The Fourier Series of $f(x)= \int_{0}^{x} \log\sqrt{|\sin(\frac{t}{2})|} dt , -\pi < x < \pi $

Question

This was a question on my exam in Fourier Analysis and I couldn't do it. I had to find the Fourier Series of $$f(x)= \int_{0}^{x} \log\sqrt{\left|\sin\left(\frac{t}{2}\right)\right|} dt , -\pi < x < \pi $$

How?

Answer 1

Hint:
The fourier series of $$-\log\bigg(2\sin\big(\frac{x}{2}\big)\bigg)$$ for $0<x<π,$ is :$$\sum_{n\in\mathbb N} \frac{\cos(nx)}{n}$$ $$\text{Hence : }\frac{1}{2}\log\bigg(\sin\big(\frac{x}{2}\big)\bigg)=-\frac{1}{2}\sum_{n=1}^{\infty}\bigg(\frac{\cos(nx)}{n}\bigg)-\frac{\log(2)}{2} \qquad (1) $$ $$\text{Let } f(x)=\int_0^x \log\sqrt{|\sin(t/2)|}dt$$ $$f'(x)=\log\sqrt{|\sin(\frac{x}{2})|}$$ $$=(1)$$ $$\sum_n\frac{\cos(nx)}{n}\text{ does converge uniformly in }(0,\pi)$$ $$\text{So, }f(x)=\int_0^xf'(t)dt=-\frac{1}{2}\sum_{n=1}^{\infty}\bigg(\int_0^x\frac{\cos(nt)}{n}dt\bigg)-\int_0^x\frac{\log(2)}{2}dt$$ $$=-\frac{1}{2}\sum_{n=1}^{\infty}\bigg(\frac{\sin(nx)}{n^2}\bigg)-x\frac{\log(2)}{2}.$$