I am trying to compute the inverse Mellin transform of : $$\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)\zeta(n+1+s)}{\zeta(s)n!}\left(-\omega\right)^{n}$$ w.r.t. the complex number $s$. $\omega$ being a real parameter.
My Attempt :
it can be easily verified that the function $\phi(s,\omega)$ given by : $$\phi(s,\omega)=\frac{(s-1)}{\Gamma(s)^{2}}\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)\zeta(n+1+s)}{n!}\left(-\omega\right)^{n}$$
is entire in $s$. Thus, the Mellin inverse may be written as : $$\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\Gamma(s)^{2}\phi(s,\omega)}{(s-1)\zeta(s)}x^{-s}ds$$ which can be computed using the residue theorem. The problem now is to find $\phi(s,\omega)$, and it's derivatives at negative integers, and the non-trivial zeros of the Riemann zeta function. hence my question.
$\phi(s,\omega)$ is complicated, it grows very fast on vertical lines, it won't help.
For $\Re(s) > 1$
$$\zeta(s)\zeta(s+1) = \sum_{n=1}^\infty a_n n^{-s}, a_n = \sum_{d | n} d^{-1}$$
$$F(s)=\Gamma(s)\zeta(s)\zeta(s+1) = \sum_{n=1}^\infty a_n \int_0^\infty x^{s-1} e^{-nx}dx=\int_0^\infty x^{s-1} \sum_{n=1}^\infty a_n e^{-nx}dx$$ For $|\omega| < 1$ $$\sum_{k=0}^\infty \frac{1}{k!} F(s+k) (-\omega)^k =\sum_{k=0}^\infty \frac{1}{k!} (-\omega)^k \int_0^\infty x^{s+k-1} \sum_{n=1}^\infty a_n e^{-nx}dx $$ $$= \int_0^\infty x^{s-1} \sum_{k=0}^\infty \frac{1}{k!} (-\omega x)^k \sum_{n=1}^\infty a_n e^{-nx}dx = \int_0^\infty x^{s-1} e^{-\omega x} \sum_{n=1}^\infty a_n e^{-nx}dx$$ Note the latter is analytic in $\omega, \Re(\omega) > -1$ even if the power series in $\omega$ diverges.
Can you finish from there ?