$f(x)=\log|2\sin(\frac{x}{2})| \in L^2(-\pi,\pi)$

Question

I need to prove that $f(x)=\log|2\sin(\frac{x}{2})| \in L^2(-\pi,\pi)$. I tried to integrate $f^2(x)$ but it seems too difficult...

Thank you for any help!

Answer 1

Hint:

This may help you deal with a simpler integral. Notice that $$\log|2\sin(x/2)|=\log\Big|\frac{\sin(x/2)}{x/2}x\Big|=\log\Big|\frac{\sin(x/2)}{x/2}\Big| + \log|x|$$

Now, the function $f_1:x\mapsto\frac{\sin(x/2)}{x/2}$ is away from $0$ in the $[-\pi,\pi]$ so the $\log\circ f_1$ is nice and square integrable.

The problem is now reduced to see whether $\log|x|$ is square integrable in $[0,\pi]$ (this function is even), for your initial integral is the sum of two functions and so, the square integrability of both implies the square integrability of the sum.

For $\int^\pi_0\log^2|x|\,dx$, you may try using integration by parts.