I am working on the FMM method implementation. One of the important equation is the integral representation of the inverse distance $$ \frac{1}{r} = \frac{1}{2 \pi} \int_{0}^{\infty} e^{-\lambda z} \int_{0}^{2\pi} e^{i \lambda (x \cos \alpha + y \sin \alpha)}\, d\alpha d\lambda. $$
I want to understand, where this equation comes from? How to derive this formula? May be use Fourier transform?
I think I found the solution.
Based on the thread Simpler proof of an integral representation of Bessel function of the first kind $J_n(x)$ the
$$ \frac{1}{2\pi}\int_{0}^{2 \pi} e^{i\lambda(x \cos \alpha + y \sin \alpha)} d\alpha = J_0(\lambda\sqrt{x^2 + y^2}), $$ where $J_0$ is the Bessel function order zero.
It is known property of the Bessel function [Watson 1922][1]: $$ \int_0^\infty e^{-at}J_0(bt)dt = \frac{1}{\sqrt{a^2 + b^2}} $$
Now we have $$ \frac{1}{2 \pi}\int_0^{\infty}e^{-\lambda z}\int_0^{2\pi}e^{i\lambda(x\cos\alpha + y\sin\alpha})d\alpha d\lambda = \int_0^\infty e^{-\lambda z}J_0(\lambda\sqrt{x^2+y^2})d\lambda = \frac{1}{x^2+y^2+z^2} = \frac{1}{r}. $$
[1]: Watson G. N. Theory of Bessel functions, 1922. Page 384.