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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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    $\begingroup$ 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
    Feb 2, 2023 at 10:00
  • $\begingroup$ Ok, given the answer it seems I was wrong. Apologies, I am retrieving my previous comment. $\endgroup$ Feb 2, 2023 at 10:08
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    $\begingroup$ @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
    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
    Feb 2, 2023 at 13:46
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    $\begingroup$ @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) $\endgroup$ Feb 2, 2023 at 13:56

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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 $$F(0,z)=\frac{\pi z \coth (\pi z)-1}{2 z^2}.$$

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