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?
Best Answer
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?