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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?

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    Note that $$\cos \theta_x \cos \theta_y + \sin \theta_x \sin \theta_y = \cos(\theta_y - \theta_x)$$ If you are going to work with trig functions, you have to learn to recognize that one and the similar one for sine. – 2017-02-19
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    Your integral is a 'line integral' but $y \in \mathbb{R}^{2}$. I don't understand your integral !!!. – 2017-02-19
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    @FelixMarin $y = (y_1, y_2)$ is a vector $\in \mathbb{R}^2$ and its components are given by the region of integration which is the unit circle. So in polar coordinates, $y$ becomes $(\cos \theta_y, \sin \theta_y)$. $x = (x_1, x_2)$ is a particular fixed point on the unit circle so in polar coordinates it is $(\cos \theta_x, \sin \theta_x)$. – 2017-02-19

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

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    Ok great thanks! – 2017-02-19
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    @eurocoder Thanks. I'm glad it was useful. – 2017-02-19