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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 singularities of the integrand $\left( s=-\frac{1}{n}\right)$ , but this proves to be very hard to do !!

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Should the integral be taken along a line away from the imaginary $s$-axis? – Ron Gordon Jan 20 '13 at 21:01
No, the integral is along the line $\Re(s)=0$ – Mohammad Al Jamal Jan 20 '13 at 21:06
The points $s = -1/n$ are not singularities of the integrand; rather, the exponential goes to zero at those points. Further, the problems with this integral all lie at $s=0$. – Ron Gordon Jan 20 '13 at 21:19
at the reciprocals of each -ive integer, there are two different limits of the exponential, one of which is $\infty$. hence, the claim ! – Mohammad Al Jamal Jan 20 '13 at 21:24
Aha. But, in that case, I believe they are essential singularities, not poles. – Ron Gordon Jan 20 '13 at 21:26
up vote 2 down vote accepted

Your integral, if it existed, would be equal to $$\frac{1}{2\pi i} \int_{-\infty}^{\infty} e^{-i (\pi/2) \!\coth(\pi/s)} \frac{x^{is}}{s}\,ds.$$


$$ \lim_{s \to 0^+} e^{-i (\pi/2) \!\coth(\pi/s)} = -i $$


$$ \lim_{s \to 0^-} e^{-i (\pi/2) \!\coth(\pi/s)} = i, $$

so that near $s=0$ the integrand looks like

$$ \frac{-i}{|s|}. $$

This is not integrable in the standard sense and, further, the integral does not admit a principal value.

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using$$\frac{\pi}{2}\cot\left(\frac{\pi}{s} \right)=\frac{s}{2}+\sum_{k=1}^{\infty}\frac{1}{2k(1-ks)}-\frac{1}{2k(1+ks)}$$we have around each -ive singularity of the integrand the expansion : $$\frac{e^{-\pi/2\cot(\pi/s)}e^{1/2n(ns+1)}}{s}=\sum_{k=0}^{\infty}c_{k}\left(s+‌​\frac{1}{n}\right)^{k}$$. Using the definition of the Bessel function of the first kind, we have $$x^{-1/n}e^{1/2n(ns+1)}x^{s+1/n}=\sum_{m=-\infty}^{\infty}I_{m}\left(\frac{\sqr‌​t{\ln x^{2}}}{n}\right)\left( \sqrt{\ln x^{2}}(ns+1)\right )^{m}$$ if i'm right about closing the contour to the left, the residues can be computed. – Mohammad Al Jamal Jan 20 '13 at 23:00
Tex is failing me. the expansion above is $$x^{s}e^{-1/2n(ns+1}=x^{-1/n}e^{-1/2n(ns+1)}x^{s+1/n}=x^{-1/n}\;\sum_{m=-\infty‌​}^{\infty}I_{m}\left(-i\frac{\sqrt{\ln x^{2}}}{n}\right)\left( i\sqrt{\ln x^{2}}(ns+1)\right )^{m}$$ – Mohammad Al Jamal Jan 20 '13 at 23:11
@Mohammad, if you are going to ignore the singularity at $s=0$, you must devise a way to do that, for example by adding a small "bump" in your contour. But if you close your contour on the left (say with a semicircle), then any small bump will exclude infinitely-many singularities, so that this new integral will result in a sum of only finitely-many residues. How do you choose which singularities you want? And, even if you could do this in some way, how do you relate this new integral over the closed contour avoiding the origin with the original integral over the whole imaginary axis? – Antonio Vargas Jan 20 '13 at 23:16
You can compute all the residues you want, but I don't see any meaningful way you can use them to assign this integral a value. The integral simply does not exist. – Antonio Vargas Jan 20 '13 at 23:20

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