Have trouble with this proof

Question

Don´t know how to start, or apply the theorem. Aprecciate your help.

Answer 1

By comparison test to $p$-series with $p=2$ we know the sum converges for all $x$ since $|\sin(x)| \leq 1$. If we call it's sum $f(x)$, and use the theorem you have, this allows us to integrate term by term. i.e

$$ \int_0^\pi f(x) = \int_0^\pi \sum_{n=1}^{\infty} \frac{ \sin(n x)}{n^2}dx =\sum_{n=1}^\infty \frac{1}{n^2} \int_0^\pi \sin(nx) dx $$

I think you can take it from here. If you have issues, let me know.

Answer 2

Sketch: $$f(x)=\sum_{j=1}^\infty \frac{\sin{jx}}{j^2} $$ $$f_n(x)=\sum_{j=1}^n \frac{\sin{jx}}{j^2}$$

First proof $f_n$ converges uniformly to $f$.

Then by the theorem:

$$\lim_{n \to \infty}\int_0^\pi f_n(x)=\int_0^\pi f(x)$$

And, as you can see:

$$\int_0^\pi f_n(x)=\frac{1-\cos{n\pi}}{n^3}$$

So the sum $$\sum_{j=1}^\infty \frac{1-\cos{j\pi}}{j^3}=\sum_{j=1}^\infty \frac{1-\cos{(2j-1)\pi}}{(2j-1)^3}+\sum_{j=1}^\infty \frac{1-\cos{2j\pi}}{(2j)^3}$$ $$=\sum_{j=1}^\infty \frac{1-(-1)}{(2j-1)^3}+\sum_{j=1}^\infty \frac{1-1}{(2j)^3}$$ $$=\sum_{j=1}^\infty \frac{2}{(2j-1)^3}$$ The even terms disappear and the odd remains and give your result