In this paper about the electrodynamcis of a spiral resonator, the authors write
$$\frac{e^{-ikR}}{R}=\int_{0}^{\infty} \frac{xdx}{4\pi\sqrt{x^2-k^2}}J_0(Dx)e^{-\sqrt{x^2-k^2}|z|}$$
with $R=\sqrt{z^2+D^2}$
The reference given is Handbook of Mathematical Function with Formulas, but still, I have spent hours trying to derive this equation unsuccessfully. Some help would be appreciated.
Let $G$ denote the Green (or Green's) function for the Helmholtz equation
$$(\nabla^2+k^2)G(x,y,z)=- \delta(x)\delta(y)\delta(z)\tag1$$
Assuming a $e^{i\omega t}$ time convention, we know that the solution to $(1)$ is $G(x,y,z)=\frac{e^{-ikR}}{4\pi R}$ (i.e., to represent outward traveling waves), where $R=\sqrt{x^2+y^2+z^2}$. We wish to show that $G$ can be represented by the integral
$$G(x,y,z)=\frac1{4\pi}\int_0^\infty \frac{k_\rho}{\sqrt{k_\rho^2-k^2}} e^{-\sqrt{k_\rho^2-k^2}|z|} J_0(k_\rho \sqrt{x^2+y^2})\,dk_\rho$$
To do so, we denote by $g(k_x,k_y,z)$ the two-dimensional Fourier transform of $G(x,y,z)$ as given by
$$g(k_x,k_y,z)=\int_{-\infty}^\infty \int_{-\infty}^\infty G(x,y,z)e^{i(k_xx+k_yy)}\,dx\,dy\tag2$$
Then taking the Fourier transform of $(1)$, we find that
$$\frac{d^2g(k_x,k_y,z)}{dz^2}-(k_x^2+k_y^2-k^2)g(k_x,k_y,z)=- \delta(z)\tag3$$
Solution to $(3)$, along with the condition $\lim_{|z|\to \infty}g=0$, is given by
$$g(k_x,k_y,z)=\frac{e^{-\sqrt{k_x^2+k_y^2-k^2}|z|}}{2\sqrt{k_x^2+k_y^2-k^2}} \tag4$$
Taking the inverse Fourier transform yeilds
$$\begin{align} G(x,y,z)&=\frac1{(2\pi)^2}\int_{-\infty}^\infty \int_{-\infty}^\infty g(k_x,k_y,z)e^{-i(k_xx+k_yy)}\,dk_x\,dk_y\\\\ &=\frac1{(2\pi)^2}\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{e^{-\sqrt{k_x^2+k_y^2-k^2}|z|}}{2\sqrt{k_x^2+k_y^2-k^2}}e^{-i(k_xx+k_yy)}\,dk_x\,dk_y\\\\ &=\frac1{(2\pi)^2} \int_0^\infty \int_0^{2\pi}\frac{e^{-\sqrt{k_\rho^2-k^2}|z|}}{2\sqrt{k_\rho^2-k^2}}e^{ik_\rho \sqrt{x^2+y^2}\cos(\phi)}\,k_\rho\,dk_\phi\,dk_\rho\\\\ &=\frac1{4\pi}\int_0^\infty \frac{k_\rho}{2\sqrt{k_\rho^2-k^2}} e^{-\sqrt{k_\rho^2-k^2}|z|} \frac1\pi\int_{-\pi}^{\pi} e^{ik_\rho \sqrt{x^2+y^2}\cos(\phi)}\,d\phi\,dk_\rho\\\\ &=\frac1{4\pi}\int_0^\infty \frac{k_\rho}{\sqrt{k_\rho^2-k^2}} e^{-\sqrt{k_\rho^2-k^2}|z|} \frac1\pi\int_{0}^{\pi/2} \cos(k_\rho \sqrt{x^2+y^2}\sin(\phi))\,d\phi\,dk_\rho\\\\ &=\frac1{4\pi}\int_0^\infty \frac{k_\rho}{\sqrt{k_\rho^2-k^2}} e^{-\sqrt{k_\rho^2-k^2}|z|} J_0(k_\rho \sqrt{x^2+y^2})\,dk_\rho\\\\ \end{align}$$
as was to be shown!
