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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

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  • How did you compute the Fourier coefficients? You should obtain a result with $n^3$ in the denominator. – Daniel Fischer Apr 07 '15 at 18:43
  • Ok, let me post the computation :) – user162343 Apr 07 '15 at 18:44
  • In the second integral, you should have $\theta^2$, since $\theta(\pi -\theta) = \pi\theta - \theta^2$. And I think you should have used the interval $[-\pi,\pi]$ for the Fourier series, i.e. compute the Fourier series of the $2\pi$-periodic function with values $$f(\theta) = \theta(\pi - \lvert\theta\rvert)$$ on $[-\pi,\pi]$. I haven't done the calculations, though, so I may think wrong here. But the mention of "odd function" suggests $[-\pi,\pi]$. – Daniel Fischer Apr 07 '15 at 19:11
  • I am sorrry I have the theta squred in my notes but I forgot tu put it, but then what else can be done, because Is the fourier series of the new funtion is equal to the old one? – user162343 Apr 07 '15 at 19:15

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