Residue Theorem for Gamma Function times Hurwitz Zeta function

Question

I am trying to evaluate an integral which is a product of three functions:

$$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} ds \,\left(\tilde{\beta}\sqrt{2r}\right)^{-s}\, \Gamma(s)\,\zeta_{H}\left(\dfrac{s}{2},\dfrac{1}{2r}\right) $$ We know that $\Gamma$ has simple poles at the non-positive integers $-n$. The residue is well known and it is given by, $$ 2\pi i \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} $$ we notice that the condition for the convergence of the series $s/2 > 1 $ implies $ c > 2$. It can be analytically continued to the whole complex plane with one singularity, a simple pole with residue 1 at $s = 1$. I end up to something like this $$ -\frac{1}{2}-\frac{1}{2r} + \frac{1}{2r\tilde{\beta}^2}\sum_{n=0}^{\infty} \frac{(-\sqrt{2r})^n}{n!} \zeta_{H}\left(\dfrac{-n}{2},\dfrac{1}{2r}\right) $$ I am not sure I got the correct result. Can someone help and check my work. Thanks

Answer 1

Mathematica gives the following result where the first term is the residue as $s=2$:

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$$\frac{1}{\beta ^2 r}+\sum\limits_{n=0}^\infty \frac{ \left(\beta \left(-\sqrt{2 r}\right)\right)^n}{n!}\zeta_H\left(-\frac{n}{2},\frac{1}{2 r}\right)$$

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Here's a table of evaluations which is the basis for the formula above.

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$$\begin{array}{cc} n & \text{Residue[($\beta $}\sqrt{2r})^{-s}\text{Gamma[s]HurwitzZeta[}\frac{s}{2},\frac{1}{2r}\text{],$\{$s,n$\}$]} \\ -10 & \frac{\beta ^{10} r^5 \zeta \left(-5,\frac{1}{2 r}\right)}{113400} \\ -9 & -\frac{\beta ^9 r^{9/2} \zeta \left(-\frac{9}{2},\frac{1}{2 r}\right)}{11340 \sqrt{2}} \\ -8 & \frac{\beta ^8 r^4 \zeta \left(-4,\frac{1}{2 r}\right)}{2520} \\ -7 & -\frac{\beta ^7 r^{7/2} \zeta \left(-\frac{7}{2},\frac{1}{2 r}\right)}{315 \sqrt{2}} \\ -6 & \frac{1}{90} \beta ^6 r^3 \zeta \left(-3,\frac{1}{2 r}\right) \\ -5 & -\frac{\beta ^5 r^{5/2} \zeta \left(-\frac{5}{2},\frac{1}{2 r}\right)}{15 \sqrt{2}} \\ -4 & \frac{1}{6} \beta ^4 r^2 \zeta \left(-2,\frac{1}{2 r}\right) \\ -3 & -\frac{1}{3} \sqrt{2} \beta ^3 r^{3/2} \zeta \left(-\frac{3}{2},\frac{1}{2 r}\right) \\ -2 & \beta ^2 r \zeta \left(-1,\frac{1}{2 r}\right) \\ -1 & -\sqrt{2} \beta \sqrt{r} \zeta \left(-\frac{1}{2},\frac{1}{2 r}\right) \\ 0 & \zeta \left(0,\frac{1}{2 r}\right) \\ 1 & 0 \\ 2 & \frac{1}{\beta ^2 r} \\ 3 & 0 \\ 4 & 0 \\ 5 & 0 \\ 6 & 0 \\ 7 & 0 \\ 8 & 0 \\ 9 & 0 \\ 10 & 0 \\ \end{array}$$

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Note the formula and table above both use $\beta$ instead of $\tilde{\beta}$ which would have required extra work to edit the results above.

Answer 2

You can use (or prove with the residue theorem or the Fourier inversion theorem) that $$\frac1{2i\pi} \int_{c-i\infty}^{c+i\infty} \Gamma(s)x^{-s}ds = e^{-x}$$ to get that for $c > 2,x>0,a>0$ $$\frac1{2i\pi} \int_{c-i\infty}^{c+i\infty} \Gamma(s)x^{-s}\sum_{n\ge 1} (n+a)^{-s/2}ds = \sum_{n\ge 1} e^{-x \sqrt{n+a}}$$ This is a complicated function, but it extends analytically and most asymptotics should be tractable.