Consider the odd function $f(\theta)=\theta (\pi - \theta)$, then I need to show that:
$f(\theta)=\frac{8}{\pi} \sum_{k \;odd \ge 1} \frac{sin(k \theta)}{k^3}$
then I computed the Fourier coefficients and I have get:
$\widehat{f}(n)=\frac{(-\pi)^{n+1}-(-2\pi)^{n+1}}{in}$
and
$\widehat{f}(0)=\frac{2}{3}\pi^3$
But then I wanted to use the formula given here Writing a Fourier series of a $2\pi$-periodic function. but I dont know how I am going to get the $\frac{8}{\pi k^3}$ from those integrals.
Can you tell me if I am going in the correct way?, if not can you help me to fix it please, thanks a lot.
My computation of the Fourier coefficients:
$\widehat{f}(n)= \frac{1}{2 \pi} \int_{0}^{2 \pi}\theta (\pi - \theta)e^{-in \theta}d\theta=\frac{1}{2 \pi}\int_{0}^{2 \pi}\theta \pi e^{-in \theta}d\theta - \frac{1}{2 \pi}\int_{0}^{2 \pi}\theta e^{-in \theta}d\theta $
then for the first integral we do the substitution $u=\theta$ and $v=e^{-in \theta}$, therefore:
$$\frac{1}{2 \pi}\int_{0}^{2 \pi}\theta \pi e^{-in \theta}d\theta = \frac{1}{2 } [-\frac{\theta}{in}e^{-in \theta}]_{0}^{2\pi}+\int_{0}^{2\pi}\frac{e^{-in \theta}}{in}d\theta=\frac{1}{2}[\frac{-2\pi}{in}e^{-2\pi in}]=\frac{(-\pi)^{n+1}}{in}$$
Similarly the othe integral