$\sum_{n=-\infty}^\infty\left(\frac1{n^2+a^2}\right)\left(\frac1{n^2+b^2}\right)$

Question

I have to solve this sum, but I can't seem to finish it. $$\sum_{n=-\infty}^\infty\left(\frac1{n^2+a^2}\right)\left(\frac1{n^2+b^2}\right)$$

Here's my ideas: I've been suggested to use the Poisson summation formula as follows: $\sum_{n=-\infty}^\infty f(n)=\sum_{m=-\infty}^\infty\hat{f}(2\pi m),$ where $\hat{f}$ is the Fourier transform of $f$. But I am not entirely sure how to proceed afterwards.

Answer 1

We first show that, for $a \in \mathbb{R}$,

$$ S(a) := \sum_{n=-\infty}^{\infty} \frac{1}{n^2 + a^2} = \frac{\pi \coth(\pi a)}{a}. $$

Since both sides of the equality are even functions of $a$, it suffices to prove this for $a \geq 0$ and we do so.

Starting off, we realize the sum $S(a)$ in the form $\sum_{n=-\infty}^{\infty} f(n)$. Obviously, we can choose $f(x) = \frac{1}{x^2+a^2}$. Then, in order to invoke Poisson summation formula, we need to compute

\begin{align*} \hat{f}(2\pi m) &= \int_{-\infty}^{\infty} \frac{e^{-2\pi i m x}}{x^2+a^2} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\cos(2\pi m x)}{x^2+a^2} \, \mathrm{d}x, \end{align*}

where the last step holds becaue the imaginary term $\frac{i\sin(2\pi m x)}{x^2+a^2}$ is an odd integrable function and hence its integral over $\mathbb{R}$ vanishes by symmetry. Then, applying the substitution $x = at$ and invoking the formula $\int_{-\infty}^{\infty}\frac{\cos(at)}{1+t^2}\,\mathrm{d}t=\pi e^{-|a|}$, we get

\begin{align*} \hat{f}(2\pi m) &= \frac{1}{a} \int_{-\infty}^{\infty} \frac{\cos(2\pi m a t)}{t^2 + 1} \, \mathrm{d}t = \frac{\pi}{a} e^{-2\pi |m|a}. \end{align*}

Plugging this to Poisson summation formula, we get

\begin{align*} S(a) &= \sum_{m=-\infty}^{\infty} \hat{f}(2\pi m) = \sum_{m=-\infty}^{\infty} \frac{\pi}{a}e^{-2\pi|m|a} = \frac{\pi}{a} \left( 1 + 2 \sum_{m=1}^{\infty} e^{-2\pi ma} \right) \\ &= \frac{\pi}{a} \left( 1 + \frac{2e^{-2\pi a}}{1 - e^{-2\pi a}} \right) = \frac{\pi \coth (\pi a)}{a}, \end{align*}

proving the claim. Then, finally, the original sum can be written as

\begin{align*} \sum_{n=-\infty}^{\infty} \frac{1}{(n^2 + a^2)(n^2 + b^2)} &= \sum_{n=-\infty}^{\infty} \frac{1}{b^2 - a^2} \left[ \frac{1}{n^2 + a^2} - \frac{1}{n^2 + b^2} \right] \\ &= \frac{1}{b^2 - a^2}[S(a) - S(b)] \\ &= \frac{1}{b^2 - a^2}\left( \frac{\pi \coth(\pi a)}{a} - \frac{\pi \coth(\pi b)}{b}\right). \end{align*}

Answer 2

For the fun of extending the problem to partial sums.

$$S_p=\sum_{n=-p}^{p} \frac{1}{(n^2 + a^2)(n^2 + b^2)} = \frac{1}{b^2 - a^2}\sum_{n=-p}^{p} \left( \frac{1}{n^2 + a^2}- \frac{1}{n^2 + b^2}\right)$$ $$\frac{1}{n^2 + c^2}=\frac{1}{(n+i c)(n-ic)}=\frac i{2c} \left(\frac{1}{n+i c}-\frac{1}{n-i c}\right)$$ Using the digamma function $$\sum_{n=-p}^{p}\frac{1}{n+i c}=-\psi ^{(0)}(p+1-ic)+\psi ^{(0)}( p+1+i c)+\frac{i}{c}+\psi ^{(0)}(-i c)-\psi ^{(0)}(i c)$$ Converting to harmonic numbers $$\sum_{n=-p}^{p}\frac{1}{n+i c}=-H_{p-i c}+H_{p+i c}-i \pi \coth (\pi c)$$ $$\sum_{n=-p}^{p}\frac{1}{n-i c}=H_{p-i c}-H_{p+i c}+i \pi \coth (\pi c)$$ $$\sum_{n=-p}^{p}\frac{1}{n^2 + c^2}=\frac{-i H_{p-i c}+i H_{p+i c}+\pi \coth (\pi c)}{c}$$ Using asymptotics and Taylor series for large values of $p$. $$S_p=\frac \pi {b^2-a^2}\left(\frac{\coth (\pi a)}{a}-\frac{\coth (\pi b)}{b}\right)-\frac{2}{3 p^3}+\frac{1}{p^4}+O\left(\frac{1}{p^5}\right)$$

Answer 3

There is a way to find the value of the sum that does not use the Poisson summation formula.

Note that the function $$\frac{2\pi i}{e^{2\pi i z}-1}$$ has simple poles at integer points with residues 1. Then the complex integral $$\oint_{\partial \Gamma}f(z) \frac{2\pi i}{e^{2\pi i z}-1}dz$$ where $f(z)$ is a meromorphic function with no poles at integer numbers equals $$2\pi i \left(\sum_{n \in \Gamma \cap \mathbb{Z}}f(n) + \sum_{\text{poles of }f(z) \ z_{0}\in \Gamma} \text{Res}_{z = z_0}f(z) \frac{2 \pi i }{e^{2\pi i z} - 1} \right)$$.

