We have the following complex integral :
$$\frac{1}{2\pi i}\int\_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds$$
Where $x\in\mathbb{R}:x>1$. i tried closing the contour to the left, and computing the residues at the essential singularities of the integrand $\left( s=-\frac{1}{n}\right)$ in the following manner :
Using the partial fraction expansion of $\cot(\pi z)$, we have:
$$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right )=\frac{s}{2}-\sum\_{k=1}^{\infty}\frac{1}{2k(ks-1)}+\frac{1}{2k(ks+1)}$$
And around the reciprocal of each negative integer $n$ we have the Taylor expansion:
$$e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{e^{\frac{1}{2n(ns+1)}}}{s}=\sum\_{m=0}^{\infty}c\_{n,m}\left(s+\frac{1}{n}\right)^{m}$$
Furthermore, using the definition of the Bessel function of the first kind, we have:
$$x^{s}e^{-\frac{1}{2n(ns+1)}}=x^{-\frac{1}{n}}\exp\left[\frac{\sqrt{\ln x}}{n\sqrt{2}}\left(\sqrt{\ln x^{2}}(ns+1)-\frac{1}{\sqrt{\ln x^{2}}(ns+1)}\right)\right]$$
$$=x^{-\frac{1}{n}}\sum\_{i=-\infty}^{\infty}J\_{i}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)\left(\sqrt{\ln x^{2}}(ns+1)\right)^{i}$$
From which we obtain:
$$\underset{s=-n^{-1}}{\text{Res}}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}=x^{-\frac{1}{n}}\sum\_{k=0}^{\infty}c\_{n,k}\frac{J\_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$
And i claim that:
$$\frac{1}{2\pi i}\int\_{-i\infty}^{i\infty}e^{-\frac{\pi}{2}\cot\left(\frac{\pi}{s}\right)}\frac{x^{s}}{s}ds=\sum\_{n=1}^{\infty}\sum\_{k=0}^{\infty}c\_{n,k}x^{-\frac{1}{n}}\frac{J\_{k+1}\left(\frac{\sqrt{\ln x^{2}}}{n}\right)}{n^{k+1}(\ln x^{2})^{(k+1)/2}}$$
However, when i posted this question on [MS](https://math.stackexchange.com/questions/283013/how-to-evaluate-this-complex-integral), someone pointed out that this scheme is meaningless, since the original integral doesn't exist !! i find myself not content with the answer!! hence the thread. 



How to evaluate this complex integral !?
Mohammad, I do believe your integral is divergent. Taking for granted the computations that led you to your last comment, I can assure you that the integral


$\int\_{1}^{+\infty}\sin\left(\frac{\ln{x}}{s}-\frac{\pi}{2}\coth{(\pi{s})}\right)\frac{ds}{s}$


does not converge. Indeed for big values of $s$ the integrand is equivalent to $\frac{-1}{s}$, since $\coth(\pi{s})$ tends to $1$.


I took the liberty to answer you (eventhough I strongly believe the place this question belongs to is the original thread at MS) so that you won't keep posting here about this. I'm not critiscizing the question, only its relevance to this research-level site.


