In the Digital Library of Mathematical Functions they list some analytic continuations of modified Bessel functions.
I'm highly interested in the analytic continuation of this function:
$$ \Psi(s)=\sum_{n=1}^\infty 2\sqrt{ns}K_1(2\sqrt{ns}) $$
which converges for real $s>0.$
Why am I interested in finding this analytic continuation?
Because: $$ \frac{1}{w\Gamma(w)^2}\mathcal{M}(\Psi)(w)=\zeta(w) $$
which converges for real $s>1,$ where $\zeta(w)$ is the Riemann zeta function, $\Gamma(w)$ is the Gamma function, and $\mathcal{M}$ is the Mellin transform of $\Psi.$
In this spirit I tried writing: $$ \Psi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} z\Gamma(z)^2\zeta(z)s^{-z}~dz$$
and examining the residues, but I have become quite stuck at this point. This integrand tends to $0$ rapidly enough when $z\to \infty.$
The equation $\Psi(s) = \int_C z \Gamma(z)^2 \zeta(z) s^{-z} dz$ is correct. The inverse Mellin transform equation is about picking up the residues, there's nothing essential about using a straight-line contour. In fact, as you have probably noticed, a straight-line contour won't work for this function. Here's a graph of integrand $z \Gamma(z)^2 \zeta(z) s^{-z}$ for negative $s$
(This was generated by the very helpful complex plotter tool which you can find here)
With this in mind, you can take a contour like the following
I'll go back and check for any computational mistakes later, but you should get something like:
$$\frac{1}{s} + -\frac{1}{2} +\sum_{n=1}^{\infty}\frac{s^{n}}{n!^{2}}\left(\zeta(-n)\left(1+n\ln(s)-2n\psi(1+n)\right)-n\zeta'(-n)\right)$$