# Mellin inverse of $\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)\zeta(n+1+s)}{\zeta(s)n!}\left(-\omega\right)^{n}$

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 ?

• It might be the case that $\phi(s,\omega)$ grows very fast on vertical lines, but we have the $\Gamma(s)^{2}$ factor, which decays extremely fast in the half-plane $\Re(s)<0$ in any vertical strip. won't that cancel the growth of $\phi(s,\omega)$ ? – Mohammad Al Jamal Feb 18 at 8:11
• @MohammadAlJamal Do you follow my answer ? Can you finish it or not ? If it was easier to use the residue theorem I would use it, but it is not because $\phi$ grows too fast on vertical lines and because there is a much simpler solution. – reuns Feb 19 at 5:03
• i do follow your answer, and i can finish. the answer will be an infinite sum in terms of the Mobius function and the log of the Euler function - related to the Dedekind eta - but i am looking for an answer in terms of the non-trivial zeros of the zeta function. – Mohammad Al Jamal Feb 19 at 6:58
• @MohammadAlJamal What do you get for $h_\omega$ the inverse Mellin transform of $H_\omega(s)=\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)\zeta(n+1+s)}{\zeta(s)n!}\left(-\omega\right)^{n}$ then what do you want to know about $h$ and $H$. – reuns Feb 19 at 7:03
• If $U$ is entire and $\frac{U(s) \Gamma(s)}{\zeta(s)}$ decreases fast enough as $Im(s) \to \infty$ and $\Re(s) \to \infty$ and the zeros of $\zeta(s)$ are all simple then what do you get for $\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty} \frac{U(s)\Gamma(s)}{\zeta(s)} x^sds$ using the residue theorem. – reuns Feb 19 at 7:13