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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.

0xbadf00d
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  • 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. $\frac{\pi }{\sqrt a}$ – Crostul Nov 25 '19 at 08:14
  • the series extends to a periodic function, is this perhaps related to a known fourier series of some function...? – Calvin Khor Nov 25 '19 at 08:32
  • @Crostul Have would this approximation look like? – 0xbadf00d Nov 25 '19 at 08:35
  • @CalvinKhor The series is, up to a constant factor, the density of the wrapped normal distribution $\sum_{k\in\mathbb Z}\mathcal N_{x-k,:\sigma^2}$ ($a=2\sigma^2$). – 0xbadf00d Nov 25 '19 at 08:36

2 Answers2

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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 $x\in[-1,1]$.

Claude Leibovici
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  • What is $\psi$? – 0xbadf00d Nov 25 '19 at 09:02
  • @0xbadf00d. The digamma function. – Claude Leibovici Nov 25 '19 at 09:05
  • Thanks for clarifying. Can we give a nice upper bound $c(x)$? – 0xbadf00d Nov 25 '19 at 10:50
  • @0xbadf00d. First, please, simplify the expression and add the result to the post. We will see later. – Claude Leibovici Nov 25 '19 at 10:52
  • 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.) – 0xbadf00d Nov 25 '19 at 13:38
  • I need to admit that I don't see how we can simplify the series using your identities. Do you've got $$\forall x,y\in\mathbb R:\coth(x+{\rm i}y)+\coth(x-{\rm i}y)=-\frac{2\sinh(2x)}{\cos(2y)-\cosh(2x)}\tag1$$ in mind? This yields $$\sum_{k\in\mathbb Z}\frac1{a+(k+x)^2}=-\frac\pi{\sqrt a}\frac{\sinh(2\pi\sqrt a)}{\cos(2\pi x)-\cosh(2\pi\sqrt a)}\tag3.$$ If that's what you've got in mind, how can we proceed for the bound? – 0xbadf00d Nov 25 '19 at 14:11
  • @0xbadf00d The expression is simple. Do we need a upper bound? What kind of bounds do we need? – River Li Nov 25 '19 at 14:16
  • @ClaudeLeibovici Okay, the best bound we can obtain should be $$\sum_{k\in\mathbb Z}\frac1{a+(k+x)^2}=\frac\pi{\sqrt a}\frac{\overbrace{\sinh(2\pi\sqrt a)}^{\ge:0}}{\underbrace{\cosh(2\pi\sqrt a)}{>:1}-\underbrace{\cos(2\pi x)}{\in[-1,:1]}}\le\frac\pi{\sqrt a}\frac{\sinh(2\pi\sqrt a)}{\cosh(2\pi\sqrt a)-1}\tag4.$$ Do you agree? That's unfortunate, since this bound tends to $\infty$ as $a\to0+$. – 0xbadf00d Nov 25 '19 at 19:30
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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}

Jacky Chong
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