How to integrate $\int_{0}^{1}\text{log}(\text{sin}(\pi x))\text{d}x$ using complex analysis

Let $I$ be the integral given by

$$I=\int_0^1 \log(\sin(\pi x))\,dx \tag 1$$

Enforcing the substitution $ x \to x/\pi$ in $(1)$ reveals

$$\begin{align} I&=\frac{1}{\pi}\int_0^\pi \log(\sin(x))\,dx \\\\ &=\frac{1}{2\pi}\int_0^\pi \log(\sin^2(x))\,dx\\\\ &=\frac{1}{2\pi}\int_0^\pi \log\left(\frac{1+\cos(2x)}{2}\right)\,dx \tag 2 \end{align}$$

Next, enforcing the substitution $x\to x/2$ in $(2)$ yields

$$\begin{align} I&=\frac{1}{4\pi}\int_0^{2\pi} \log\left(\frac{1+\cos(x)}{2}\right)\,dx \\\\ &=-\frac{\pi}{2}\log(2)+\frac{1}{4\pi}\int_0^{2\pi}\log(1+\cos(x))\,dx\tag 3 \end{align}$$

We now move to the complex plane by making the classical substitution $z=e^{ix}$. Proceeding, $(3)$ becomes

$$\begin{align} I&=-\frac{\pi}{2}\log(2)+\frac{1}{4\pi}\oint_{|z|=1}\log\left(1+\frac{z+z^{-1}}{2}\right)\,\frac{1}{iz}\,dz \\\\ &=-\frac{\pi}{2}\log(2)+\frac{1}{4\pi i}\oint_{|z|=1}\frac{2\log(z+1)-\log(z)}{z}\,dz \tag 4 \end{align}$$

Note that the integrand in $(4)$ has branch points at $z=0$ and $z=-1$, and a first-order pole at $z=0$. We choose to cut the plane with branch cuts from $z=0$ to $z=-\infty$, and from $z=-1$ to $z=-\infty$, along the non-positive real axis.

We can deform the contour $|z|=1$ around the branch cuts and the pole and write

$$\begin{align} \oint_{|z|=1}\frac{2\log(z+1)-\log(z)}{z}\,dz &=\lim_{\epsilon \to 0^+}\left(\int_{-\epsilon}^{-1} \frac{2\log(x+1)-\log|x| -i\pi}{x}\,dx \right.\\\\ &-\int_{-\epsilon}^{-1} \frac{2\log(x+1)-\log|x|+i\pi}{x}\,dx\\\\ &\left. +\int_{-\pi}^{\pi}\frac{2\log(1+\epsilon e^{i\phi})-\log(\epsilon e^{i\phi})}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\right)\\\\ &=0 \tag 5 \end{align}$$

Finally, using $(5)$ in $(4)$, we obtain the coveted equality

$$I=-\frac{\pi}{2}\log(2)$$

as was to be shown!