Mysterious Inverse Mellin transform using residue theorem

The origin of this problem lies in the explanation of the evaluation of the series $\sum_{n\geq1}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{2\pi}{4}+\frac{\pi^2}{6}$

see this link ( Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ )

In the proposed solution a complex integral needs to be evaluated, which is a inverse mellin transform. This is done using the residue theorem.

Let $Q(s)=-\Gamma(s-2)\zeta(s)\cos(\frac{\pi s}{2})$.

The question is how to evaluate $\int_{\frac{5}{2}-i\infty}^{\frac{5}{2}+i\infty} Q(s)/x^s \, ds$

The author states that he integrates over the left plane, I suppose he uses a semi circle as a contour, which includes the 3 poles and if $R\rightarrow +\infty$ the integral over the arc vanishes and the part where $\operatorname{Re}(s)>\frac{5}{2}$ is covered. But how can I prove this? I tried to apply Jensens lemma which didn't work. What am I missing?

• This was an early post of mine where I neglected to prove that the integral along the left boundary of the rectangle vanishes. Consult this MSE link to see how to prove this rigorously. – Marko Riedel Nov 16 '15 at 19:54

Note that for any $\sigma$, the absolute convergence of the integral

$$\int_{\sigma-i\infty}^{\sigma+i\infty} Q(s)/x^s ds$$

is guaranteed by Stirling's formula and the bound for Riemann zeta function in vertical strips, provided that we stay away from the poles of $Q(s)$. This is because we know the growth properties of $\Gamma(s-2)$, $\cos(\pi s/2)$ and $\zeta(s)$ as $\text{Im}(s)\rightarrow \infty$.

In detail, let $s=\sigma+it$, then for $\sigma<0$, as $t\rightarrow \infty$,

$$\zeta(s)\ll (2\pi)^{-\sigma} |t|^{1/2-\sigma}$$ $$\Gamma(s-2)\sim \exp (-\frac{\pi}{2}|t|)|t|^{\sigma-2-1/2}$$ $$\cos(\pi s/2)\sim \exp (\frac{\pi}{2}|t|)$$

So their product is far less than$$(2\pi)^{-\sigma} |t|^{-2}$$

After this is justified, we can just construct a box to be our contour, with vertex $5/2-iT,5/2+iT,-N-iT,-N+iT$ for large $T$ and $N$. As long as $|x/2\pi|<1$, the integral goes to $0$ as $\sigma=-N\rightarrow -\infty$.

• But if $Im(s)\rightarrow \infty$, than $\cos(\frac{\pi s}{2})$ is unbounded and also exponentially increasing, right? – J. Spel Nov 18 '15 at 10:28
• @J.Spel Sorry i was sloppy there. As the comment of Marko suggests that the integrand is $|t|^{-2}$ as $t=Im(s)\rightarrow \infty$. So the convergence is no problem. The key is to know the convexity bound for $\zeta$. – AlgRev Nov 18 '15 at 16:50
• do you know an elementary proof for this $\zeta$ bound? – tired Jun 13 '18 at 21:52