It would be great if someone can help me do these integrals - using numerical integration on Mathematica it seems that these converge - in what follows $a \in \mathbb{R}$ and $q \in \mathbb{N}$ and $n \in \mathbb{Z}$

  • $\int _0 ^\infty dx\text{ } tanh (\pi \sqrt{x} )[ \frac{1}{x + a^2 + (\frac{n}{q})^2 } - \frac{1}{x + a^2 + n^2 } ] $

  • $\int_0 ^\infty dx \text{ }\frac{tanh(\pi \sqrt{x})}{\sqrt{x + a^2 } } [ coth (\pi q \sqrt{x + a^2 } ) - coth (\pi \sqrt{x + a^2 } ) ] $

I am wondering if there is some complex analysis trick that can help here...

  • $\begingroup$ Have you tried to use $$\sum_{k\in\mathbb Z}\frac{y^2}{y^2+(k+\frac12)^2}=\pi y\tanh(\pi y)$$? $\endgroup$ – მამუკა ჯიბლაძე Mar 10 '14 at 8:59
  • $\begingroup$ For $a=n=1$ and $q=2$, the value of the first integral is $0.27092733012243642067462062834169051560736001350324$, which is not recognized by Maple's identify() command or by the inverse symbolic calculator site (oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html). That makes me dubious that there is any useful closed form. $\endgroup$ – Neil Strickland Mar 10 '14 at 11:58

Here is my attempt at the first one: \begin{multline*} \int\limits_0^\infty\tanh(\pi\sqrt x)\left(\frac1{x+c^2}-\frac1{x+d^2}\right)dx\\=2\int\limits_0^\infty y\tanh(\pi y)\left(\frac1{y^2+c^2}-\frac1{y^2+d^2}\right)dy\\=\frac2\pi\int\limits_0^\infty\sum_{k\in\mathbb Z}\frac{y^2}{y^2+(k+\frac12)^2}\left(\frac1{y^2+c^2}-\frac1{y^2+d^2}\right)dy\\=\frac2\pi\int\limits_0^\infty\sum_{k\in\mathbb Z}\frac{(d^2-c^2)y^2}{(y^2+(k+\frac12)^2)(y^2+c^2)(y^2+d^2)}dy\\=\frac{2(d^2-c^2)}\pi\sum_{k\in\mathbb Z}\int\limits_0^\infty\frac{y^2}{(y^2+(k+\frac12)^2)(y^2+c^2)(y^2+d^2)}dy\\=(|d|-|c|)\sum_{k\in\mathbb Z}\frac1{(|k+\frac12|+|c|)(|k+\frac12|+|d|)}\\=\sum_{k\in\mathbb Z}\left(\frac1{|k+\frac12|+|c|}-\frac1{|k+\frac12|+|d|}\right)\\=2(\psi(|d|+\frac12)-\psi(|c|+\frac12)) \end{multline*} I believe the second integral can be treated similarly by expanding $\coth$ into elementary fractions too.

Just don't ask me whether all this was legitimate :)

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  • $\begingroup$ (1) Thanks! This is out-of-the-blue! :) Can you kindly explain/define the $\psi$ function that you used and the integration that you did in the step just before introducing it? $\endgroup$ – user6818 Mar 11 '14 at 8:57
  • $\begingroup$ (2) Also I believe that the summation identity you have used is kind of the derivative of two other identities that I have been using to get these integrals in the first place! - $\sum_{n \in \mathbb{Z}} log (\frac{n^2}{q^2} + a^2 ) = 2log [ 2 sinh (\pi q \vert a\vert ) ] $ and $\sum_{n \in \mathbb{Z}+ \frac{1}{2}} log (\frac{n^2}{q^2} + a^2 ) = 2log [ 2 cosh (\pi q \vert a\vert ) ] $ - but these identities are zeta-function regularized identities - so I would be surprised if mathematicians also use them as legitimate ones! $\endgroup$ – user6818 Mar 11 '14 at 9:02
  • $\begingroup$ So you are suggesting that I revert these coth functions back to these summations to do these integrals? $\endgroup$ – user6818 Mar 11 '14 at 9:02
  • 1
    $\begingroup$ $\psi$ is the logarithmic derivative of the $\Gamma$; in Mathematica, $\psi(z)$ is PolyGamma[z] - you can see it e.~g. at the Wolfram functions site, where it is defined through$$\psi(z)=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z-1}\right).$$ $\endgroup$ – მამუკა ჯიბლაძე Mar 11 '14 at 9:06
  • $\begingroup$ As for the integration, I've just decomposed $\frac{y^2}{(y^2+b^2)(y^2+c^2)(y^2+d^2)}$ into elementary fractions and then integrated separately each summand using $$ \int\limits_0^\infty\frac{dy}{y^2+e^2}=\frac\pi{2|e|}. $$ $\endgroup$ – მამუკა ჯიბლაძე Mar 11 '14 at 9:14

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