Here is the function: $f(x) = \frac{x^3}{3}+x$, $-\pi \le x \le \pi$, period: $2\pi$
I tried to compute the coefficients, but noticed that the integral for coef. $a_n$ looks labourous and there might be some catch. Is it computable the usual way or somehow else? ( I am not asking for integral computations) $$a_0 = \frac{1}{2\pi}\int^{\pi}_{-\pi}f(x) dx = \frac{1}{2\pi} \left[\frac{x^4}{12}+\frac{1}{2}x^2 \right]^{\pi}_{-\pi} = 0 \\ a_n = \frac{1}{\pi}\int^{\pi}_{-\pi} \left(\sin (\frac{\pi x n}{\pi}) \right)(\frac{1}{3}x^3+x)dx$$