I'm trying to evaluate the following integral using complex function theory: \begin{equation} \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 \end{equation}
I though that it is possible if i can calculate: \begin{equation} \int_{-\infty}^{\infty}\frac{e^{i(ap+b\sqrt{k^2-p^2})}}{\sqrt{k^2-p^2}}dp \end{equation}
I'm trying to go around the singularity as follows:
- Substitute $p=z$ to work in the complex plane.
- 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.
- 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: \begin{equation} \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 \end{equation}
But when I integrate the contour around the singularity I seem to get zero, which isn't right I think.
This question is related to Evaluating the integral $\int_{-\infty}^{\infty}\frac{1}{\sqrt{k-p}\sqrt{k+p}}\,\mathrm dp$ and Help solving $\int_{-\infty}^{\infty}\frac{e^{i(ap+b\sqrt{k^2-p^2})}}{\sqrt{k^2-p^2}}dp$
Kind Regards.