Calculate the sum $\sum_{n=1}^\infty {(-1)^n\over 1+n^2}$ The question:

Calculate the sum $$I:=\sum_{n=1}^\infty {(-1)^n\over 1+n^2}$$

My attempt:


*

*Notation: In a previous question I have calculated $$\sum_{n=1}^\infty{1\over n^2+1}={1\over 2}\left(\pi{e^\pi+e^{-\pi}\over e^\pi-e^{-\pi}}-1\right)$$ and if possible, I would like to use it.


On the one hand:
$$\sum_{-\infty}^\infty {(-1)^n\over 1+n^2}=1+2I$$
On the other hand:
$$\sum_{-\infty}^\infty {(-1)^n\over 1+n^2}=-Res((-1)^z\cdot{\pi \cot(\pi z)\over 1+z^2},i)-Res((-1)^z\cdot{\pi\cot(\pi z)\over 1+z^2},-i)
\\ Res((-1)^z\cdot{\pi \cot(z\pi)\over 1+z^2},i)={(-1)^i\over 2i}\cdot\cot(\pi i)
\\ Res((-1)^z\cdot{\pi \cot(z\pi)\over 1+z^2},-i)={(-1)^{-i}\over -2i}\cdot\cot(-\pi i)
$$
And in general:
$$
\sum_{-\infty}^\infty={\pi\over 2i}((-1)^i\cot(-\pi i)-(-1)^i\cot(\pi i))
$$
But I don't know how to keep evaluate it.
 A: If you know how to compute $\sum_{n\geq 1}\frac{1}{n^2+a^2}$ (for instance, through the Poisson summation formula) then you may just exploit
$$ \sum_{n\geq 1}\frac{(-1)^n}{n^2+1}=-\sum_{n\geq 1}\frac{1}{n^2+1}+2\sum_{m\geq 1}\frac{1}{(2m)^2+1}=\frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2+\left(\frac{1}{2}\right)^2}-\sum_{n\geq 1}\frac{1}{n^2+(1)^2}.$$
Since $\int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-ax}\,dx = \frac{1}{n^2+a^2}$ we also have
$$ \sum_{n\geq 1}\frac{1}{n^2+a^2} = \int_{0}^{+\infty}W(x) e^{-ax}\,dx = \frac{1}{1-e^{-2\pi a}}\int_{0}^{2\pi}\frac{\pi -x}{2}e^{-ax}\,dx$$
where $W(x)$ is $2\pi$-periodic and piecewise-linear sawtooth wave, which equals $\frac{\pi-x}{2}$ over $(0,2\pi)$.
This immediately leads to
$$ \sum_{n\geq 1}\frac{(-1)^n}{n^2+1} = \frac{\pi}{e^{\pi}-e^{-\pi}}-\frac{1}{2}.$$
A: $$\sum_{n=1}^\infty {(-1)^n\over 1+n^2}=\sum_{n=1}^\infty {1\over 1+(2n)^2}-\sum_{n=1}^\infty {1\over 1+(2n-1)^2}$$
we know that 
$$\frac{\pi x\coth(\pi x)-1}{2x^2}=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}=\frac{1}{x^2}\sum_{n=1}^{\infty}\frac{1}{1+(\frac{n}{x})^2}$$
$$\frac{\pi x\coth(\pi x)-1}{2}=\sum_{n=1}^{\infty}\frac{1}{1+(\frac{n}{x})^2}$$
let$x=\frac{1}{2}$
$$\frac{\frac{\pi}{2}\coth(\frac{\pi}{2})-1}{2}=\sum_{n=1}^{\infty}\frac{1}{1+(2n)^2}\tag1$$
and we have
$$\frac{\pi \tanh(\pi x/2)}{4x}=\sum_{n=1}^{\infty}\frac{1}{x^2+(2n-1)^2}$$
let $x=1$
$$\frac{\pi \tanh(\pi /2)}{4}=\sum_{n=1}^{\infty}\frac{1}{1+(2n-1)^2}\tag2$$
so
$$\sum_{n=1}^\infty {(-1)^n\over 1+n^2}=\frac{\frac{\pi}{2}\coth(\frac{\pi}{2})-1}{2}-\frac{\pi \tanh(\pi /2)}{4}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} =
\Im\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n - \ic} =
\Im\sum_{n = 0}^{\infty}
\pars{{1 \over 2n + 2 - \ic} - {1 \over 2n + 1 - \ic}}
\\[5mm] = &\
{1 \over 2}\,\Im\sum_{n = 0}^{\infty}
\pars{{1 \over n + 1 - \ic/2} - {1 \over n + 1/2 - \ic/2}} =
{1 \over 2}\,\Im\bracks{\Psi\pars{{1 \over 2} - {\ic \over 2}} -
\Psi\pars{1 - {\ic \over 2}}}
\end{align}
where $\ds{\Psi}$ is the Digamma Function. See
$\ds{\mathbf{\color{black}{6.3.16}}}$ in A & S Table

Then,
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} =
{1 \over 2}\braces{{\bracks{\Psi\pars{1/2 - \ic/2} -
\Psi\pars{1/2 + \ic/2}} \over 2\ic} -
{\bracks{\Psi\pars{1 - \ic/2} -
\Psi\pars{1 + \ic/2}} \over 2\ic}}
\end{align}
With $\ds{\Psi}$-Recurrence and Euler Reflection Formula:
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} =
-\,{1 \over 4}\,\ic\braces{%
\pi\cot\pars{\pi\bracks{{1 \over 2} + {\ic \over 2}}} -
\pi\cot\pars{\pi\,{\ic \over 2}} + {1 \over \ic/2}}
\\[5mm] = &
\bbx{\pi\,\mrm{csch}\pars{\pi} - 1 \over 2} \approx -0.3640
\end{align}
