TITLE: Solution of Definite integral:$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i(k_{1}x+k_{2}y)}dxdy$
QUESTION [6 upvotes]: I'm trying to evaluate the following two dimensional integral:
$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i(k_{1}x+k_{2}y)}dxdy$
The paper that i'm following reports the following solution:
$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i(k_{1}x+k_{2}y)}dxdy = \frac{1}{\sqrt{k_{1}^{2}+k_{2}^{2}}}e^{-\sqrt{k_{1}^{2}+k_{2}^{2}}z}$
it might be useful using the following definitions:
$r = \sqrt{x^{2}+y^{2}+z^{2}}$
and
$k = \sqrt{k_{1}^{2}+k_{2}^{2}}$
I think this should follow from some kind of substitution;
anyway i tried to solve it with mathematica but i'm not getting any solution...
Any idea on how to obtain the result shown by the authors?
Thanks in advance

REPLY [4 votes]: In my first answer, I proceeded by trying to directly compute the momentum integrals. In doing so, I'd forgotten that there is a rather less direct approach which yet has the advantage of not requiring Bessel function identities.
First, note that our task is to take the $2$D Fourier transform of the function $G(\mathbf{r})=(x^2+y^2+z^2)^{-1/2}=r^{-1}$. My choice of $G$ as symbol for this function is not done idly, for this is in fact the free-space Green's function of the $3$D  Laplacian i.e. $\nabla^2 G(\mathbb{r})=-4\pi \delta^{3}(\mathbf{r})$. This is a standard Dirac-delta identity which can be verified via the divergence theorem and direct differentiation.
The $3$D Fourier transform of this is readily computed to be $(k_x^2+k_y^2+k_z^2)G(\mathbf{k})=4\pi.$ To get the planar Fourier transform, we can then take the inverse transform with respect to $k_z$. That iss,
\begin{align}
G(k_x,k_y,z)=\mathcal{F}^{-1}[G(\mathbf{k})](z)
&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-i k_z z}G(\mathbf{k})\,dk_z\\
&=\int_{-\infty}^\infty \frac{2e^{-i k_z z}}{k_x^2+k_y^2+k_z^2}\,dk_z\\
&=\frac{2}{k}\int_{-\infty}^\infty \frac{e^{-i k z u}}{1+u^2}\,du
\end{align}
where in the last line I have introduced $k=\sqrt{k_x^2+k_y^2}$ and $k_z=k u$. This last integral can be evaluated via methods of contour integration (see Wikipedia of such) to obtain the desired result
$$\boxed{G(k_x,k_y,z)=2\pi\dfrac{e^{-k |z|}}{k}=2\pi\left(\dfrac{e^{-|z|\sqrt{k_x^2+k_y^2}}}{\sqrt{k_x^2+k_y^2}}\right)}$$