{{\rm i}(ax+by)}}{x^2-y^2 + sy+q} {\rm d}x\,{\rm d}y, $$ where both $s$ and $q$ are complex numbers (for example $s=-2{\rm i}$, $q=4-{\rm i}$).

I have reasons to believe that this integral can be calculated in a closed form using Bessel $J_0$ function.

Answer 1

Let us consider a function

$$ f(x,y) = \frac{1}{x^2+ay^2+q}. $$

If $a$ and $b$ is a positive real values then the Fourier transform is $$ \hat{f}(k_x,k_y) = \frac{2 \pi}{\sqrt{a}} K_0\left(\sqrt{q (k_x^2+k_y^2/a)}\right). $$

It can be easily derived as in your link. Now you can try to analytically continue the function to other values of $a$ and $q$. I.e. just substitute desired values of $a$ and $q$ to the formula above (and maybe choose the right branch of square roots). I numerically verify it and it works! See my post on mathematica SE for details.

Answer 2

I'll assume the exponential is supposed to be $\exp\left[-i\left(ax+by\right)\right]$ instead of $\exp\left[+i\left(ax+by\right)\right]$.

Here's what I would try. Write this as: $$ \int_{-\infty}^{\infty} dy \ e^{-iby} \left[\int_{-\infty}^{\infty} dx \ e^{-iax} \frac{1}{x^2 + \left(q-y^2\right)}\right] $$ The expression in brackets is calculated here (or Google a bit for "lorentzian fourier transform"), though you'll have to be careful since you said $q$ is complex.

Once you've evaluated the integral in brackets, do the $y$ integral.