# Compute integral $\int_0^{2\pi} \ln|R e^{i\phi}-a|d\phi$

How to evaluate this integral without using Jensen's formula? $\int_0^{2\pi} \ln|R e^{i\phi}-a|d\phi$, where $R>0$ and $a \in \mathbb{C}$. Thank you!

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If $a=0$, then he value of the integral is $2\,\pi\ln R$. From now on I will assume $a=\alpha+i\,\beta\ne0$, $\alpha,\beta\in\mathbb{R}$.

First case: $|a|>R$. Then the function $$f(x,y)=\frac12\ln\bigl((x-\alpha)^2+(y-\beta)^2\bigr)=\ln|x+i\,y-a|$$ is harmonic on an open set containing the closed disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2\le R^2\}$. By the mean value property of harmonic functions $$\int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi=2\,\pi\,f(0,0)=\frac12\ln\bigl(\alpha^2+\beta^2\bigr)=2\,\pi\,\ln|a|.$$

Second case: $|a|<R$. Let $a=|a|\,e^{i\psi}$. Then $$\int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi=\int_0^{2\pi} \ln|R\,e^{-i\psi}-|a|e^{-i\phi}|\,d\phi=\int_0^{2\pi} \ln||a|e^{i\phi}-R\,e^{-i\psi}|\,d\phi.$$ By the result for the first case, this integral is equal to $2\,\pi\ln R$.

Third case: $|a|=R$. Let $a=|a|\,e^{i\psi}$. Then \begin{align*} \int_0^{2\pi} \ln|R\,e^{i\phi}-a|\,d\phi&=2\,\pi\ln R+\int_0^{2\pi} \ln|e^{i\phi}-e^{i\psi}|\,d\phi\\ &=2\,\pi\ln R+\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi\\ &=2\,\pi\ln R. \end{align*}

The las equality follows from \begin{align*} \int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi&=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_\pi^{2\pi} \ln|e^{i\phi}-1|\,d\phi\\ &=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_0^{\pi} \ln|e^{i(\phi+\pi)}-1|\,d\phi\\ &=\int_0^{\pi} \ln|e^{i\phi}-1|\,d\phi+\int_0^{\pi} \ln|e^{i\phi}+1|\,d\phi\\ &=\int_0^{\pi} \ln|e^{2i\phi}-1|\,d\phi\\ &=2\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi, \end{align*} which implies $$\int_0^{2\pi} \ln|e^{i\phi}-1|\,d\phi=0.$$

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+1.    – Did Apr 13 '12 at 11:45
@Julián Aguirre: Very nice proof, Julian. Thank you! – Tran Apr 14 '12 at 3:25