Fourier series for $e^x$ over $[0,\pi)$ I am trying to solve the following,

Find the Fourier series of $h(x) = \text{e}^x, x \in [0,\pi)$.

I'm not sure how to approach it since the question does not specify whether to use an even or an odd extension.
Any help would be much appreciated!
 A: You may use an even extension or an odd extension. Since:
$$ \int_{0}^{\pi}e^{x}\cos(nx)\,dx = \frac{1}{n^2+1}\left(-1+(-1)^n e^{\pi}\right), $$
$$ \int_{0}^{\pi}e^{x}\sin(nx)\,dx = \frac{n}{n^2+1}\left(1-(-1)^n e^{\pi}\right) $$
that are both consequences (take the real or imaginary part) of
$$ \int_{0}^{\pi} e^{x} e^{nix}\,dx = \left.\frac{e^{(1+ni)x}}{1+ni}\right|_{0}^{\pi},$$
by considering the Fourier cosine series of $e^{|x|}$ over $(-\pi,\pi)$ we have:
$$ e^x = \frac{e^\pi-1}{\pi}+\frac{2}{\pi}\sum_{n\geq 1}\left((-1)^n e^\pi-1\right)\frac{\cos(nx)}{n^2+1}$$
and by considering the Fourier sine series of $\text{sign}(x)\,e^{|x|}$ over $(-\pi,\pi)$ we have:
$$ e^x = \frac{2}{\pi}\sum_{n\geq 1}\left(1-(-1)^n e^{\pi}\right)\frac{n\sin(nx)}{n^2+1}.$$
However, the last series converges to zero for $x=0$, so, if we want a pointwise convergent representation over $[0,\pi)$, it is best to take the even one (also because it converges faster, since $e^{|x|}$ is a way more regular function than $\text{sign}(x)\,e^{|x|}$).
