Integral $\int_{0}^{2\pi}\log|e^{i \theta}-1|d \theta$
Let's try a result based on Cauchy's theorem. Let $z=e^{i \theta}$, $d\theta = -i dz/z$. Also note that $\log{|\zeta|} = \Re{[\log{\zeta}]}$. Therefore the integral is equal to
$$\Re{\left [ -i \oint_{C} dz \frac{\log{(z-1)}}{z}\right ]}$$
where $C$ is the unit circle with a small semicircular indentation into the unit circle at the branch point $z=1$. As the contribution to the integral about this branch point is zero, we need not consider this indentation further.
The integral is equal to $i 2 \pi$ times the residue at the pole $z=0$, or
$$i 2 \pi (-i) \log{(-1)} = i 2 \pi^2$$
As the original integral is the real part of this, that integral is in fact zero. Note that we could have used any possible value of $\log{(-1)}$ and come to the same conclusion.
$$|e^{ix}-1|=|\cos x+i\sin x-1|=|(\cos x-1)+i\sin x|=\sqrt{(\cos x-1)^2+\sin^2x}=$$
$$=\sqrt{(\underline{\cos}^2x-2\cos x+\underline1)+\underline{\sin}^2x}=\sqrt{2-2\cos x}=\sqrt{4\frac{1-\cos x}2}=2\cdot|\sin\tfrac x2|$$
$$\int_0^{2\pi}\ln\Big(2\cdot|\sin\tfrac x2|\Big)dx=\int_0^{2\pi}\Big(\ln2+\ln|\sin\tfrac x2|\Big)dx=2\pi\ln2+\underbrace{\int_0^{2\pi}\ln|\sin\tfrac x2|dx}_I$$
$$I=2\int_0^\pi\ln|\sin t|dt=2\int_0^\pi\ln(\sin t)dt=2\int_0^1\frac{\ln u}{\sqrt{1-u^2}}du=2\int_0^1\frac{\ln\sqrt y}{\sqrt{1-y}}\cdot\frac{dy}{2\sqrt y}=$$
$$=\frac12\int_0^1(\ln y)\cdot y^{^{-\frac12}}(1-y)^{^{-\frac12}}dy=\frac12\cdot\bigg[\frac d{dn}\int_0^1y^n(1-y)^{^{-\frac12}}dy\bigg]_{n=-\frac12}=\frac{\dfrac d{dn}B\Big(n+1;\ \tfrac12\!\Big)}2$$
$$\frac d{dn}B\Big(n+1,\tfrac12\!\Big)=\frac d{dn}\circ\frac{\Gamma(n+1)\cdot\overbrace{\Gamma\Big(\tfrac12\Big)}^\sqrt\pi}{\Gamma\Big(n+\tfrac32\Big)}=\sqrt\pi\cdot\bigg[\frac{\Gamma'(n+1)}{\Gamma\Big(n+\frac32\Big)}-\frac{\Gamma(n+1)\Gamma'\Big(n+\frac32\Big)}{\Gamma^2\Big(n+\frac32\Big)}\bigg]$$
$$\Gamma(m+1)=m!\quad,\quad n=-\frac12:\quad I=\frac{\sqrt\pi}2\cdot\bigg[\Gamma'\Big(\tfrac12\Big)-\sqrt\pi\cdot\Gamma'(1)\bigg];\quad\Gamma'(m)=H_{_{m-1}}-\gamma$$ where $H_m=\displaystyle\sum_{k=1}^m\frac1k=\int_0^1\frac{1-x^m}{1-x\quad}dx$ is the m$^\text{th}$ harmonic number and $\gamma$ is Euler-Mascheroni's constant.
$$H_{_{-\frac12}}=\int_0^1\frac{1-x^{^{-\frac12}}}{1-x\quad}dx=2\int_0^1\frac{1-t^{^{-1}}}{1-t^2\ }t\cdot dt=2\int_0^1\frac{t-1}{1-t^2}dt=-2\int_0^1\frac{dt}{1+t}=$$
$=-2\ln(1+t)|_0^1=-2\ln2.\quad$ Also, $H_0=0.\quad$ Putting it all together, we (finally!) arrive at the $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ desired result.