$\displaystyle Let\ c=\ 0\ ,\ c+2l=a\ ;\ \\[2ex] \displaystyle \ \ \therefore{}\ l=\frac{a}{2}. \\[2ex] \displaystyle C=\frac{1}{2l}\int_c^{c+2l}f(x)e^{\frac{-in\pi{}x}{l}}dx \\[2ex] \displaystyle C=\frac{1}{a}\int_0^ae^{ax}e^{\frac{-in\pi{}x}{\frac{a}{2}}}dx \\[2ex] \displaystyle C=\frac{1}{a}\ \int_0^ae^{ax-\frac{2\pi{}inx}{a}}dx=\ \frac{1}{a}\ \int_0^ae^{x(a-\frac{2\pi{}in}{a})}\ dx \\[2ex] \displaystyle C=\frac{1}{a}\frac{{\left[e^{x(a-\frac{2\pi{}in}{a})}\right]}_0^a\ }{(a-\frac{2\pi{}in}{a})}\ =\frac{1}{a}\left(\frac{e^{a\left(a-\frac{2\pi{}in}{a}\right)}-1}{a-\frac{2\pi{}in}{a}}\ \right)\\[2ex] C=\frac{1}{a^2-2\pi{}in}(e^{a^2}.\ e^{-2\pi{}in\ }-1) \\[5ex] \displaystyle We\ know,\ e^{\pm{}2in\pi{}}=1 \\[2ex] \displaystyle C=\frac{1}{a^2-2\pi{}in}(e^{a^2}-1) \\[5ex] \displaystyle By\ Complex\ Fourier\ Series, \\[2ex] \displaystyle F(x)=\sum_{-\infty{}}^{\infty{}}Cn\ e^{\frac{in\pi{}x}{l}} \\[2ex] \displaystyle e^{ax}=\ 1(e^{a^2}-1)\ \sum_{n=\ -\infty{}}^{\infty{}}\frac{1\ }{(a^2-2\pi{}in)e^{inx}} \\[2ex] $