$f(x) = e^{-x}, x \in (0, 2\pi) $ Then The Fourier Series formula is given by    The values of a0 and are given as follows $a_0 = \dfrac{1}{\pi}\int_{-\pi}^{\pi} e^{-x}\ dx$ = $\dfrac{2\sinh \pi}{\pi} a_{n} $ = $\dfrac{1}{\pi}\int_{0}^{2\pi} e^{-x}\cos (nx)\ dx$  $a_1$$= \dfrac{1}{\pi}[{ \dfrac{1}{n}\ e^{-x} \sin (nx)]_{0}^{2\pi} - \dfrac{1}{n}\int_{0}^{2\pi} e^{-x}\sin (nx)\ dx} ]$ $= \dfrac{(-1)^{n+1}(e^{2\pi} - e^{0})}{\pi(1+n^2)} b_{n} $ $= \dfrac{1}{\pi}\int_{0}^{2\pi}e^{x}\sin (nx)\ dx$ $b_0= \dfrac{1}{\pi}\int_{0}^{2\pi}e^{x}\sin (nx)\ dx$ =$ \dfrac{1}{\pi}[ \dfrac{1}{n}\ e^-{x}\cos (nx)]_{0}^{2\pi} -\dfrac{1}{n}\int_{0}^{2\pi} e^{-x}\cos(nx)\ dx}$  = $\dfrac{n(-1)^{n}(e^{2\pi} - e^{0})}{\pi(1+n^2)}$  $b_1=\dfrac{n(-1)^{n}(e^{2\pi} - e^{0})}{\pi(1+n^2)}$ Hence adding the values of a0,a1,b0 and b1 we have the final answer as =$\dfrac{1} {\pi} sinh(\pi) +\Sigma^{\infty}_ {n=1} \dfrac{2 sinh(\pi)(-1)^n} {\pi(1 + n^2)} [cos(nx) +n sin(nx)]. $