How to prove that $\log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq 2 k\pi)$, $k$ is an integer

Question

This question number 11.15(a) Tom M Apostol's MAthematical Analysis From Chapter Fourier analysis.

Prove that $ \log |\sin(x/2)| = -\log 2 -\sum_{n=1}^{\infty} \frac{\cos nx} {n} $ ( if $ x\neq 2 k\pi)$.

So, I tried by using $a_0 = 2/\pi \int_{0}^{\pi} \log|\sin(x/2)| dx$ and I used $\sin(2x) =2\sin x\cos x$.

So Using it I get $a_0 /2 =\log 2 + 2/\pi [\int_{0}^{\pi}\log|\sin(x/4)|dx +\int_{0}^{\pi}\log|\cos (x/4) |dx]$.

But the problem is how to compute $\int_{0}^{\pi}\log|\cos (x/4) |dx$. The limit of integral can be taken from $0$ to $2 \pi $

Similar problem will also arise when I will compute $a_n = \int_{0}^{\pi} \log|\sin(x/2)| \times\cos(nx) dx$.

Can you please tell how to compute this integral OR If there is alternate way of computing the fourier series.

Thank you.

Answer 1

Note that$$\ln|2\sin(x/2)|=\ln|1-e^{ix}|=\tfrac12(\ln(1-e^{ix})+\text{c.c.})=-\tfrac12\sum_{n\ge1}\tfrac{e^{inx}+\text{c.c.}}{n}=-\sum_n\tfrac{\cos nx}{n},$$where $z+\text{c.c.}$ is short for $z+\bar{z}$.

Answer 2

$$S=\Re \sum_{n=1}^{\infty} \frac{e^{inx}}{n}=-\Re\ln(1-e^{ix})=-\Re \ln[1-\cos x+i\sin x]=-\frac{1}{2}\ln (2-2\cos x)$$$$ \implies S=-\frac{1}{2}\ln 4\sin^2(x/2)=- \ln 2- \ln{|\sin(x/2)|}=-\ln 2-\ln |\sin (x/2)|$$ Here we have used $-\ln(1-z)=\sum_{n=0}^{\infty} \frac{z^n}{n}$ and $\ln(x+iy)=\frac{1}{2}\ln(x^2+y^2)+i\operatorname{atan2}(y,\,x)$