Find Fourier series of $\cosh(ax)$

Question

Prove that in the range $-\pi < x < \pi$, $$\cosh(ax) = \frac{2a^2 \sinh(a \pi)}{\pi} \left(\frac1{2a^2} + \sum_{n=1}^\infty(-1)^n \frac{1}{n^2 + a^2}\cos (nx) \right) $$

Now, I have tried to get the Fourier series of $\cosh(ax)$.
I got $$ a_0 = 2\frac{\sinh(a \pi)}{\pi a}\\ a_n = - \frac{2a\sinh(a \pi)}{\pi(n^2 - a^2)} $$ Probably I got $$ f(x) = \frac{\sinh(a \pi)}{\pi} \left( \frac1a- 2a \sum_{n=1}^\infty {(-1)^n \frac1{n^2 - a^2} \cos nx}\right) $$ I tried twice and again I got this answer which is not matching with the question. Please tell me what is my fault and how to solve this?

Answer 1

Note that we have

$$\begin{align} \int_{-\pi}^\pi \cosh(ax)\cos(nx)\,dx&=2\text{Re}\left(\int_0^\pi \cosh(ax)\,e^{inx} \,dx\right)\\\\ &= \text{Re}\left(\int_0^\pi \left(e^{(a+in)x}+e^{-(a-in)x}\right)\,dx\right)\\\\ &=\text{Re}\left((-1)^n\left(\frac{e^{a\pi}}{a+in}-\frac{e^{-a\pi}}{a-in}\right)\right)\\\\ &=(-1)^n\,\frac{2a\sinh(a\pi)}{a^2+n^2}\\\\ \end{align}$$

Can you finish now?

Answer 2

This is the solution that I've got. The answer matches the required solution, except that it's 2a and not 2a^2.