Hankel transform of $exp(-a\sqrt{r^2+z^2})/\sqrt{r^2+z^2}$ We know the Hankel transform of order 0 is defined as
\begin{equation}
{\displaystyle F_{0}(k)=\int _{0}^{\infty }f(r)J_{0}(kr)\,r\,\mathrm {d} r}.
\end{equation}
In this regard, I am now trying to calculate the Hankel transform of the function
\begin{equation}
f(r)=\frac{e^{-a\sqrt{r^2+z^2}}}{\sqrt{r^2+z^2}},
\end{equation}
with $a\in \mathbb{R}$. Unfortunately, I was only able to obtain the solution for $a=0$. Any thoughts on how to solve thi? PS: I did not manage to find this integral within Hankel transform tables.
Thanks!
 A: $$ \tag{1}F(k,a,z):=\int_0^\infty \frac{\exp{\big(\!-a\sqrt{r^2+z^2}\big)}}{\sqrt{r^2+z^2}} J_0(k \ r) \  r dr  = \frac{\exp{\big(\!-z\sqrt{a^2+k^2}\big)}}{\sqrt{a^2+k^2}} $$
Proof sketch:  Gradshteyn and Ryzhik 6.616.2 states
$$\frac{ \exp{\big(\!-a\sqrt{r^2+z^2}\big)}}{a \sqrt{r^2+z^2}} = \int_1^\infty e^{-a \ r \ t} J_0(a \ z \sqrt{t^2-1}) dt $$
Insert into left-hand side of (1) and interchange $\int.$  The innermost integral has a closed form,
$$ \int_0^\infty e^{-a \ r \ t} J_0(k \ r) r dr  = \frac{a \ t}{\big( (at)^2 + k^2 \big)^{3/2} }$$
which Mathematica knows.  So we now have
$$ \tag{2a}F(k,a,z)=a^2 \int_1^\infty  J_0(a \ z \sqrt{t^2-1}) \frac{t \ dt}{\big( (at)^2 + k^2 \big)^{3/2} }$$
$$ \tag{2b}=\frac{1}{2a}\int_1^\infty \frac{ J_0(a \ z \sqrt{t-1}) }{\big( t + (k/a)^2 \big)^{3/2} }dt = \frac{1}{2a} \int_0^\infty \frac{ J_0(a \ z \sqrt{t}) }{\big( t + p \big)^{3/2} } dt \text{ with } p=1+(k/a)^2$$
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
$$ \int_0^\infty J_0(c\sqrt{u})(u+p)^{-3/2} du = 2 \frac{\exp{\big(\!-c\sqrt{p}\big)}}{\sqrt{p}} .$$
Mathematica knows this integral with $c=1,$ and it is easy to work in the scaling factor.  Algebra completes the proof.
