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Let $a>0$ and $x\in(-1,1)$. Can we give a closed form expression for $\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}$? Note that the series is convergent. There should be an expression in terms of the cotangent.

My goal is to find a (sharp) upper bound $c(x)$ of the series.

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  • $\begingroup$ Have you tried to approximate it with the integral $\int_{- \infty}^{\infty} \frac{1}{a+(t+x)^2} dt$? This is equal to a finite value, i.e. $\frac{\pi }{\sqrt a}$ $\endgroup$
    – Crostul
    Nov 25, 2019 at 8:14
  • $\begingroup$ the series extends to a periodic function, is this perhaps related to a known fourier series of some function...? $\endgroup$ Nov 25, 2019 at 8:32
  • $\begingroup$ @Crostul Have would this approximation look like? $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 8:35
  • $\begingroup$ @CalvinKhor The series is, up to a constant factor, the density of the wrapped normal distribution $\sum_{k\in\mathbb Z}\mathcal N_{x-k,\:\sigma^2}$ ($a=2\sigma^2$). $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 8:36

2 Answers 2

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Assuming that you enjoy special functions $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{-\psi \left(-x-\sqrt{-a}+1\right)+\psi \left(-x+\sqrt{-a}+1\right)-\psi \left(x-\sqrt{-a}\right)+\psi \left(x+\sqrt{-a}\right)}{2 \sqrt{-a}}$$ wich can simplify as $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=-\frac{\pi \left(\cot \left(\pi \left(\sqrt{-a}-x\right)\right)+\cot \left(\pi \left(\sqrt{-a}+x\right)\right)\right)}{2 \sqrt{-a}}$$ and, since $a >0$ $$\color{blue}{\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{\pi \left(\coth \left(\pi \left(\sqrt{a}-i x\right)\right)+\coth \left(\pi \left(\sqrt{a}+i x\right)\right)\right)}{2 \sqrt{a}}}$$ what you can simplify using $$\coth(A+B)+\coth(A-B)=\frac{2 \sin (2 A)}{\cos (2 B)-\cos (2 A)}$$

Edit

In order to keep the results in the answer, I reproduce here what you wrote in comments.

So, we have finally $$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac \pi {\sqrt a}\,\,\frac{\sinh \left(2 \pi \sqrt{a}\right)}{\cosh \left(2 \pi \sqrt{a}\right)-\cos(2 \pi x)}$$ but I think that you cannot make at the same time $a \to 0$ and $x\to \pm 1$ without avoiding $\infty$. In the post, remember that you did precise $x\in(-1,1)$ and not $x\in[-1,1]$.

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  • $\begingroup$ What is $\psi$? $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 9:02
  • $\begingroup$ @0xbadf00d. The digamma function. $\endgroup$ Nov 25, 2019 at 9:05
  • $\begingroup$ Thanks for clarifying. Can we give a nice upper bound $c(x)$? $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 10:50
  • $\begingroup$ @0xbadf00d. First, please, simplify the expression and add the result to the post. We will see later. $\endgroup$ Nov 25, 2019 at 10:52
  • $\begingroup$ Note that there is a typo in your second displayed equation. The $\pi$ should precede the fraction. (And the same in the third displayed equation.) $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 13:38
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Observe \begin{align} \sum^\infty_{k=-\infty}\frac{1}{a+(x+k)^2} =&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &+\sum^\infty_{k=1} \frac{1}{a+(k+(1+x))^2}+\sum^{\infty}_{k=1} \frac{1}{a+(k+(1-x))^2}\\ \leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &\ +\int^\infty_0\frac{dk}{a+(k+(1+x))^2}+ \int^\infty_{0} \frac{dk}{a+(k+(1-x))^2}\\ \leq&\ \frac{1}{a+(1+x)^2}+\frac{1}{a+x^2}+\frac{1}{a+(1-x)^2}\\ &\ +\frac{1}{\sqrt{a}}\left(\tan^{-1}\left( \frac{\sqrt{a}}{1+x}\right)+\tan^{-1}\left( \frac{\sqrt{a}}{1-x}\right)\right) \end{align}

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  • $\begingroup$ How does the first equality follow? Clearly, the series is equal to $\frac1{a+x^2}+\sum_{k\in\mathbb Z\setminus\{0\}}\frac1{a+(k+x)^2}$. $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 8:40
  • $\begingroup$ By playing around with the indices. $\endgroup$ Nov 25, 2019 at 8:41
  • $\begingroup$ Yes, trivial. I see. $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 8:44
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    $\begingroup$ Also, you have wolframalpha.com/input/… $\endgroup$ Nov 25, 2019 at 8:44
  • $\begingroup$ Okay, everything is clear to me, but how do we obtain the last inequality? $\endgroup$
    – 0xbadf00d
    Nov 25, 2019 at 10:11

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