I want to find the fourier coefficients for $ f(x)= \sin^k (x) , k \in \mathbb{N} $.
Calculating $$ \tilde{f}(n)= \frac{1}{2 \pi } \int_{0}^{ 2 \pi} \sin^k(x) e^{-inx} dx$$
If I rewrite $\sin$ in terms of complex exponentials
$$ \tilde{f} (n) = \frac{1}{2 \pi } \int_{0}^{ 2 \pi} \left(\frac{e^{ix} -e^{-ix}}{2i} \right)^k e^{-inx} dx $$
How to proceed from here?
Edit for expanding with binomial theorem:
$$sin^k(x) = \frac{1}{(2i)^k} ( e^{ix}-e^{-ix})^k $$ $ = \frac{1}{(2i)^k} \sum_{j=0}^k $ $ k \choose{j} $ $(e^{ix})^{k-j}(-e^{-ix})^j) $
$= \frac{1}{(2i)^k} \sum_{j=0}^k $ $ k \choose{j} $ $e^{ixk-j}(-e^{-ixj})$
so it gets to :
$ \tilde{f} (n) = \frac{1}{2 \pi } \int_{0}^{ 2 \pi} \frac{1}{(2i)^k} \sum_{j=0}^k $ $ k \choose{j} $ $e^{ixk-j}(-e^{-ixj}) e^{-inx} dx $
is this even right so far or have I made a mistake? how can I calculate an integral like this though..o.o