Let $c=-\pi$ and $c+2l=-\pi$

$\therefore -\pi + 2l = \pi$

$\therefore 2l = 2 \pi$

$\therefore l=\pi$

Let f(x)=sin ax

$\displaystyle C_n = \dfrac {1}{2l} \int^{c+2l}_{c} f(x)e^{-in\pi x/l} dx$

$\displaystyle = \dfrac {1}{2\pi} \int^\pi_{-\pi} \sin ax\cdot e^{-in\pi x/\pi}dx$

$= \dfrac {1}{2\pi} \displaystyle \int^\pi_{-\pi} e^{-inx} \sin ax \ dx$

$= \dfrac {1}{2\pi} \left [ \dfrac {e^{-inx}}{(-in)^2 + a^2} (-in\sin ax - a\cos ax) \right ] ^{\pi}_{-\pi}$

$= \dfrac {1}{2\pi (i^2 n^2 + a^2)} \left \{ \left [ e^{-in\pi}(-in \sin a\pi - a\cos a\pi ) \right ] - \left [ e^{in\pi}(-in\sin (-a\pi)-a\cos (-a\pi)) \right ] \right \} $

Consider,

$e^{\pm in \pi} = \cos n \pi \pm i \sin n \pi = (-1)^n \pm i0 = (-1)^n$

$\therefore C_n = \dfrac {1}{2\pi (-n^2 + a^2)} \Big \{ (-1)^n [ - in \sin a\pi - a \cos a\pi] - (-1)^n [in \sin a\pi - a \cos ]\Big\} $

$= \dfrac {(-1)^n}{2\pi (n^2 + a^2)} \Big \{ -in \sin a\pi - a\cos a\pi - in \sin a\pi + a\cos a\pi \Big\}$

$= \dfrac {(-1)^n}{2\pi (a^2 - n^2)} (-2 in \sin a\pi)$

$= \dfrac {in (-1)^{n+1} \sin a\pi}{\pi (a^2 - n^2)}$

In Complex Fourier series,

$f(x) = \displaystyle \sum^\infty_{n=\infty} C_n e^{in\pi x / l} $

$\therefore \sin ax = \displaystyle \sum^{\infty}_{n=\infty} \dfrac {in (-1)^{n+1} \sin a\pi}{\pi (a^2 - n^2)} \cdot e^{in\pi x/\pi }$

$\therefore \sin ax = \dfrac {i\sin a\pi}{\pi} \displaystyle \sum^{\infty} _{n=\infty} \dfrac {(-1)^{n+1}}{a^2 - n^2} e^{inx}$