# How to evaluate this complex integral !?

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, 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.

• Is there something wrong with the answer you got on MathStackexchange? After all, if the integral doesn't exist, then it makes little sense to produce formulas for it. – Andreas Blass Jan 22 '13 at 22:11
• i'm not convinced by the argument there !! – mohammad-83 Jan 23 '13 at 16:25
• we can rewrite the integral as : $$\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin\left(s\ln x-\frac{\pi}{2}\coth\left(\frac{\pi}{s} \right ) \right )}{s}ds$$ Now we split the integral at $s=1$ : $$\int_{0}^{1}+\int_{1}^{\infty}$$ We make the substitution $s\rightarrow s^{-1}$in the leftmost integral, to obtain : $$\int_{1}^{\infty}\frac{\sin\left(s\ln x-\frac{\pi}{2}\coth\left(\frac{\pi}{s} \right ) \right )}{s}ds+\int_{1}^{\infty}\frac{\sin\left(s^{-1}\ln x-\frac{\pi}{2}\coth\left(\pi s \right ) \right )}{s}ds$$ Now this integral exists. But i don't know how to do it !! – mohammad-83 Jan 24 '13 at 23:22

$\int_{1}^{+\infty}\sin\left(\frac{\ln{x}}{s}-\frac{\pi}{2}\coth{(\pi{s})}\right)\frac{ds}{s}$
does not converge. Indeed for big values of $s$ the integrand is equivalent to $\frac{-1}{s}$, since $\coth(\pi{s})$ tends to $1$.