Question : obtain the complex form of fourier series for f(x) = eax  in ( l,-l)

$ Complex \ form \ of\ f(x) = \displaystyle\sum_{-\infty}^{\infty} C_n e^{in\pi x/l} \\Where\ C_n= \dfrac{1}{2l} \int_{-l}^{l} e^{-ax}.e^{-in\pi x/l}.dx \\ = \dfrac{1}{2l} \int_{-l}^{l} e^{-ax}.e^{-in\pi x/l}.dx\\=\dfrac{1}{2l} \int_{-l}^{l} e^{(a-in\pi/l)x}.dx\\ \\ =\dfrac{1}{2l} [\dfrac{e^{a-in\pi x/l}} {a-in\pi / l}]_{-l}^{l}.dx \\= \dfrac{1}{2l}( \dfrac{e^{(a-in\pi l)/l}-e^{-(a-in\pi l)/l}}{(a-in\pi/l)} ) $

$\dfrac{1}{2} [ \dfrac{e^{al}e^{-in\pi}-e^{-al}e^{in\pi}}{(al-in\pi)}] \\ e^{\pm in\pi }=cos(\pm in\pi)+isin(\pm in\pi) = (-1)^n$

$\therefore C_n = \dfrac{e^{al}(-1)^n-e^{-al}(-1)^n}{ 2(al-in\pi)}\\ =\dfrac{(-1)^n(e^{al}-e^{-al})}{ 2(al-in\pi)}\\ $

$C_n =\dfrac{ (-1)^n \ sinh al}{al-in\pi}.\dfrac{al+in\pi}{al+in\pi} by\ rationalising\ the\ numerator \\ =\dfrac{ (-1)^n \ sinh al}{a^2l^2+n^2\pi^2} .(al+in\pi)\\ \therefore Complex\ form \ of \ e^{ax} =\displaystyle\sum_{-\infty}^\infty \dfrac{ (-1)^n \ sinh al. (al+in\pi) }{a^2l^2+n^2\pi^2} .e^{in\pi/l} $