Find the odd Fourier series for the periodic function whose period is $2\pi$, and which is equal to $\cosh{x}$ in the range $0<x<\pi$. Hence show that
$\frac{\pi}{4}sech{(\frac{\pi}{2}})=\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)+(2n+1)^{-1}}$.
$a_n=\frac{1}{2\pi}\int^{\pi}_0(e^x+e^{-x})\cos{nx}$ $dx$
$=\frac{sinh (\pi)}{\pi(1+n^2)}(-1)^n$
$a_0=\frac{1}{\pi} sinh(\pi)$
$b_n=\frac{1}{2\pi}\int^{\pi}_0cosh(x). \sin{nx}$ $dx$
$b_n=\frac{\cosh(\pi)}{\pi}[\frac{2n}{1+n^2}[1-(-1)^n]]$
$1-(-1)^n =$
\begin{cases}
0, & \text{if $n$ is even} \\[2ex]
2, & \text{if $n$ is odd}
\end{cases}
$\therefore b_n=\frac{4(2k+1) cosh(\pi)}{\pi[1+(2k+1)^2]}$
For Odd series,
$cosh(x)=\frac{4}{\pi}cosh(\pi) \sum^{\infty}_{n=1}\frac{(2k+1)}{1+(2k+1)^2}\sin{nx}$
Putting $x=\frac{\pi}{2}$
$\frac{\pi}{4}[\frac{cosh(\frac{\pi}{2})}{\cosh(\pi)}=\sum^{\infty}_{n=1}\frac{1}{(2k+1)+(2k+1)^{-1}}$
I am a bit far from what is required.
P.S I need to know if all my Fourier constants are correct, because I have got other similar problems to solve, this time with $a_n$ and $a_0$
Best Answer
$$\int_{0}^{\pi}\cosh(x)\sin(nx)\,dx = \frac{n}{n^2+1}\left(1-(-1)^n \cosh\pi\right)\tag{1}$$ leads to $$ \cosh(x) = \frac{2}{\pi}\sum_{n\geq 1}\frac{n\sin(nx)}{n^2+1}(1-(-1)^n\cosh \pi) \tag{2} $$ for any $x\in(0,\pi)$. By evaluating both sides of $(2)$ at $x=\frac{\pi}{2}$, $$ \cosh\frac{\pi}{2}=\frac{2}{\pi}\sum_{m\geq 0}\frac{(2m+1)(-1)^m}{(2m+1)^2+1}(1+\cosh \pi)\tag{3}$$ follows, and by rearranging both sides of $(3)$, we get $$\sum_{m\geq 0}\frac{(-1)^m}{(2m+1)+(2m+1)^{-1}} = \frac{\pi\cosh\frac{\pi}{2}}{2(1+\cosh\pi)}=\color{red}{\frac{\pi}{4\cosh\frac{\pi}{2}}}\tag{4} $$ as wanted.