Showing $ \int_0^\pi{f(x)}dx=\sum_{k=0}^\infty{\frac{2}{(2k+1)^4}} $ where $ f(x)=\sum_{n=0}^\infty{\frac{\sin(n\cdot x)}{n^3}} $

Question

I have to show that the two terms below are equivalent $$ \int_0^\pi{f(x)}dx=\sum_{k=0}^\infty{\frac{2}{(2k+1)^4}} $$ where $f(x)$ is the sum function: $$ f(x)=\sum_{n=0}^\infty{\frac{\sin(n\cdot x)}{n^3}} $$

I know there is a theorem that shows that the below is true $$ \sum_{n=1}^\infty{\int_a^b{f_n(x)}dx} = \int_a^b{\sum_{n=1}^\infty{f_n(x)}}dx $$ and I think I have to use that theorem, but I dont know how to use it. I hope I can get some help.

Thanks in advance

Answer 1

hint

$$\int_0^\pi \frac{\sin(nx)}{n^3}dx=\Bigl[\frac{-\cos(nx)}{n^4}\Bigr]_0^\pi$$

$$=\frac{1-(-1)^n}{n^4}$$

$$=\frac{2}{(2k+1)^4} \text{ if } n=2k+1$$

Answer 2

The series$$\sum_{n=1}^\infty\frac{\sin(nx)}{n^3}$$converges uniformly, by the Weierstrass $M$-test (since $\left|\frac{\sin(nx)}{n^3}\right|\leqslant\frac1{n^3}$) and if a series $\sum_{n=1}^\infty f_n(x)$ of integrable function from $[a,b]$ into $\Bbb R$ converges uniformly, then its is integrable and$$\int_a^b\sum_{n=1}^\infty f_n(x)\,\mathrm dx=\sum_{n=1}^\infty\int_a^bf_n(x)\,\mathrm dx.$$