Fourier transform of Bessel Function Second Kind

Question

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.

Answer 1

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.