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What is the closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$? We can use Fourier series of $e^{-bx}$ ($|x|<\pi$) to evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{n^2+b^2}$. But this one, it seems to me it is tough to get the closed form.

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xpaul
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4 Answers4

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In my opinion, this is a rather nice example of application of the Poisson summation formula: $$ \sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\hat{f}(k),\qquad \hat{f}(\nu)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i \nu x}dx.$$ Namely, setting $f(x)=\frac{1}{(x-a)^2+b^2}$, we find \begin{align} \hat{f}(\nu)=\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{(x-a)^2+b^2}=e^{-2\pi i \nu a}\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{x^2+b^2}= \frac{\pi}{b}e^{-2\pi i \nu a-2\pi |\nu| b}, \end{align} where we have assumed that $b>0$ and calculated the last integral using residues. Therefore, the sum we are trying to calculate reduces to geometric series: $$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi}{b}\sum_{k\in\mathbb{Z}}e^{-2\pi i k a-2\pi |k| b}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}.$$

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19

This one can also be done with the standard technique of using the $\pi \cot(\pi z)$ multiplier, integrating $$f(z) = \frac{\pi \cot(\pi z)}{(z-a)^2+b^2}$$ along a circle using the same technique as here.

We integrate $f(z)$ along a circle of radius $R$ with $R$ going to infinity and the integral disappears in the limit so that the residues sum to zero (actually a square with vertices $$(\pm(N+1/2),\pm(N+1/2))$$ $N$ a positive integer is easier to handle computationally). The poles of $f(z)$ other than at the integers are at $$z_{0,1} = a\pm ib$$ and the residues are $$\operatorname{Res}(f(z); z=z_0) = \left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_0} = \frac{\pi\cot(\pi(a+bi))}{2bi} = \frac{\pi}{2bi} i \frac{e^{i\pi(a+bi)}+ e^{-i\pi(a+bi)}}{e^{i\pi(a+bi)}- e^{-i\pi(a+bi)}}$$ and $$\operatorname{Res}(f(z); z=z_1) = \left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_1} = \frac{\pi\cot(\pi(a-bi))}{-2bi} = \frac{\pi}{-2bi} i \frac{e^{i\pi(a-bi)}+ e^{-i\pi(a-bi)}}{e^{i\pi(a-bi)}- e^{-i\pi(a-bi)}}.$$ Now put $x=e^{i\pi a}$ and $y=e^{-\pi b}.$ The first residue becomes $$\operatorname{Res}(f(z); z=z_0) = \frac{\pi}{2b}\frac{xy+1/x/y}{xy-1/x/y}$$ and the second residue is $$\operatorname{Res}(f(z); z=z_1) = -\frac{\pi}{2b}\frac{x/y+y/x}{x/y-y/x}.$$

Adding the two contributions and simplifying with Euler's formula yields $$-\frac{\pi x^2 (y^4-1)}{b (x^2y^2-1)(x^2-y^2)} = -\frac{\pi (y^2-1/y^2)}{b (x^2-1/y^2)(1-y^2/x^2)} \\= -\frac{\pi (y^2-1/y^2)}{b (x^2+1/x^2-y^2-1/y^2)} = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)}.$$

Now with $S$ being our sum we have by the Cauchy Residue Theorem that $$ S + \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)} = 0$$ so that finally $$ S = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}.$$

This MSE link contains a similar computation, this one including the computation of the relevant bounds.

Marko Riedel
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{n\ =\ -\infty}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} \\[5mm] = &\ \sum_{n\ =\ -\infty}^{-1}{1 \over \pars{n - a}^{2} + b^{2}} +{1 \over a^{2} + b^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}} \\[5mm]&=\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n + a}^{2} + b^{2}} +{1 \over a^{2} + b^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}} \\[5mm]&=-\,{1 \over a^{2} + b^{2}} +\bracks{\color{#c00000}{% \sum_{n\ =\ 0}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} + \pars{a \to -a}} \end{align}

\begin{align}&\color{#c00000}{% \sum_{n\ =\ 0}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} =\sum_{n\ =\ 0}^{\infty} {1 \over \bracks{n - \pars{a + b\ic}}\bracks{n - \pars{a - b\ic}}} \\[5mm]&={\Psi\pars{a + b\ic} - \Psi\pars{a - b\ic}\over \pars{a + b\ic} - \pars{a - b\ic}} ={2\ic\,\Im\Psi\pars{a + b\ic} \over 2b\ic}={\Im\Psi\pars{a + b\ic} \over b} \end{align}

\begin{align}&\color{#66f}{\large% \sum_{n\ =\ -\infty}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} \\[5mm] = &\ \color{#66f}{\large -\,{1 \over a^{2} + b^{2}} + {\Im\Psi\pars{a + b\ic} + \Im\Psi\pars{-a + b\ic} \over b}} \end{align}

$\ds{\Psi\pars{z}}$ is the Digamma Function .

Felix Marin
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4

Hints:

  • Differentiate the natural logarithm of Euler's infinite product formula for the sine function. The logarithm of a product is the sum of logarithms, and the derivative of a sum is the sum
    of derivatives.

  • Then, returning to your series, factor the denominator, using complex numbers if needed, and expand the summand with the help of partial fractions. Then break up the sum into two series, both of which will be expressible in closed forms using the previous result.

  • Employ Euler's formula, linking trigonometric functions of complex argument to hyperbolic ones, in order to furbish the final result.


Euler : $\quad\displaystyle\sin\pi x~=~\pi x~\prod_{\large n\in\mathbb Z}^{\large n\neq0}\bigg(1-\frac xn\bigg)\quad=>\quad\ln\sin\pi x~=~\ln\pi x~+~\sum_{\large n\in\mathbb Z}^{\large n\neq0}\ln\bigg(1-\frac xn\bigg)$.

Differentiating, we have $~\displaystyle\pi\cot\pi x~=~\sum_{\large n\in\mathbb Z}~\frac1{x-n}.~$ At the same time, $~(n-a)^2+b^2~$ can be

factored into $\Big[(n-a)-{\bf i}b\Big]\Big[(n-a)+{\bf i}b\Big]=\Big[n-(a+{\bf i}b)\Big]\Big[n-(a-{\bf i}b)\Big]=(n-z)(n-\bar z)$,

allowing us after partial fraction decomposition to write the sum as $~\displaystyle\frac{\bf i}{2b}~\sum_{\large n\in\mathbb Z}\bigg[\frac1{n-\bar z}-\frac1{n-z}\bigg]$

which then becomes $\dfrac{{\bf i}\pi}{2b}\Big(\cot\pi z-\cot\pi\bar z\Big).~$ Now all that's left to do is employing the formula

for $\cot\Big(A\pm B\Big)~=~\dfrac{\cot A\cot B\mp1}{\cot B\pm\cot A},~$ where $A=a\pi$ and $B={\bf i}b\pi,~$ and $\cot{\bf i}x=-{\bf i}\coth x$.

Lucian
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