3

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.

Community
  • 1
bcp
  • 485
  • 3
  • 18
  • You could try a change of variables $x_1=x+y$ and $x_2=x-y$, so you get something like $\int e^{i(ax_1+bx_2)}/(x_1x_2+q),dx$ (constants change). Then after some of algebra and another change of variable you end up with $\int e^{i(bx_2-qa/x_2)}/x_2,dx_2\int e^{iax_1}/x_1,dx_1$. I hope you can with both of them using complex integration. – user90189 Oct 04 '14 at 07:37
  • Be careful with any singularity. – user90189 Oct 04 '14 at 07:44

2 Answers2

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.

Community
  • 1
ybeltukov
  • 296
  • Thank you for an answer. And can you imagine an approach to calculating Forier transform of a bit more general function: $\frac1{x^2-y^2+sy+q}$, where both $s$ and $q$ are complex? – bcp Oct 05 '14 at 20:56
  • @bcp After applying the same idea I get $\frac{2 \pi}{\sqrt{a}} e^{-i s k_y/2 a} K_0\Big(\sqrt{(q-s^2/4 a^2) (k_x^2+k_y^2/a)}\Big)$. In your case $a=-1$. Your general function even more difficult from numerical point of view because it has poles. However, the result seems to be valid. – ybeltukov Oct 05 '14 at 21:30
  • I think it's not so easy: the Fourier transfrom of this new function (see sample parameters in the updated question) looks quite different: it is non-zero at $|k_x|<-k_y$ and is strictly zero at the other points. (I used your numerical approach. Note that at $s=−2{\rm i}, q=4−{\rm i}$ the function has no poles). – bcp Oct 05 '14 at 21:39
  • @bcp Indeed. Everything simple have a certain limit :) – ybeltukov Oct 05 '14 at 22:33
0

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.

Community
  • 1
Eric Angle
  • 1,837
  • The problem will be with the y-integral: you'll have something like $\displaystyle\int \frac{{\rm exp}\left{-|a|\sqrt{q-y^2}\right}}{\sqrt{q-y^2}} {\rm e}^{i b y},{\rm d}y$ – bcp Sep 30 '14 at 13:45