Function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion 
Find the function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion

$$
f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))
$$
where $$\frac{a_0}{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)dx,\\{a_n}=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(n x)dx,\\{b_n}=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)dx$$
$$
\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}=\sum_{n=1}^\infty \frac{\cos(n\pi x)}{n^2+\alpha^2}
$$
How do I proceed further ?
 A: You may use the Poisson summation formula:
If $f\in L_1(\mathbb{R})$ and $\widehat{f}\in L_1(\mathbb{R})$, then $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n)$ converges uniformly,  $f\in\mathcal{C}(\mathbb{S}^1)$, and
$$Pf(x)=\sum_{n\in\mathbb{Z}} \widehat{f}(n) e^{2\pi kx}$$
where $\widehat{f}$ is the Fourier transform of $f$.
Using the fact that $\frac12 e^{-|2\pi t|}=\frac{1}{2\pi}\int^\infty_{-\infty} \frac{e^{-i2\pi tx}}{1+x^2}\,dx$, (this requires a little bit of Fourier analysis on $\mathbb{R}$, or some knowledge on characteristic functions of certain probability distributions), we have that
$$\sum_{n\in\mathbb{Z}} e^{-2\pi a |n|}e^{-2\pi kx}=Pf(x)=\frac{1}{\pi}\sum_{n\in\mathbb{Z}}\frac{a}{a^2+|x+n|^2}$$
At $x=0$ one gets
\begin{align}
\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}&=\sum^\infty_{n=1}\frac{1}{a^2+n^2} +\frac{1}{2a^2}=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac{1}{2a^2} +\frac14\sum^{\infty}_{n=1}\frac{1}{\big(\tfrac{a}{2}\big)^2+ n^2}\\
&=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2} +\frac{1}{2a^2} + \frac{1}{4}\Big(\frac{\pi}{a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{2}{a^2} \Big)\\
&=\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac{\pi}{4a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}
\end{align}
Hence
$$
\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2} =\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} -\frac{\pi}{4a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}
$$
The series in the OP can be then expressed as
$$\begin{align}
\sum^\infty_{n=1}\frac{(-1)^n}{a^2+n^2}&=-\sum^\infty_{n=1}\frac{1}{a^2+(2n-1)^2}+\frac14\sum^{\infty}_{n=1}\frac{1}{\big(\tfrac{a}{2}\big)^2+ n^2}\\
&=-\frac{\pi}{2a}\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} +\frac{\pi}{2a}\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{1}{2a^2} \\
&=\frac{\pi}{2a}\Big(\frac{1+e^{-\pi a}}{1-e^{-\pi a}}-\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}\Big)-\frac{1}{2a^2} \\
&=\frac{\pi}{2a}\operatorname{csch}(\pi a)-\frac{1}{2a^2}
\end{align}
$$
A: As far as I can understand, the task is to find such a function that its Fourier series coincides with the given one at some point.
$$
\frac{a_0}{2}+\sum_{n=1}^\infty \frac{(-1)^n\cos(n x)}{n^2+\alpha^2} +b_n\sin(nx)
$$
Then at $x=0$, we will get the series subtracting $a_0$ if it is not zero.
Thus we need to find such an $f$ that
$$
\frac{1}{\pi} \int_{-\pi}^\pi f(x)\cos(nx) dx = \frac{1}{n^2+\alpha^2}.
$$
This is guessing work. Let's keep it simple and try the first thing which comes to a head. Consider $f(x)=\cos(\alpha x)$. This gives
$$
\frac{1}{\pi} \int_{-\pi}^\pi f(x)\cos(nx) dx =\frac{2 \alpha  (-1)^n \sin (\pi  \alpha )}{\pi  \alpha ^2-\pi  n^2}
$$
This is not exactly what we wanted, but let's change $\cos(\alpha x) \rightarrow \cos(i \alpha x)+\cos(-i \alpha x)  $, the result is:
$$
\frac{4 \alpha  (-1)^n \sinh (\pi  \alpha )}{\pi  \left(\alpha ^2+n^2\right)}.
$$
This is better. Thus the function we are interested in is
$$
f=(\cos(i\alpha x)+\cos(-i\alpha x))\frac{\pi}{4\alpha \sinh(\pi\alpha)}=\frac{1}{2\alpha^2}+\sum_{n=1}^\infty \frac{(-1)^n\cos(n x)}{n^2+\alpha^2} +b_n\sin(nx).
$$
It gives the answer for $x=0$
$$
 \frac{\pi}{2\alpha \sinh(\pi\alpha)}-\frac{1}{2\alpha^2}=\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}
$$
