We know the Hankel transform of order 0 is defined as F 0 ( k...

$\begin{array}{}\text{(1)}& F(k,a,z):={\int }_0^{\mathrm{\infty }}\frac{\exp (-a\sqrt{r^2+z^2})}{\sqrt{r^2+z^2}}{J}_0(kr)rdr=\frac{\exp (-z\sqrt{a^2+k^2})}{\sqrt{a^2+k^2}}\end{array}$
$\frac{\exp (-a\sqrt{r^2+z^2})}{a\sqrt{r^2+z^2}}={\int }_1^{\mathrm{\infty }}{e}^{-art}{J}_0(az\sqrt{t^2-1})dt$
Insert into left-hand side of (1) and interchange $\int .$ The innermost integral has a closed form,
${\int }_0^{\mathrm{\infty }}{e}^{-art}{J}_0(kr)rdr=\frac{at}{((at{)}^2+k^2{)}^{3/2}}$
which Mathematica knows. So we now have
$\begin{array}{}\text{(2a)}& F(k,a,z)=a^2{\int }_1^{\mathrm{\infty }}{J}_0(az\sqrt{t^2-1})\frac{tdt}{((at{)}^2+k^2{)}^{3/2}}\end{array}$
$\begin{array}{}\text{(2b)}& =\frac{1}{2a}{\int }_1^{\mathrm{\infty }}\frac{J_0(az\sqrt{t-1})}{(t+(k/a{)}^2{)}^{3/2}}dt=\frac{1}{2a}{\int }_0^{\mathrm{\infty }}\frac{J_0(az\sqrt{t})}{(t+p{)}^{3/2}}dt\text{ with }p=1+(k/a{)}^2\end{array}$
where from line 2a to 2b we substituted $t^2\to t$ and in the last step the limits of the integrand have been shifted with a subsequent change of the parameter. Now use the integral relationship
${\int }_0^{\mathrm{\infty }}{J}_0(c\sqrt{u})(u+p{)}^{-3/2}du=2\frac{\exp (-c\sqrt{p})}{\sqrt{p}}.$
Mathematica knows this integral with $c=1,$ and it is easy to work in the scaling factor. Algebra completes the proof.