Compute the sum of the series. 
I just see the equality in my textbook, but I really have no idea how it arises (maybe it is obvious to the author), and it seems Fourier methods are not applicable. I would appreciate if someone could show me some hints or background of it.
 A: Let $f(x)=\cosh{\alpha x}$, its Fourier series on $\left[-\pi,\pi\right]$
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx + b_n \sin nx$$
has $b_n =0$ for all $n$ since $f(x)$ is symmetric, so
$$a_n=\frac{2}{\pi}\int_0^\pi\cosh \alpha x \cos nx \;\mathrm{d}x$$
This integral can be computed via per partes :
$$\frac{\pi a_n}{2}=\frac{1}{\alpha}\sinh\alpha x\cos nx\bigg{|}_0^\pi+\frac{n}{\alpha^2}\cosh\alpha x\sin nx\bigg{|}_0^\pi-\frac{n^2}{\alpha^2}\int_0^\pi\cosh\alpha x\cos nx\;\mathrm{d}x=\\ \frac{1}{\alpha}\sinh\alpha\pi\cos\pi n-\frac{n^2}{\alpha^2}\frac{\pi a_n}{2}$$
So
$$a_n=\frac{2\alpha}{\pi}\frac{ \sinh\pi \alpha}{\alpha^2+n^2}\cos{\pi n}$$
hence
$$\cosh \alpha x= \frac{\sinh\pi \alpha}{\pi\alpha}+\frac{2\alpha\sinh\pi \alpha}{\pi}\sum_{n=1}^\infty\frac{\cos{\pi n}\cos n x}{\alpha^2+n^2}$$
For $x=\pi-y$ we get :
$$\cosh \alpha y\cosh \pi \alpha-\sinh \alpha y\sinh \pi \alpha= \frac{\sinh\pi \alpha}{\pi\alpha}+\frac{2\alpha\sinh\pi \alpha}{\pi}\sum_{n=1}^\infty\frac{\cos n y}{\alpha^2+n^2}$$
Rearanging :
$$\frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha \cos n x}{\alpha^2+n^2}= \cosh \alpha x\coth \pi\alpha-\sinh \alpha x-\frac{1}{\pi \alpha} \quad ;\text{for}\quad x\in\left[0,2\pi\right]\,\tag{1}$$
Denote
$$S(x)= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha\cos n x}{\alpha^2+n^2}$$
splitting into odd and even terms :
$$S(\alpha,x)= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha\cos (2n) x}{\alpha^2\!+\!(2n)^2}\!+\!\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n\!+\!1) x}{\alpha^2\!+\!(2n\!+\!1)^2}=\frac{1}{2}S\left(\frac{\alpha}{2},2x\right)\!+\!\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n\!+\!1) x}{\alpha^2\!+\!(2n\!+\!1)^2}$$
ergo, by $(1)$
$$\begin{align}W(\alpha,x) &=\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n+1) x}{\alpha^2+(2n+1)^2}=S(\alpha,x)-\frac{1}{2}S\left(\frac{\alpha}{2},2x\right) = \\
&=\cosh \alpha x \left(\coth \pi\alpha-\frac{1}{2}\coth \frac{\pi\alpha}{2}\right)-\frac{1}{2}\sinh \alpha x \quad ;\text{for}\quad x\in\left[0,\pi\right]\end{align}$$
this can be further simplified to :
$$W(\alpha,x)=\frac{1}{2}\frac{\sinh \alpha(\pi-2x)/2}{\cosh \alpha\pi/2}$$
Therefore

$$\sum_{n=0}^\infty\frac{\cos\left[(2n+1) x\right]}{\alpha^2+(2n+1)^2}= \frac{\pi}{4\alpha}\frac{\sinh \alpha(\pi-2x)/2}{\cosh \alpha\pi/2} \quad\quad\quad ;\text{for}\quad x\in\left[0,\pi\right] $$

Tracing back to the original series is done by taking $\alpha=\sqrt{24y}$ and $x=\frac{\pi}{6}$
Ergo :
$$\frac{48}{\sqrt3}\sum_{n=0}^\infty \frac{\cos\left[(2n+1)\pi/6\right]}{(2n+1)^2+24y}= \pi\sqrt{\frac{2}{y}}\frac{\sinh \pi\sqrt{8y/3}}{\cosh \pi\sqrt{6}}$$
