1. Jun 6, 2012

### MichealM

I'm trying to evaluate the following intergral using complex function theory:

\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{i(ap+aq+b\sqrt{k^2-p^2-q^2})}}{\sqrt{k^2-p^2-q^2}}dpdq

I though that it is possible if i can calculate:

\int_{-\infty}^{\infty}\frac{e^{i(az+b\sqrt{k^2-p^2})}}{\sqrt{k^2-p^2}}dp

I'm trying to go around the singularity as follows:

1. Substitute p=z to work in the complex plane
2. Move one pole up to k+iγ and after the integration add a limit of γ going to zero and similarly move the other singularity downward.
3. Which results in two contour integrals in the complex plane one around the singularity in the upper plane plus an over the singularity in lower plane.

This enables me to express the integration into the following integral in the complex plane:

\lim_{\gamma \rightarrow 0} \int_{-\infty}^{\infty}\frac{e^{i(ap+b\sqrt{k^2-z^2})}}{\sqrt{k+i \gamma+z}\sqrt{k-i \gamma-z}}dz

But when I integrate the contour around the singularity I seem to get zero, which isn't right I think.

Kind Regards,
Micheal

Last edited: Jun 6, 2012
2. Jun 6, 2012

### mathman

Re: Very tricky Fourier transform plz help!

I suggest that you switch to polar coordinates.

3. Jun 6, 2012

### MichealM

Re: Very tricky Fourier transform plz help!

I already tried that and I got

\int_{0}^{\infty}\int_{0}^{2\pi}\frac{e^{i(a r\ cos\theta +ar\ sin \theta +b\sqrt{k^2-r^2})}}{\sqrt{k^2-r^2}}r sin \theta drd\theta

But this doesn’t bring me further, because of the integration over $\theta$... I think

4. Jun 6, 2012

### Charles49

How is this a Fourier Transform?

5. Jun 6, 2012

### Charles49

I don't think this integral converges. Here's my analysis:
$$\iint _{\mathbb{R}^2}\!{\frac {{{\rm e}^{i \left( x+y+\sqrt {k-{x }^{2}-{y}^{2}} \right) }}}{\sqrt {k-{x}^{2}-{y}^{2}}}}{dx}{dy} =\iint_{\mathbb{R}^2}\!{\frac {\cos \left( x+y+\sqrt {k-{x}^{2}-y^2} \right) }{\sqrt {k^2-{x}^{2}-{y}^{2}}}}{dx}{dy}+i\iint_{\mathbb{R}^2}\!{\frac {\sin \left( x+y+\sqrt {k-{x}^{2}-{y}^2} \right) }{\sqrt {k^2-{x}^{2}-{y}^{2}}}}{dx}{dy}.$$
The neither of the above integrals converge...

Last edited: Jun 6, 2012