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!
$$ \tag{1}F(k,a,z):=\int_0^\infty \frac{\exp{\big(\!-a\sqrt{r^2+z^2}\big)}}{\sqrt{r^2+z^2}} J_0(k \ r) \ r dr = \frac{\exp{\big(\!-z\sqrt{a^2+k^2}\big)}}{\sqrt{a^2+k^2}} $$ 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.