Suppose $z=x+iy$. Consider a sequence of square contours $\partial \Gamma_{n}$ outlined by the lines $x=\pm n+\frac{1}{2}$ and $y= \pm n \pm \frac{1}{2}$. Note that the length of the $n$-th contour is linear in $n$. Note also that the absolute value of $\frac{2\pi i}{e^{2\pi i z}-1}$ tends to $2\pi$ as $y \rightarrow \infty$ and to $0$ as $y \rightarrow -\infty$. At the same time, the absolute value of $\frac{2\pi i}{e^{2\pi i z}-1}$ is bounded from above on the vertical lines passing through semi-integer real points, and $\left| \frac{1}{z^2+a^2} \right|$ decays as $|z|^{-2}$ at infinity, let alone $\left| \frac{1}{(z^2 + a^2)(z^2 + b^2)} \right|$ which decays as $|z|^{-4}$. Hence, the value of the contour integral for $n$ sufficiently large for the contour to encompass both $\pm ia$ and $\pm ib$ is $$\oint_{\partial \Gamma_{n}} \frac{1}{(z^2+a^2)(z^2+b^2)}\frac{2\pi i}{e^{2\pi i z}-1}dz=2\pi i \left(\sum_{-n}^{n} \frac{1}{(n^2+a^2)(n^2+b^2)} + \text{Res}_{z=\pm ia, \pm ib} \frac{1}{(z-ia)(z+ia)(z-ib)(z+ib)} \frac{2\pi i}{e^{2\pi i z}-1} \right) \rightarrow 0$$ as $n \rightarrow \infty$. Hence, in the limit we have $$\sum_{n=-\infty}^{\infty} \frac{1}{(n^2+a^2)(n^2+b^2)}=-\sum_{z=\pm ia, \pm ib}\text{Res}\frac{1}{(z-ia)(z+ia)(z-ib)(z+ib)}\frac{2\pi i}{e^{2\pi i z}-1}$$

The four residues on the right hand side are at the poles of degree 1 and have the form $g(z)/(z-z_0)$, and their values are thus $g(z_0)$ where $g(z)$ is the product $\frac{1}{(z-ia)(z+ia)(z-ib)(z+ib)} \frac{2 \pi i }{e^{2\pi i z} - 1}$ without one multiplier $\frac{1}{z-z_0}$, and $z_0 = \pm ia, \pm ib$.

Answer 4

Use the result $$\sum_{n=-\infty}^\infty\frac1{n^2+a^2}=\frac{\pi}a\coth(\pi a)$$To prove this, apply the summation theorem to get that $$\sum_{n=-\infty}^\infty\frac1{n^2+a^2}=\mathrm{Res}_{z=ia}\frac{\pi\cot(\pi z)}{z^2+a^2}+\mathrm{Res}_{z=-ia}\frac{\pi\cot(\pi z)}{z^2+a^2}$$The poles $\pm ia$ are first order, so $$\mathrm{Res}_{z=ia}\frac{\pi\cot(\pi z)}{z^2+a^2}=\lim_{z\rightarrow ia}\frac{\pi\cot(\pi z)(z-ia)}{z^2+a^2}=\frac{\pi\cot(\pi ai)}{2ai}$$The second residue is equivalent to the first one, and so the very first sum is equivalent to $$\sum_{n=-\infty}^\infty\frac1{n^2+a^2}=\frac{\pi\cot(\pi ai)}{ai}=\frac{\pi\coth(\pi a)}a$$ By Partial fraction decomposition of the original summand in the OP, we get $$\sum_{n=-\infty}^\infty\frac1{n^2+a^2}\frac1{n^2+b^2}=\frac1{b^2-a^2}\left(\frac{\pi\coth(\pi a)}a-\frac{\pi\coth(\pi b)}b\right)$$Partial fraction decomposition is always a great tool to solving sums. It's not only integrals that use this trick.

Answer 5

In this answer is a residue proof and in this answer is a real proof of $$ \sum_{k=-\infty}^\infty\frac1{k+z}=\pi\cot(\pi z)\tag1 $$ which gives $$ \begin{align} \sum_{n=-\infty}^\infty\frac1{n^2+a^2} &=\frac1{2ia}\sum_{n=-\infty}^\infty\left(\frac1{n-ia}-\frac1{n+ia}\right)\tag{2a}\\ &=\frac1{2ia}\left(\pi\cot(-\pi ia)-\pi\cot(\pi ia)\right)\tag{2b}\\[6pt] &=\frac\pi{a}\coth(\pi a)\tag{2b} \end{align} $$ Explanation:
$\text{(2a):}$ partial fractions
$\text{(2b):}$ apply $(1)$
$\text{(2c):}$ $\cot(iz)=-i\coth(z)$

Therefore, $$ \begin{align} \sum_{n=-\infty}^\infty\frac1{n^2+a^2}\frac1{n^2+b^2} &=\frac1{b^2-a^2}\sum_{n=-\infty}^\infty\left(\frac1{n^2+a^2}-\frac1{n^2+b^2}\right)\tag{3a}\\ &=\frac1{b^2-a^2}\left(\frac\pi{a}\coth(\pi a)-\frac\pi{b}\coth(\pi b)\right)\tag{3b} \end{align} $$ Explanation:
$\text{(3a):}$ partial fractions
$\text{(3b):}$ apply $(2)$