$\displaystyle f(x) = \sum_{-\infty{}}^{\infty{}}C_n\ e^{\dfrac{in\pi{}x}{l}} \[2ex] \displaystyle C_n=\frac{1}{2l}\int_{-l}^lf\left(x\right)e^{\dfrac{-in\pi{}x}{l}}\ dx\ \[2ex] \displaystyle=\frac{1}{2l}\int_{-l}^le^{ax}.e^{\dfrac{-in\pi{}x}{l}}\ dx\ \[2ex] \displaystyle =\frac{1}{2l}\int_{-l}^le^{(\dfrac{a-in\pi{}}l)x}dx \[2ex]$ $=\dfrac{1}{2l}{\left[\frac{e^{\left(\frac{a-in\pi{}}{l}\right)x}}{\left(\frac{a-in\pi{}}{l}\right)}\right]}_{-l}^l \[2ex] =\dfrac{1}{2l}\left[\dfrac{e^{\left(\frac{a-in\pi{}}{l}\right)l}\ -\ e^{-\left(\frac{a-in\pi{}}{l}\right)l}\ }{\left(\frac{a-in\pi{}}{l}\right)}\right] \[2ex] =\dfrac{1}{2}\left[\dfrac{e^{al}\ e^{-in\pi{}}-\ e^{-al}e^{in\pi{}}\ }{\left(al-in\pi{}\right)}\right] \ $ $But \ e^{\pm{}in\pi{}} = cos (\pm{}n\pi{}) + i\ sin (\pm{}n\pi{})\ =(-1)^n+i(0)\ =(-1)^n$ $\therefore C_n=\dfrac{e^{al}\ {(-1)}^n-\ e^{-al}{(-1)}^n\ }{2\left(al\ -\ in\pi{}\right)}$ $=\dfrac{{(-1)}^n\ }{\left(al\ -\ in\pi{}\right)}\dfrac{e^{al}\ -\ e^{-al}\ }{2}$ $=\dfrac{{(-1)}^n\ \sinh{al}}{al\ -\ in\pi{}}.\ \dfrac{al\ +\ in\pi{}}{al\ +\ in\pi{}}$ $=\dfrac{{(-1)}^n\ \sinh{al\ \left(al\ +\ in\pi{}\right)}}{a^2l^2+n^2{\pi{}}^2}$ $\displaystyle \therefore{}e^{ax}=\ \sum_{-\infty{}}^{\infty{}}\frac{{(-1)}^n\ \sinh{al\ \left(al\ +\ in\pi{}\right)}}{a^2l^2+n^2{\pi{}}^2}\ e^{\dfrac{in\pi{}x}{l}}$