Showing $\int_C \ln |x - y| dy = 0$ where $C$ is the unit circle? I saw it mentioned in a proof that
$$\int_C \ln |x - y| dy = 0,$$
where $x, y \in \mathbb{R}^2$ and they are both on the unit circle $C$. I'm trying to figure out the calculation but I'm not sure how to show it. First of all I switch to polars and get
$$\int_0^{2\pi} \ln \sqrt{(\cos \theta_x - \cos\theta_y)^2 - (\sin \theta_x - \sin\theta_y)^2} d\theta_y \\ = \frac{1}{2}\int_0^{2\pi} \ln (\cos^2 \theta_x - 2 \cos \theta_x \cos \theta_y + \cos^2 \theta_y + \sin^2 \theta_x - 2 \sin \theta_x \sin \theta_y + \sin^2 \theta_y) d\theta_y \\ = \frac{1}{2}\int_0^{2\pi} \ln (2 - 2 \cos \theta_x \cos \theta_y - 2 \sin \theta_x \sin \theta_y) d\theta_y \\ = \frac{1}{2}\int_0^{2\pi} \ln 2 d\theta_y  + \frac{1}{2}\int_0^{2\pi} \ln (1 - (\cos \theta_x \cos \theta_y + \sin \theta_x \sin \theta_y) d\theta_y$$
What steps are needed to get this expression equal to zero?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Your last integral is, indeed, $\ds{\theta_{x}}$-independent $\ds{\pars{~Why\ ?~}}$. So,

\begin{align}
&{1 \over 2}\int_{0}^{2\pi}\ln\pars{1 - \cos\pars{\theta_{y}}}\,\dd\theta_{y} =
{1 \over 2}\int_{0}^{2\pi}\ln\pars{2\sin^{2}\pars{\theta_{y} \over 2}}
\,\dd\theta_{y}
\\[5mm] = &
{1 \over 2}\int_{0}^{\pi}\bracks{\ln\pars{2\sin^{2}\pars{\theta_{y} \over 2}} +
\ln\pars{2\cos^{2}\pars{\theta_{y} \over 2}}}\,\dd\theta_{y}
\\[5mm] = &\
\int_{0}^{\pi}\ln\pars{\sin\pars{\theta_{y}}}\,\dd\theta_{y} = \bbx{\ds{-\pi\ln\pars{2}}}
\end{align}

It cancels the first integral $\ds{{1 \over 2}\int_{0}^{2\pi}\ln\pars{2}\,\dd\theta_{x}}$ !!!.

