The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series $$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2}\sin\left(\frac{2\pi x}{l} \right) + \frac{1}{3}\sin\left(\frac{3\pi x}{l} \right) - \ldots \right)$$ Integrate the series again, and find the sine series for $x^3 - l^2 x$.
This question was originally started here, I am really not sure how to proceed with this. Any suggestions are greatly appreciated.
Attempted solution - (Following the same procedure I took in the link) We know that $$\frac{1}{2}x^2 = \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right) + \frac{l^2}{6}$$ or that $$\frac{1}{2}x^2 - \frac{l^2}{6} = \frac{4l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right)$$ Thus integrating the series we have \begin{align*} \frac{1}{6}x^3 - \frac{l^2}{6}x &= \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^2}\frac{l}{m\pi}\sin\left(\frac{m\pi x}{l}\right) + \xi\\ &= \frac{2l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi x}{l}\right) + \xi \end{align*} Then $$\xi = \frac{1}{2l}\int_{-l}^{l}\frac{1}{6}x^3 - \frac{l^2}{6}x dx 0$$ Therefore $$x^3 - l^2x = \frac{12l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi x}{l}\right)$$
I am not sure if this is correct, any suggestions are greatly appreciated.