I've been trying to work through a problem in which I have to prove the following is true.
$$\int_0^{\pi}\sum_{n=1}^\infty \left(\frac{\sin(nx)}{n^3}\right)dx = 2 \sum_{n=1}^\infty \left(\frac{1}{(2n-1)^4}\right) $$ I know the right hand side is equal to $\left(\frac{\pi^4}{48}\right)$ by reciprocals of perfect squares.
But I can't figure out how to deal with the left hand side, any help would be appreciated. Thanks.
Assuming you meant, $n=1$. Since for all $k$, $\sum\limits_{n=1}^{k} \frac{\sin(nx)}{n^3}\leq \sum\limits_{n=1}^{k} \frac{1}{n^3}\leq \sum\limits_{n=1}^{\infty} \frac{1}{n^3}$ which is some number, we may apply Dominated convergence Theorem. Thus $$\int_0^{\pi} \sum\limits_{n=1}^{\infty} \frac{\sin(nx)}{n^3} dx= \sum\limits_{n=1}^{\infty} \int_0^{\pi}\frac{\sin(nx)}{n^3} dx$$ $$\int_0^{\pi} \sum\limits_{n=1}^{\infty} \frac{\sin(nx)}{n^3} dx= \sum\limits_{n=1}^{\infty} \frac{1-\cos(n\pi)}{n^4}$$ $$\int_0^{\pi} \sum\limits_{n=1}^{\infty} \frac{\sin(nx)}{n^3} dx= \sum_{even\ n \in N} 0 +\sum_{odd\ n \in N} \frac{2}{n^4}$$ $$\int_0^{\pi} \sum\limits_{n=1}^{\infty} \frac{\sin(nx)}{n^3} dx= \sum_{n=0}^{\infty} \frac{2}{(2n-1)^4}$$