I wonder if any of you knows how to find the value of the series $$\sum_{n=1}^{\infty}\frac{e^{-bn}}{n^2+z^{2}}.$$
This function shows up while solving a magnetostatic problem with complex-valued scalar potential.
Greetings
Oscar
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- 1$\begingroup$ Oh, c'mon @IgorSikora, your you should show your approach is offensive to every real mathematician, from children to the most advanced ones. It's highly wrong that this attitude prevails in mathematical education, and extra frustratingly wrong in the case of Math Stack Exchange. $\endgroup$Wlod AA– Wlod AA2023-02-02 10:00:10 +00:00Commented Feb 2, 2023 at 10:00
- $\begingroup$ Ok, given the answer it seems I was wrong. Apologies, I am retrieving my previous comment. $\endgroup$Igor Sikora– Igor Sikora2023-02-02 10:08:33 +00:00Commented Feb 2, 2023 at 10:08
- 1$\begingroup$ @IgorSikora, it's not about the research level of the OP Question but about the wide-spread patronizing/condescending treatment of the students of mathematics -- after all, every mathematician is a student. Mathematics is beautiful but it is surrounded by ugliness (hm, poetic justice :) ). $\endgroup$Wlod AA– Wlod AA2023-02-02 10:15:12 +00:00Commented Feb 2, 2023 at 10:15
- $\begingroup$ Please clarify by what you mean by "limiting value". If you mean a "limit", please indicate which variable tends to where. $\endgroup$GH from MO– GH from MO2023-02-02 13:46:06 +00:00Commented Feb 2, 2023 at 13:46
- 1$\begingroup$ @GHfromMO --- the OP wrote "convergence value", I changed that into "limiting value", but I presume that simply "value" is more accurate (and I have changed it accordingly) $\endgroup$Carlo Beenakker– Carlo Beenakker2023-02-02 13:56:11 +00:00Commented Feb 2, 2023 at 13:56
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1 Answer
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0 The series has no expression in terms of elementary functions, but it does represent a special function (either the incomplete beta function $B$ or the Lerch transcendent $\Phi$): $$F(b,z)=\sum_{n=1}^\infty \frac{e^{-bn}}{n^2+z^2}=-\frac{1}{z}\,{\rm Im}\,\sum_{n=1}^\infty \frac{e^{-bn}}{n+iz}$$ $$\qquad=-\frac{1}{z}{\rm Im}\,e^{-b} \Phi (e^{-b},1,i z+1)=-\frac{1}{z}{\rm Im}\,e^{ibz}B_{e^{-b}}(i z+1,0).$$ (I'm assuming real $z$ and $b\geq 0$.) Two limits are: $$F(b,0)=\text{Li}_2\left(e^{-b}\right),$$ a polylog, while $$F(0,z)=\frac{\pi z \coth (\pi z)-1}{2 z^2}.$$