Compute the fourier coefficients, and series for $\log(\sin(x))$ I posted a similar question with a bad response, so I am retrying with hopes of better knowledge. The fourier series is in the form:
$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n\cos(nx) + \sum_{n=1}^{\infty} b_n\sin(nx)$$
Where:
$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) dx$$
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) dx$$
The problem is computing the coefficients. The goal of trying to derive the series is to solve:
$$\int_{0}^{\pi} \log(\sin(x)) dx$$
$\displaystyle a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \log(\sin(x))\cos(nx) dx$
Which is very difficult to compute.
What can be done?
The series representation is:
$$\log(\sin(x)) = -\log(2) - \sum_{k=1}^{\infty} \frac{\cos(2kx)}{k}$$
 A: Use equation $\text{(1d)}$ below and the fact that
$$
\int_0^\pi\cos(2kx)\,\mathrm{d}x=0
$$
to get
$$
\int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2)
$$

$$
\begin{align}
\log(\sin(x))
&=\log\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\tag{1a}\\
&=+ix-\log(2)-i\pi/2+\log(1-e^{-2ix})\tag{1b}\\
&=-ix-\log(2)+i\pi/2+\log(1-e^{+2ix})\tag{1c}\\
&=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1b)}$: bring $e^{+ix}$ and $+\frac1{2i}$ out of the log
$\text{(1c)}$: bring $e^{-ix}$ and $-\frac1{2i}$ out of the log
$\text{(1d)}$: use $\log(1-x)=-\sum\limits_{k=1}^\infty\frac{x^k}k$ while averaging $\text{(1b)}$ and $\text{(1c)}$  

$$
\begin{align}
\log(\cos(x))
&=\log\left(\frac{e^{ix}+e^{-ix}}2\right)\tag{2a}\\
&=+ix-\log(2)+\log(1+e^{-2ix})\tag{2b}\\
&=-ix-\log(2)+\log(1+e^{+2ix})\tag{2c}\\
&=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}{k}\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$\text{(2b)}$: bring $e^{+ix}$ and $\frac12$ out of the log
$\text{(2c)}$: bring $e^{-ix}$ and $\frac12$ out of the log
$\text{(2d)}$: use $\log(1+x)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac{x^k}k$ while averaging $\text{(2b)}$ and $\text{(2c)}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
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\begin{align}&\dsc{\sum_{k\ = 0}^{n}\exp\pars{2ky\ic}}
=\sum_{k\ = 0}^{n}\bracks{\exp\pars{2y\ic}}^{k}
={1 - \exp\pars{\bracks{n + 1}\bracks{2y\ic}} \over 1 - \exp\pars{2y\ic}}
\\[5mm]&={\exp\pars{-y\ic} - \exp\pars{\bracks{2n + 1}y\ic}\over
\exp\pars{-y\ic} - \exp\pars{y\ic}}
=\dsc{\half\,\ic\,{\exp\pars{-y\ic} - \exp\pars{\bracks{2n + 1}y\ic} \over \sin\pars{y}}}
\end{align}

We'll take the imaginary part in both members:
  \begin{align}&\sum_{k\ = 1}^{n}\sin\pars{2ky}
=\half\,\cot\pars{y} - {\cos\pars{\bracks{2n + 1}y} \over 2\sin\pars{y}}
\end{align}

With $\ds{0 < x \leq {\pi \over 2}}$, integrate in both members over
$\ds{\pars{x,{\pi \over 2}}}$:
\begin{align}&\sum_{k\ = 1}^{n}
{-\cos\pars{2k\bracks{\pi/2}} + \cos\pars{2kx} \over 2k}
=\left.\half\,\ln\pars{\sin\pars{y}}\right\vert_{x}^{\pi/2}
-\half\int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y
\end{align}

Then,
  \begin{align}-\ \overbrace{\sum_{k\ = 1}^{n}{\pars{-1}^{k}\over k}}
^{\dsc{-\ln\pars{2}}}
+\sum_{k\ = 1}^{n}{\cos\pars{2kx} \over k}
=-\ln\pars{\sin\pars{x}}
-\int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y
\end{align}

