Integral of $\log(\sin(x))$ using contour integrals I know the integral is possible with a simple fourier series expansion of $-\log(\sin(x))$
But I am interested in complex analysis, so I want to try this.
$$I = \int_{0}^{\pi} \log(\sin(x)) dx$$
The substitution $x = \arcsin(t)$ first comes to mind. 
But that substitution isnt valid as, 
The upper and lower bounds would both be $0$ because $\sin(\pi) = \sin(0) = 0$
What is a workaround using inverse trig or some other way?
Inverse sine is good, because that gives us a denominator from which we will be able to find poles.
$$x = arctan(t)$$
Is also good, but it would hard to do.
Idea?
 A: We have
\begin{align}\int_0^{\pi} \log(\sin x)\, dx &= \int_0^{\pi} \log\left|\frac{e^{ix}-e^{-ix}}{2i}\right|\, dx\\
&= \int_0^{\pi} \log\left|\frac{e^{2ix}-1}{2}\right|\, dx\\
&= \int_0^{\pi} \log|e^{2ix}-1|\, dx - \int_0^{\pi} \log 2\, dx\\
&= \frac{1}{2}\int_0^{2\pi} \log|1 - e^{ix}|\, dx - \pi\log 2\\
&= \lim_{r\to 1^{-}} \frac{1}{2}\int_0^{2\pi} \log|1 - re^{ix}|\, dx - \pi\log 2\\
&= \lim_{r\to 1^{-}}\bigl(\pi\log|1 - z||_{z = 0}\bigr) - \pi \log 2 \quad (\text{by Gauss's mean value theorem})\\
&= -\pi\log 2.
\end{align}
A: Another solution is as follows.
Let 
$$ I = \int_0^{\pi} \log(\sin x)\, dx. $$
First note by symmetry, 
$$ I = 2\int_0^{\pi/2} \log(\sin x)\, dx. $$
Now letting $x = 2u$, $dx = 2du$, and $\sin{2u} = 2\sin{u}\cos{u}$,
\begin{align}
2\int_0^{\pi/2} \log(\sin x)\, dx &= 4\int_0^{\pi/4} \log(\sin 2u)\, du\\
&= 4\int_0^{\pi/4} \log(2\sin{u}\cos{u})\, du\\
&= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_0^{\pi/4} \log(\cos u)\, du.
\end{align}
In the last integral, letting $z = \pi/2 - u$, gives $dz = -du$ and $\cos u = \sin(z)$ and hence
\begin{align}
I &= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du - 4\int_{\pi/2}^{\pi/4} \log(\sin z)\, dz\\
&= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_{\pi/4}^{\pi/2} \log(\sin z)\, dz\\
&= \pi \log{2} + 2I.
\end{align}
Solving for $I$ shows
$$ I = -\pi \log{2}. $$
A: Consider $f(z)=\ln(1-z)$. Then $f$ is analytic in $|z|<r<1$. Thus by Cauchy's Formula, we have
$$\int_{|z|=r}\frac{\ln(1-z)}{z}dz= 2\pi i 0=0.$$
Letting $z=re^{i\theta}$, we have
$$ \int_0^{2\pi}r\ln\left(1-re^{i\theta}\right) \,d\theta=0. $$
Letting $r\to0^-$, we have
$$ \int_0^{2\pi}\ln\left(1-e^{i\theta}\right) \,d\theta=0 $$
so that
$$ \Re\int_0^{2\pi}\ln\left(1-e^{i\theta}\right) \,d\theta = \int_0^{2\pi}\Re\ln\left(1-e^{i\theta}\right) \,d\theta = \int_0^{2\pi}\ln\left|1-e^{i\theta}\right| \,d\theta=0. $$
But
\begin{eqnarray}
|1-e^{i\theta}|&=&|1-\cos\theta-i\sin\theta|=4\sin^2\frac{\theta}{2}
\end{eqnarray}
and hence
$$ \int_0^{2\pi}\ln\left(4\sin^2\frac{\theta}{2}\right) \,d\theta=0.$$
Changing variable $\theta\to\frac{\theta}{2}$, we have
$$\int_0^\pi\ln(\sin\theta) \,d\theta=-\pi\ln2. $$
