I'm asked to find the complex Fourier series of $cosh(t)$ , $L^2 [-\pi,\pi]$ space,using exponentials only,that equals $\frac{sinh (\pi)}{\pi}\sum_{n=-\infty}^{n=\infty}\frac{(-1)^n}{n^2+1} \cdot e^{int}$.
This is the first complex series that I do.
$FS[x(t)]= \sum_{n=-\infty}^{n=\infty} = x[n]e^{int}.$
The $x[n]$ coefficients are equal to $x[n] =\frac{1}{2\pi} \int_{-\pi}^{\pi} x(t)e^{-int} dt $
$cosh(t)=\frac{e^t+e^{-t}}{2} $
$x[n] =\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^t+e^{-t}}{2}e^{-int} dt =$
$\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^te^{-int}+e^{-t}e^{-int}}{2}dt$ =
$\frac{1}{4\pi} \int_{-\pi}^{\pi} e^{t(1-in)}+e^{-t(1+in)} dt =$
$\frac{1}{4\pi} \left \{ \right. _{-\pi}^{\pi} \frac{e^{t(1-in)}}{1-in}+\frac{e^{-t(1+in)}}{1+in}$ =
$\left ( \frac{1}{4\pi} \right )$ $\frac{e^{\pi(1-in)}}{1-in}+\frac{e^{-\pi(1+in)}}{1+in} -\frac{e^{-\pi(1-in)}}{1-in}-\frac{e^{\pi(1+in)}}{1+in}$
Here I wasn't sure how to go on. I tried with
$\left ( \frac{1}{4\pi} \right )$ $\frac{e^{\pi(1-in)}-e^{-\pi(1-in)}}{1-in}-\frac{e^{\pi(1+in)}-e^{-\pi(1+in)}}{1+in} =$
$\left ( \frac{1}{4\pi{(1+n^2)}} \right )$$\left [{(1+in)}\left ({e^{\pi(1-in)}-e^{-\pi(1-in)}} \right ) - {(1-in)}\left ({e^{\pi(1+in)}-e^{-\pi(1+in)}} \right ) \right ]=$
$\left ( \frac{1}{4\pi{(1+n^2)}} \right ) \cdot$ $\left [ e^{\pi(1-in)} + in \cdot e^{\pi(1-in)} - e^{-\pi(1-in)} - in \cdot e^{-\pi(1-in)} -e^{\pi(1+in)} +e^{-\pi(1+in)} + in \cdot e^{\pi(1+in)} -in \cdot e^{-\pi(1+in)} \right]$
I got stuck here . I don't know how to keep on or whether I should've simplify something earlier.
You are on the right track. Note that $e^{in\pi}=(-1)^n$ and therefore we have
$$\begin{align} \frac1{2\pi}\int_{-\pi}^\pi \left(\frac{e^t+e^{-t}}{2}\right)\,e^{-int}\,dt&=\frac1{4\pi}\left(\frac{e^{(1-in)\pi}-e^{-(1-in)\pi}}{1-in}+\frac{e^{(1+in)\pi}-e^{-(1+in)\pi}}{1+in}\right)\\\\ &=\frac{(-1)^n}{4\pi}\left(\frac{2\sinh(\pi)}{1-in}+\frac{2\sinh(\pi)}{1+in}\right)\\\\ &=\frac{(-1)^{n}\sinh(\pi)}{\pi(n^2+1)} \end{align}$$
as expected!