If the interval is (-1 , 1)

We have $f(x) = \sum\limits_{-\infty}^{\infty}C_{n}e^{in\pi x/l}$

Where $C_{n} = \frac{1}{2l}\int\limits_{-l}^{l}f(x)e^{-in\pi x/l}dx$

Given $f(x) = e^{ax}$

$\therefore C_{n} \frac{1}{2l}\int\limits_{-l}^{l}e^{ax}e^{-in\pi x/l}dx$

$\therefore C_{n} = \frac{1}{2l}\int\limits_{-l}^{l}e^{(a - in\pi /l)x}dx$

$\therefore C_{n} = \frac{1}{2l}\bigg[\frac{e^{(a - (in\pi/l))l} - e^{-(a-(in\pi / l))l}}{(a - (in\pi /l))}\bigg]$

$\therefore C_{n} = \frac{1}{2l}\bigg[\frac{e^{al}*e^{-(in\pi)} - e^{-al}*e^{in\pi}}{(al - in\pi)/l}\bigg]$

$\therefore C_{n} = \frac{1}{2}\bigg[\frac{e^{al}*e^{-(in\pi)} - e^{-al}*e^{in\pi}}{al - in\pi}\bigg]$

$e^{\pm in\pi} = cosn\pi \pm isinn\pi = (-1)^n$

$\therefore C_{n} = \frac{1}{2}\bigg[\frac{e^{al}*(-1)^n - e^{-al}(-1)^n}{al - in\pi}\bigg]$

$\therefore C_{n} = \frac{(-1)^n}{al - in\pi}\bigg[\frac{e^{al} - e^{-al}}{2}\bigg]$

$\therefore C_{n} = \frac{(-1)^n sinhal}{al - in\pi} * \frac{al + in\pi}{al + in\pi}$

Now $f(x) = \sum\limits_{-\infty}^{\infty}C_{n}e^{in\pi x/l}$

$f(x) = \sum\limits_{-\infty}^{\infty}\frac{(-1)^{n} sinhal * al + in\pi}{(al)^2 + (n\pi)^2}*e^{in\pi x/l}$