How to find $A_n$ such that $\sum_{n=1}^\infty A_n\sqrt{2} n \sin (nx)=1$

Question

I meet a trouble to find $A_n$ such that the following equality holds.

$$\sum_{n=1}^\infty A_n\sqrt{2} n \sin (nx)=1, \ \ \ \ 0<x<\pi$$

I am not sure if I can really find such $A_n$ since the LHS could be divergent.

Can someone could give me a hint on this?

Answer 1

Hint :

Multiply both sides by $\sin(mx)$ where $m \in \mathbb Z$ and $m \neq n$. Then :

$$\sum_{n=1}^\infty A_n\sqrt{2} n \sin (nx)\sin(mx)=\sin(mx) $$ $$\implies$$ $$ \sum_{n=1}^\infty A_n\sqrt{2}n \int_0^\pi\sin(nx)\sin(mx)\mathrm{d}x = \int_0^\pi \sin(mx)\mathrm{d}x$$ Now note that the terms vanish for $n \neq m$. Thus the only one withstanding is for $n =m$. Can you proceed ?

Answer 2

Hint: the equation is equivalent to $ \sum\limits_{n=1}^{\infty} A_n\sqrt 2 n \sin (nx)=f(x)$ for $-\pi \leq x \leq \pi$ where $f(x)=1$ for $0<x\leq \pi$, $-1$ for $-\pi \leq x<0$ and $0$ for $x=0$. The coefficients of the series are just the Fourier coefficients of the sine series of this odd function.