Residue theorem integration with Bessel function in half real line I'm trying to calculate the following integral:
$$
\int_0^\infty dk\frac{J_0(k\rho)e^{-kz}}{\sqrt{k^2+\lambda^2}},
$$
where $J_0(y)$ is the Bessel function, and $\rho$, $z$ and $\lambda$ are real and positive constants.
I wish to compute it by using the residue theorem but I'm not sure what kind of contour I have to chose. Moreover, I'm confused about branches (if there is any), given that the function is not even in $k$. Do you guys can give me some advice?
Thanks.
 A: Too long for a comment.
Using
$$J_0(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2 }\left(\frac{x}{2}\right)^{2 n}$$ would lead to some monster for any $\lambda \neq 0$ since the integrand would be
$$\sum_{n=0}^\infty(-1)^n \frac {\rho^{2n}} {4^n\,(n!)^2}\int_0^\infty \frac{k^{2 n} }{\sqrt{k^2+\lambda ^2}}e^{-k z}\,dk$$
In the case where $\lambda=0$, we have
$$\int_0^\infty \frac{k^{2 n} }{\sqrt{k^2+\lambda ^2}}e^{-k z}\,dk < \int_0^\infty k^{2 n-1} e^{-k z}\,dk=z^{-2 n} \Gamma (2 n)$$ Even for this simple case, there is no closed form for the summation.
Just to give an idea (given by a CAS)
$$\frac 2{\sqrt \pi} z^{2 n} \int_0^\infty \frac{k^{2 n} }{\sqrt{k^2+\lambda ^2}}e^{-k z}\,dk =$$
$$2 \sqrt{\pi }\, \Gamma (2 n) \,
   _1F_2\left(\frac{1}{2};\frac{1}{2}-n,1-n;-\frac{1}{4} z^2 \lambda ^2\right)+$$
$$\Gamma (-n) \Gamma \left(n+\frac{1}{2}\right) (\lambda  z)^{2 n} \,
   _1F_2\left(n+\frac{1}{2};\frac{1}{2},n+1;-\frac{1}{4} z^2 \lambda ^2\right)-$$
$$\Gamma \left(-n-\frac{1}{2}\right) \Gamma (n+1) (\lambda  z)^{2 n+1} \,
   _1F_2\left(n+1;\frac{3}{2},n+\frac{3}{2};-\frac{1}{4} z^2 \lambda ^2\right)$$
