Define a function $f(x) =(2\cos(\pi x))^{10} $$f\in L^{1}$ so it's one-period. I would like to calculate the Fourier coefficient $\hat{f}(2)$.
So we get $\displaystyle\hat{f}(n)=\int_{0}^{1}e^{-2\pi inx}(2\cos(\pi x))^{10}dx$ , $n\in \mathbb{Z}$
I know that we can write $\cos(\pi x)^{10}$ = $((e^{i\pi x}+e^{-i\pi x})/2)^{10}$ and with the binomial formula we get
$$\sum_{k=0}^{10}\binom{10}{k}(e^{i\pi x})^{10-k}(e^{-i\pi x})^{k}$$
So the Fourier coefficient $\displaystyle \hat{f}(n) = \int_{0}^{1}e^{-2\pi inx}2^{10}\sum_{k=0}^{10}\binom{10}{k}(e^{i\pi x})^{10-k}(e^{-i\pi x})^{k}dx$ , where $n\in \mathbb{Z}$
I have no idea how to easily integrate this one. I appreciate any help.
$$ \begin{align} \int_0^1e^{-4\pi ix}(2\cos(\pi x))^{10}\,\mathrm{d}x &=\int_0^1e^{-4\pi ix}\left(e^{i\pi x}+e^{-i\pi x}\right)^{10}\,\mathrm{d}x\\ &=\int_0^1e^{-4\pi ix}\binom{10}{3}e^{7\pi ix}e^{-3\pi ix}\,\mathrm{d}x\\ &=\binom{10}{3} \end{align} $$ All the other terms in the binomial expansion of $\left(e^{i\pi x}+e^{-i\pi x}\right)^{10}$ are multiples of $e^{2k\pi ix}$ for $k\ne2$, and therefore, integrate to $0$ against $e^{-4\pi ix}$.