Prove the identity:
$$\pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$
Solution
We are invkoking the Fourier series of $\cosh ax, \; x \in [-\pi, \pi] , \; a\neq 0$. Indeed it can be expanded into Fourier Series. We expect the $b_n$ coefficients to be zero, since $\cosh ax$ is even. Indeed a straightforward calculation shows that the coefficients are zero. We are now computing $a_0, \; a_n$.
$\displaystyle {\color{gray} \blacksquare} \;\; a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \, {\rm d}x= \frac{2\sinh \pi a}{\pi a}$
$\displaystyle {\color{gray} \blacksquare} \;\; a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \cos n x\, {\rm d}x= \frac{2[a \sinh \pi a \cos \pi n+ \cancelto{0}{n \cosh \pi a \sin \pi n}]}{\pi (a^2+n^2)}$
Therefore:
$$\cosh ax = \frac{\sinh \pi a}{\pi a}+ \frac{2 a\sinh \pi a}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^n \cos nx}{n^2 +a^2} \overset{x=\pi}{\implies} \\
\overset{x=\pi}{\implies} \pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$
and the identity is proved.
$$\pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$
Solution
We are invkoking the Fourier series of $\cosh ax, \; x \in [-\pi, \pi] , \; a\neq 0$. Indeed it can be expanded into Fourier Series. We expect the $b_n$ coefficients to be zero, since $\cosh ax$ is even. Indeed a straightforward calculation shows that the coefficients are zero. We are now computing $a_0, \; a_n$.
$\displaystyle {\color{gray} \blacksquare} \;\; a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \, {\rm d}x= \frac{2\sinh \pi a}{\pi a}$
$\displaystyle {\color{gray} \blacksquare} \;\; a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \cos n x\, {\rm d}x= \frac{2[a \sinh \pi a \cos \pi n+ \cancelto{0}{n \cosh \pi a \sin \pi n}]}{\pi (a^2+n^2)}$
Therefore:
$$\cosh ax = \frac{\sinh \pi a}{\pi a}+ \frac{2 a\sinh \pi a}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^n \cos nx}{n^2 +a^2} \overset{x=\pi}{\implies} \\
\overset{x=\pi}{\implies} \pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$
and the identity is proved.
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