Fourier transform of Bessel Function Second Kind How do I prove the following equation,
$$\frac{1}{\sqrt{(x^2 + y^2)}}=\int_0^{\infty}\frac{2}{\pi}K_0(yt)(\cos(xt))\,dt  $$
This is a Fourier transform of $K$,
I proceeded as follows:
\begin{align}
& =\int_0^{\infty}\frac{2}{\pi}\frac{\pi}{2}iH_0^1(iyt)(\cos(xt))\,dt \\[10pt]
& =\int_0^{\infty}\frac{2}{\pi}\frac{\pi}{2}iH_0^1(iyt)(e^{ixt}+e^{-ixt})\frac{1}{2}\,dt \\[10pt]
& =\int_0^{\infty}iH_0^1(iyt)(e^{ixt}+e^{-ixt})\frac{1}{2}\,dt \\[10pt]
& =\int_0^{\infty}iH_0^1(iyt)(e^{ixt})\frac{1}{2}\,dt+\int_0^{\infty}iH_0^1(iyt)(e^{-ixt})\frac{1}{2}\,dt
\end{align}
At this moment I don't know where to go from here.

Since:
$$K_0(yt)=\int_{0}^{+\infty}\exp(-yt\cosh z)\,dz$$
the Fubini theorem ensures:
$$ I = \frac{2}{\pi}\int_{0}^{+\infty}\frac{y\cosh z}{x^2+y^2\cosh^2 z}\,dz$$
and the substitution $z=2\operatorname{arctanh} u$ and the residue theorem finish the proof.