We'll take the limit $\ds{n \to\ \infty}$:
\begin{align}\ln\pars{\sin\pars{x}}&=
-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k}\
-\ \underbrace{\lim_{n\ \to\ \infty}
\int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y}_{\ds{=}\ \dsc{0}}
\end{align}

$$
\color{#66f}{\large\ln\pars{\sin\pars{x}}}=
\color{#66f}{\large-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k}}
\,,\qquad x \in \left(\, 0,{\pi \over 2}\,\right]\tag{1}
$$
  Indeed $\ds{\large\it\mbox{it's still valid}}$ in $\pars{0,\pi}$
  $\ds{\pars{~\color{#00f}{\sf\mbox{see below}}~}}$

When $\ds{x \in \pars{{\pi \over 2},\pi}}$:
\begin{align}
\ln\pars{\sin\pars{x}}&=\ln\pars{\sin\pars{\pi - x}}
=-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2k\bracks{\pi - x}} \over k}
\\[5mm]&=-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k}
\end{align}
such that $\pars{1}$ is valid in $\pars{0,\pi}$.
A: Ok, here I am. The goal is to compute the Fourier series of $g(x)=\log\sin x$ over $[0,\pi]$, or the Fourier series of $h(x)=\log\sin\frac{x}{2}$ over $[0,2\pi]$, or the Fourier series of $f(x)=\log\cos\frac{x}{2}$ over $[-\pi,\pi]$. Since $f(x)$ is an even function, we have to compute:
$$ a_n = \frac{1}{\pi}\int_{-\pi}^{+\pi}\cos(nx)\log\cos\frac{x}{2}\,dx = \frac{2}{\pi}\int_{0}^{\pi}\cos(nx)\log\cos\frac{x}{2}\,dx$$
for any $n\geq 1$ to be able to state:
$$ f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx + \sum_{n\geq 1}a_n\cos(nx)=\frac{a_0}{2}+\sum_{n\geq 1}a_n\cos(nx).$$
for any $x\in(-\pi,\pi)$. Integration by parts gives:
$$ a_n = \frac{2}{\pi}\left(\frac{1}{n}\left.\sin(nx)\log\cos\frac{x}{2}\right|_{0}^{\pi}+\frac{1}{2n}\int_{0}^{\pi}\sin(nx)\tan\frac{x}{2}\,dx\right)$$
or just:
$$ a_n = \frac{1}{\pi n}\int_{0}^{\pi}\frac{\sin(nx)\sin(x/2)}{\cos(x/2)}\,dx = \frac{2}{\pi n}\int_{0}^{\pi/2}\frac{\sin(2nx)\sin x}{\cos x}\,dx.$$
Since $\cos((2n+1)x)=2\cos x\cos(2nx)-\cos((2n-1)x)$, we have:
$$ a_n = \frac{1}{\pi n}\int_{0}^{\pi/2}\sum_{k=1}^{n}\cos((2k-1)x)\,dx =\frac{(-1)^{n+1}}{n}.$$
This gives:
$$\log\cos\frac{x}{2}=\frac{a_0}{2}+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\cos(nx)$$
for any $x\in(-\pi,\pi)$. In order to find $a_0$, we can simply match $f(0)=0$ with the series on the right hand side. Since:
$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\int_{0}^{1}\frac{dx}{1+x}=\log 2,$$
we have:
$$\log\cos\frac{x}{2}=-\log 2+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\cos(nx)\qquad\forall x\in(-\pi,\pi),\tag{1}$$
and by translating the variable:
$$\log\sin\frac{x}{2}=-\log 2-\sum_{n\geq 1}\frac{1}{n}\cos(nx)\qquad\forall x\in(0,2\pi),\tag{2}$$
$$\log\sin x = -\log 2-\sum_{n\geq 1}\frac{1}{n}\cos(2nx)\qquad\forall x\in(0,\pi),\tag{3}$$
as wanted.

A little addendum, since I think it is worth mentioning the following technique. Starting with a celebrated identity:
$$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}$$
and noticing that $\log\sin x$ is a Riemann integrable function over $(0,\pi)$, we have:
$$\int_{0}^{\pi}\log\sin x\,dx = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\lim_{n\to +\infty}\frac{\pi}{n}\log\frac{2n}{2^n}=\color{red}{-\pi\log 2}.$$
