$\displaystyle Let\ c=-2\ and\ c+2l=2 \\[2ex] \displaystyle \therefore{}-2+2l=2 \\[2ex] \displaystyle \therefore{}2l=4 \\[2ex] \displaystyle \therefore{}l=2 \\[2ex] \displaystyle C_n=\frac{1}{2l}\int_c^{c+2l}f(x)e^{\frac{-in\pi{}x}{l}}dx \\[2ex] \displaystyle =\frac{1}{2(2)}\int_{-2}^2{e^{-ax}e}^{\frac{-in\pi{}x}{l}}dx \\[2ex] \displaystyle =\frac{1}{4}\int_{-2}^2e^{-\left(a+\frac{in\pi{}}{2}\right)x}dx \\[2ex] \displaystyle =\frac{1}{4}\int_{-2}^2e^{\frac{-\left(2a+in\pi{}\right)x}{2}}dx \\[2ex] \displaystyle =\frac{1}{4}{\left[\frac{e^{\frac{-\left(2a+in\pi{}\right)x}{2}}}{\frac{-\left(2a+in\pi{}\right)x}{2}}\right]}_{-2}^2 \\[2ex] \displaystyle =\frac{1}{4}\times{}\frac{-2}{\left(2a+in\pi{}\right)}\left[e^{-\left(2a+in\pi{}\right)}-e^{\left(2a+in\pi{}\right)}\right] \\[2ex] \displaystyle =\frac{-1}{2\left(2a+in\pi{}\right)}\times{}\left[{e^-}^{2a}e^{-in\pi{}}-e^{2a}e^{in\pi{}}\right]\times{}\frac{\left(2a-in\pi{}\right)}{\left(2a-in\pi{}\right)} \\[2ex] $

$\displaystyle Consider\\[2ex] e^{\pm{}in\pi{}}=\cos{n\pi{}\pm{}i\sin{n\pi{}={\left(-1\right)}^n\pm{}i0={\left(-1\right)}^n}} \\[2ex] \displaystyle \therefore{}\ C_n=\frac{-\left(2a-in\pi{}\right)}{2\left(4a^2-i^2n^2{\pi{}}^2\right)}\times{}\left[{e^-}^{2a}{\left(-1\right)}^n-e^{2a}{\left(-1\right)}^n\right] \\[2ex] \displaystyle C_n=\frac{\left(2a-in\pi{}\right){\left(-1\right)}^n}{\left(4a^2+n^2{\pi{}}^2\right)}\times{}\left[\frac{{e^-}^{2a}-e^{2a}}{2}\right] \\[2ex] \displaystyle C_n=\frac{\left(2a-in\pi{}\right){\left(-1\right)}^n}{\left(4a^2+n^2{\pi{}}^2\right)}\sinh{2a} \\[2ex] $

$\displaystyle In\ complex\ Fourier\ series, \\[2ex] \displaystyle f(x)=\ \sum_{n=-\infty{}}^{\infty{}}Cne^{\frac{in\pi{}x}{l}} \\[2ex] \displaystyle {\therefore{}e}^{-ax}=\sum_{n=-\infty{}}^{\infty{}}\frac{\left(2a-in\pi{}\right){\left(-1\right)}^n}{\left(4a^2+n^2{\pi{}}^2\right)}\sinh{2a\bullet{}e^{\frac{in\pi{}x}{2}}} \\[2ex] \displaystyle {\therefore{}e}^{-ax}=\sinh{2a\bullet{}}\sum_{n=-\infty{}}^{\infty{}}\frac{\left(2a-in\pi{}\right){\left(-1\right)}^n}{\left(4a^2+n^2{\pi{}}^2\right)}\bullet{}e^{\frac{in\pi{}x}{2}} \\[2ex] $