# Very tricky Fourier transform

I'm trying to evaluate the following integral 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: $$\int_{-\infty}^{\infty}\frac{e^{i(ap+b\sqrt{k^2-p^2})}}{\sqrt{k^2-p^2}}dp$$

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 $k+i\gamma$ and after the integration add a limit of $\gamma$ 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.

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is there a particular reason you made this with a separate account? Or would you like to merge them? Not to assume, but I'd be very surprised if those other questions weren't asked by you. – Robert Mastragostino Jun 6 '12 at 20:30
How are you defining $\sqrt{k^2-p^2-q^2}$ when $p^2+q^2>k^2$? – Thomas Andrews Jun 6 '12 at 20:35
@Micheal Why do you keep on using the substitution $p=z$? – AD. Jun 6 '12 at 20:45
1. I see no Fourier transform. 2. If you change variables you must replace each instance of the variable with the new variable. – AD. Jun 6 '12 at 20:47
@Robert: 1. yes, please merge them. 2. no, p^2+q^2 >= 0 3. I would like to fourier transform $$\frac{e^{i(b\sqrt{k^2-p^2-q^2})}}{\sqrt{k^2-p^2-q^2}}$$ – Micheal Jun 6 '12 at 20:48