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Identify $$ f(x)=\sum\limits_{k=-\infty}^{\infty} \frac{1}{(x+k)^2}.$$

Is this a known expansion of a simpler looking function? I understand that this series becomes infinite at integers.

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    $\begingroup$ This will end up being something like $\pi^2/\sin^2{\pi x} $. $\endgroup$ – Ron Gordon Sep 10 '16 at 19:07
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Note that (see here) for $x\not \in \mathbb{Z}$ $$\pi\cot(\pi x)=\sum_{k\in \mathbb{Z}}\frac{1}{x-k}=\sum_{k\in \mathbb{Z}}\frac{1}{x+k}.$$ Since it is analytic you can differentiate it term by term and we obtain, $$-\frac{\pi^2}{\sin^2{(\pi x)}}=\left(\pi\cot(\pi x)\right)'=\sum_{k\in \mathbb{Z}}\frac{-1}{(x+k)^2}.$$ Finally $$\sum_{k\in \mathbb{Z}}\frac{1}{(x+k)^2}=\frac{\pi^2}{\sin^2{(\pi x)}}.$$

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    $\begingroup$ The verb is 'differentiate' $\endgroup$ – punctured dusk Sep 10 '16 at 19:21
  • $\begingroup$ the argument for differentiating term by term is that it converges absolutely (for $x \in \mathbb{C} \setminus \mathbb{Z}$, and grouping $1/(x-k)$ with $1/(x+k)$) $\endgroup$ – reuns Sep 11 '16 at 13:25
  • $\begingroup$ Yes, it is important to group this. Conditional convergence indexed by $\mathbb Z$ needs such explanation. $\endgroup$ – GEdgar Sep 11 '16 at 19:38
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\left.\sum_{k = 0}^{\infty}{1 \over \pars{k + z}^{2}}\right\vert_{\ x\ \not\in\ \mathbb{Z}} = \Psi\,'\pars{z}}$ is a well known identity. $\ds{\Psi}$ is the Digamma Function.

\begin{align} \color{#f00}{\mrm{f}\pars{x}} & \equiv \sum_{k = -\infty}^{\infty}\,\,{1 \over \pars{x + k}^{2}} = {1 \over x^{2}} + \sum_{k = 0}^{\infty} \bracks{{1 \over \pars{k + x}^{2}} + {1 \over \pars{k - x}^{2}}} \\[5mm] & = \overbrace{{1 \over x^{2}} + \Psi\, '\pars{x}}^{\ds{\Psi\, '\pars{1 + x}}}\ +\ \Psi\, '\pars{-x}\qquad\pars{~Recurrence~} \\[5mm] & = -\pi\,\totald{\cot\pars{\pi x}}{x}\qquad \pars{~Euler\ Reflection\ Formula~} \\[5mm] & = \color{#f00}{\pi^{2}\csc^{2}\pars{\pi x}} \end{align}

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We can also use the well known summation formula $$\sum_{k\in\mathbb{Z}}f\left(k\right)=-\sum\left\{ \textrm{residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ so we have to evaluate $$\underset{z=-x}{\textrm{Res}}\frac{\pi\cot\left(\pi z\right)}{\left(z+x\right)^{2}}=\lim_{z\rightarrow-x}\frac{d}{dz}\left(\pi\cot\left(\pi z\right)\right)=-\pi^{2}\csc^{2}\left(\pi x\right)$$ hence $$\sum_{k\in\mathbb{Z}}\frac{1}{\left(k+x\right)^{2}}=\color{red}{\pi^{2}\csc^{2}\left(\pi x\right)}$$ obviously if $x$ is not an integer.

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