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.
$\begingroup$
$\endgroup$
3
-
6$\begingroup$ 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. $\endgroup$– Andreas BlassJan 22, 2013 at 22:11
-
$\begingroup$ i'm not convinced by the argument there !! $\endgroup$– mohammad-83Jan 23, 2013 at 16:25
-
$\begingroup$ 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 !! $\endgroup$– mohammad-83Jan 24, 2013 at 23:22
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
answered Jan 25, 2013 at 1:19