I've been solving some problems from my functional analysis course, and I found one where I'm close to finding the solution but still don't know how to do. I managed to solve question (a) but there's something I'm missing to solve (b). The problem says:
(a) Prove that, $\forall x\in(0,\pi)$ it's verified that $$x(\pi-x)=\frac8\pi\sum_{k=0}^\infty\frac{\sin((2k+1)x)}{(2k+1)^3}.$$ (b) Calculate, using the result obtained in (a), the value of $$\sum_{k=0}^\infty\frac{1}{(2k+1)^6}.$$
This is the work I've done:
(a) This one was easily solved. I just found the Fouries series of the odd extension of $x(\pi-x)$ inside $(-\pi,\pi)$, which is $$f(x) = \left\{x(\pi+x), x\leq0\atop x(\pi-x),x>0\right.$$ and I just found that the coefficients of $\cos$ are all $0$ and the $\sin$ ones are $\frac{8}{n^3\pi}$ for odd $n$ and $0$ otherwise. The desired identity follows then from that.
(b) This is where I'm close but don't reach the solution. I noticed that $$\iiint \frac{\sin((2k+1)x)}{(2k+1)^3}dxdxdx=\frac{\cos((2k+1)x)}{(2k+1)^6}+C,$$ and that $\cos$ can be removed evaluating the $x$ in $0$ or $\pi$. I also integrated three times $x(\pi-x)$ and got $\frac{\pi x^4}{24}-\frac{x^5}{60}+K$. I know that one of those constants $C$ or $K$ won't be zero if I try to make an equality between them. In any case, I know the final result is that the sum equals $\frac{\pi^6}{960}$, but don't know what I'm missing to get it from here.
How can I end question (b)? Any help or hint will be appreciated, thanks in advance.
For $$ f(x)=\sum_{k=0}^\infty a_k\sin[(2k+1)x] $$ Parseval's Identity says $$ \int_0^\pi|f(x)|^2dx=\sum_{k=0}^\infty a_k^2\int_0^\pi|\sin[(2k+1)x]|^2dx=\frac{\pi}{2}\sum_{k=0}^\infty a_k^2. $$ You can use it easily to get the result. In fact, since $$ \int_0^\pi[x(\pi-x)]^2dx=\frac{\pi^5}{30} $$ and $$ \frac{\pi}{2}\sum_{k=0}^\infty a_k^2=\frac{\pi}{2}\sum_{k=0}^\infty\frac{64}{\pi^2(2k+1)^6}=\frac{32}{\pi}\sum_{k=0}^\infty\frac{1}{(2k+1)^6}, $$ one has $$ \sum_{k=0}^\infty\frac{1}{(2k+1)^6}=\frac{\pi^6}{960}.$$