Let $$f(x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$.
Now let $f(x)$ be uniform convergence on $\mathbb{R}$
Show that $$\int_0^\pi f(x) \,dx=\sum_{k=0}^{\infty}\frac{2}{(2k+1)^4}$$
I'm not sure how to get past from this point. I was trying to move the index from n=0 but I can't seem to figure it out. I tried to "pull" 1 out and start at n=0, but I ended up with an undefined series (n^4 in the denominator).
$$\int_0^\pi \sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3}\,dx =\sum_{n=1}^{\infty}\frac{1}{n^3}\int_0^\pi\sin(nx) dx =\sum_{n=1}^{\infty}\frac{1-\cos(\pi n)}{n^4}$$
\begin{align*} \dfrac{1}{n^{3}}\int_{0}^{\pi}\sin(nx)dx=\dfrac{1}{n^{3}}\cdot\dfrac{-1}{n}\cos(nx)\bigg|_{x=0}^{x=\pi}=\dfrac{-1}{n^{4}}\left((-1)^{n}-1\right), \end{align*} now consider even and odd $n$ separately.