Can someone check my work on this integral? $$
\begin{align}
\int_0^{2\pi}\log|e^{i\theta} - 1|d\theta 
&= \int_0^{2\pi}\log(1-\cos(\theta))d\theta \\
&= \int_0^{2\pi}\log(\cos(0) - \cos(\theta))\,d\theta\\ 
&= \int_0^{2\pi}\log\left(-2\sin\left(\frac{\theta}{2}\right)\sin\left(\frac{-\theta}{2}\right)\right)\,d\theta\\ 
&= \int_0^{2\pi}\log\left(2\sin^2\left(\frac{\theta}{2}\right)\right)\,d\theta\\ 
&= \int_0^{2\pi}\log(2)d\theta + 2\int_0^{2\pi}\log\left(\sin\left(\frac{\theta}{2}\right)\right)\,d\theta\\ 
&= 2\pi \log(2) + 4\int_0^\pi \log\big(\sin(t)\big)\,dt\\
&=2\pi \log(2) - 4\pi \log(2) = -2\pi \log(2)
\end{align}
$$
Where $\int_0^\pi \log(\sin(t))\,dt = -\pi \log(2)$ according to this.  The first step where I removed the absolute value signs is the one that worries me the most.  Thanks.

You can use Jensen's Formula, I believe: http://mathworld.wolfram.com/JensensFormula.html
Edit: Jensen's formula seems to imply that your integral is zero...
