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

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 $\alpha=\sqrt{x}$)
$$S=\frac{\pi\alpha \coth(\alpha\pi)-1}{2\alpha^2}$$
 A: 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 $x=0$
$$\sum_{n=1}^{+\infty}\frac{\left(-1\right)^n}{n^2+\alpha^2}=
\frac{1}{2 \alpha^2}\left(\frac{\alpha \pi}{\text{sh}\left(\alpha \pi\right)}-1\right)
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
