Determine the limit of a series, involving trigonometric functions: $\sum \frac{\sin(nx)}{n^3}$ and $\frac{\cos(nx)}{n^2}$ I have $$\sum^\infty_{n=1} \frac{\sin(nx)}{n^3}.$$
I did prove convergence:
$0<\theta<1$
$$\left|\frac{\sin((n+1)x)n^3}{(n+1)^3\sin(nx)}\right|< \left|\frac{n^3}{(n+1)^3}\right|<\theta$$
Now I want to determine the limit. I found a similar proof but I need help understanding it; it goes like this. :
$$ F(x):=\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}$$
As for this series we have uniform convergence. The series of derivatives: $$-\sum^\infty_{n=1} \frac{\sin(nx)}{n}$$ converges for every $\delta >0$ on the interval $[\delta, 2\pi-\delta]$ uniform against $\frac{x-\pi}{2}$
so, for every $x \in]0,2\pi[$ : $\displaystyle F'(x) = \frac{x-\pi}{2}$$\displaystyle F(x) = \left(\frac{x-\pi}{2}\right)^2+c,c\in \mathbb{R}$.
To determine the constant we calculate:
$$ \int^{2\pi}_0F(x)dx=\int^{2\pi}_0\left(\frac{x-\pi}{2}\right)^2dx+\int^{2\pi}_0cdx=\frac{\pi^3}{6}+2\pi c$$
(Question: Why can we do this do get the constant?)
Because $\int^{2\pi}_0cos(nx)dx= 0 \forall n≥1$ we have:
$$\int^{2\pi}_0F(x)dx = \sum^\infty_{n=1}\int^{2\pi}_0\frac{\cos(nx)}{n^2}=0,$$ so $c = -\frac{\pi^2}{12}$. (Question: How does he get to that term $\frac{\pi^2}{12}$?) With that we have proven, that
$$\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}=\left(\frac{x-\pi}{2}\right)^2-\frac{\pi^2}{12}$$
If you can explain one of the questions about this proof, or if you know how to calculate the limit in my situation above, it would be cool if you leave a quick post here, thanks!
 A: We can follow the proof in the post indicated by Marko Riedel. Rewrite $$\underset{n=1}{\overset{\infty}{\sum}}\frac{\sin\left(nx\right)}{n^{3}}=x^{3}\underset{n=1}{\overset{\infty}{\sum}}\frac{\sin\left(nx\right)}{\left(nx\right)^{3}}$$
 and use the fact that the Mellin transform identity for harmonic sums with base function $g(x)$
  is$$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\, g\left(s\right)^{*}$$
 where $g\left(s\right)^{*}$
  is the Mellin transform of $g\left(x\right)$
 . So in this case we have $$\lambda_{k}=1,\,\mu_{k}=k,\, g\left(x\right)=\frac{\sin\left(x\right)}{x^{3}}$$
 and so its Mellin transform is $$g\left(s\right)^{*}=\Gamma\left(s-3\right)\sin\left(\frac{1}{2}\pi\left(s-3\right)\right).$$
 Observing that $$\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}=\zeta\left(s\right)$$
 we have $$x^{3}\underset{n=1}{\overset{\infty}{\sum}}\frac{\sin\left(nx\right)}{\left(nx\right)^{3}}=\frac{x^{3}}{2\pi i}\int_{\mathbb{C}}\Gamma\left(s-3\right)\sin\left(\frac{1}{2}\pi\left(s-3\right)\right)\zeta\left(s\right)x^{-s}ds=\frac{x^{3}}{2\pi i}\int_{\mathbb{C}}Q\left(s\right)x^{-s}ds.$$
 Note that sine term cancels poles in at odd negative integers and zeta cancels poles at even negative integers. So we have poles only at $s=0,1,2.$
  And the compute is$$\underset{s=0}{\textrm{Res}}\left(Q\left(s\right)x^{-s}\right)=\frac{1}{12}$$
 $$\underset{s=1}{\textrm{Res}}\left(Q\left(s\right)x^{-s}\right)=-\frac{\pi}{4x}$$
 $$\underset{s=2}{\textrm{Res}}\left(Q\left(s\right)x^{-s}\right)=\frac{\pi^{2}}{6x^{2}}$$
 so$$\underset{n=1}{\overset{\infty}{\sum}}\frac{\sin\left(nx\right)}{n^{3}}=\frac{\pi^{2}x}{6}-\frac{\pi x^{2}}{4}+\frac{x^{3}}{12}.$$
A: $$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= ?$$  
We know that $$\sum^\infty_{n=1}\frac{sin(nx)}{n}= \frac{x-\pi}{2}$$.
$$(\frac{sin(nx)}{n^3})''= (\frac{cos(nx)}{n^2})'=-\frac{sin(nx)}{n}$$
$$-\sum^\infty_{n=1}\frac{sin(nx)}{n}= -(\frac{x-\pi}{2})$$
$$ - \int \frac{x-\pi}{2}= -\int \frac{x}{2}-\frac{\pi}{2} = \frac{x^2}{4}+\frac{\pi x}{2}+c$$
$$\int\frac{x^2}{4}+\frac{\pi x}{2}+c=\frac{\pi x^2}{4}+\frac{x^3}{12}+xc$$
We can also determine the constant. We know that $-f'(0) = 0 + c$.
$$-\sum^\infty_{n=1} \frac{cos(n0)}{n^2}= - \sum^\infty_{n=1} \frac{1}{n^2} = -\frac{\pi^2} {6} = 0+c$$
$$-\int\frac{\pi^2} {6} = -\frac{x\pi^2} {6}$$ So 
$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= \frac{\pi x^2}{4}+\frac{x^3}{12}+(-\frac{x\pi^2}{6}) =  \frac{x\pi^2}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}$$
