Prove that $\sum_{n\geq 1} \frac{1}{x+n^2}=\frac{\pi\sqrt{x}\operatorname{coth}(\pi\sqrt{x})-1}{2x}$ for all $x>0$

Question

Wolfram Alpha gives the following equality, which seems to be valid for all $x>0$ :

$$\sum_{n\geq 1} \frac{1}{x+n^2}=\frac{\pi\sqrt{x}\operatorname{coth}(\pi\sqrt{x})-1}{2x}$$

My question is : how to prove it ?

Note that taking $x\to 0$ yields the famous $\frac{\pi^2}{6}$ on both sides, which is comforting.

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EDIT : This proof is inspired from Atmos's answer on Fourier coefficients. Let us denote $S=\sum_{n\geq 1} \frac{1}{x+n^2}$.

Define $f_\alpha:x\in]-\pi,\pi]\mapsto e^{\alpha x}$. Its complex Fourier coefficients are $$c_n(f_\alpha)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f_\alpha(t) e^{-int}\:dt\\ =\frac{(-1)^n\sinh(\alpha \pi)}{\pi}\frac{1}{\alpha-in} $$ Recall the Parseval equality $$\sum_{n=-\infty}^{\infty} |c_n(f_\alpha)|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi} |f_\alpha(x)|^2\:dx \label{eqn1}\tag{1}$$

The left-hand-side of \eqref{eqn1} is related to $S$ by $$\sum_{n=-\infty}^{\infty} |c_n(f_\alpha)|^2=\frac{\sinh^2(\alpha x)}{\pi^2}\sum_{n=-\infty}^{\infty} \frac{1}{\alpha^2+n^2}\\ =\frac{\sinh^2(\alpha x)}{\pi^2}\left(2S+\frac{1}{\alpha^2}\right)$$

The right-hand side of \eqref{eqn1} is given by $$\frac{1}{2\pi}\int_{-\pi}^{\pi} |f_\alpha(x)|^2\:dx=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{2\alpha x}\:dx=\frac{1}{2\alpha\pi}\sinh(2\alpha\pi)=\frac{1}{\alpha\pi}\cosh(\alpha\pi)\sinh(\alpha\pi)$$

Putting the pieces together gives the desired result (replace $\alpha=\sqrt{x}$)

$$S=\frac{\pi\alpha \coth(\alpha\pi)-1}{2\alpha^2}$$

Answer 1

I write it as an exercise to you.

Let $\alpha \in \mathbb{R}^{*}$ and $f : \mathbb{R} \rightarrow \mathbb{R}$ the $2\pi$-periodic function given by $$ f\left(x\right)=\text{cosh}\left(\alpha x \right) \text{ for }x \in \left]-\pi,\pi\right] $$

• Find the Fourier's coefficient $a_n$ and $b_n$ of $f$.

• Deduce from the previous question both sum $$ \sum_{n=1}^{+\infty}\frac{\left(-1\right)^n}{n^2+\alpha^2} \text{ as well as }\sum_{n=1}^{+\infty}\frac{1}{n^2+\alpha^2} $$

• Take $\alpha=\sqrt{x}$ for $x \in \mathbb{R}^{*+}$ and conclude.

EDIT : $f$ is even hence $b_n=0$ and we have with integration by part $$ \forall n \in \mathbb{N}^{*}, \ a_n=\left(-1\right)^n\frac{2 \alpha\text{sh}\left(\pi \alpha\right)}{\pi\left(n^2+\alpha^2\right)} $$ and $$ \frac{a_0}{2}=\frac{1}{2 \pi}\int_{-\pi}^{\pi}\text{cosh}\left(\alpha x\right)\text{d}t=\frac{1}{\alpha \pi}\text{sh}\left(\alpha \pi\right) $$

Then we have the equality for $x \in \left]-\pi, \pi\right[$ $$ \cosh\left(\alpha x \right)=\frac{\text{sh}\left(\alpha \pi\right)}{\alpha \pi}+\sum_{n=1}^{+\infty}\left(-1\right)^n\frac{2 \alpha\text{sh}\left(\pi \alpha\right)}{\pi\left(n^2+\alpha^2\right)} $$

For $x= \pi $ $$\sum_{n=1}^{+\infty}\frac{1}{n^2+\alpha^2}= \frac{\pi \alpha \text{coth}\left(\alpha \pi\right)-1}{2 \alpha^2} $$

For $x=0$ $$\sum_{n=1}^{+\infty}\frac{\left(-1\right)^n}{n^2+\alpha^2}= \frac{1}{2 \alpha^2}\left(\frac{\alpha \pi}{\text{sh}\left(\alpha \pi\right)}-1\right) $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\sum_{n \geq 1}{1 \over x + n^{2}}\,\right\vert_{\ x\ >\ 0} & = \sum_{n = 0}^{\infty} \pars{{1 \over n + 1 - \root{x}\ic} - {1 \over n + 1 + \root{x}\ic}} {1 \over 2\root{x}\ic} \\[5mm] & = {H_{\root{x}\ic} - H_{-\root{x}\ic}\over 2\root{x}\ic} \qquad\pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = {H_{\root{x}\ic} - \bracks{\rule{0pt}{5mm}H_{-1 - \root{x}\ic} + 1/\pars{-\root{x}\ic}} \over 2\root{x}\ic} \qquad\pars{~H_{z}-Recurrence~} \\[5mm] & = {H_{\root{x}\ic} - H_{-1 - \root{x}\ic} \over 2\root{x}\ic} - {1 \over 2x} \\[5mm] & = {\pi\cot\pars{\pi\bracks{-\root{x}\ic}} \over 2\root{x}\ic} - {1 \over 2x} \qquad\pars{~Euler\ Reflection\ Formula~} \\[5mm] & = {\pi\ic\coth\pars{\pi\root{x}} \over 2\root{x}\ic} - {1 \over 2x} = \bbx{\pi\root{x}\coth\pars{\pi\root{x}} - 1\over 2x} \end{align}