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In a PDF I am reading they say: $$\frac{1}{2\pi} \int_{0}^{2\pi}\ln|re^{i\theta}-a|d\theta=\max(\ln r,\ln|a|). $$ It is certainly a simple calculation but I can't see why. Is there someone who can explain to me. Thanks.

M. Rahmat
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3 Answers3

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What you are looking for is called Jensen's formula which gives the case when $|a|\neq r$:

Suppose that $f$ is an analytic function in a region in the complex plane which contains the closed disk $D$ of radius $r$ about the origin, $a_1$, $a_2$, ..., an are the zeros of $f$ in the interior of $D$ repeated according to multiplicity, and $f(0) \neq 0$. Jensen's formula states that $$ \log |f(0)| = \sum_{k=1}^n \log \left( \frac{|a_k|}{r}\right) + \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{i\theta})| \, d\theta. $$ This formula establishes a connection between the moduli of the zeros of the function $f$ inside the disk $D$ and the average of $\log |f(z)|$ on the boundary circle $|z| = r$, and can be seen as a generalisation of the mean value property of harmonic functions. Namely, if $f$ has no zeros in $D$, then Jensen's formula reduces to $$ \log |f(0)| = \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{i\theta})| \, d\theta, $$ which is the mean-value property of the harmonic function $ \log |f(z)|$.

When $|a|=r$, there is a singularity for the integrand and I think this case is not used in the linked paper.

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Note the following result:

If $|b|\leq 1$, then $$\int_0^{2\pi} \ln|1-be^{i\theta}| d\theta = 0$$

For $|b|< 1$, This can be proven easily by using the taylor expansion of $\ln(1-x)$ because $\int_0^{2\pi} e^{in\theta} d\theta = 0$ for each positive integer $n$. For $|b| = 1$, this follows from a continuity argument (because the integral converges).

This formula immediately implies your result by using $\ln|xy| = \ln|x|+\ln|y|$.

pisco
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{r\expo{\ic\theta} - a}}\,\dd\theta = \max\braces{\ln\pars{r},\ln\pars{\verts{a}}}:\ {\large ?}. \qquad a \in \mathbb{C}.}$

Lets $\ds{r = \verts{r}\expo{\ic\phi_{\large r}}}$ and $\ds{a = \verts{a}\expo{\ic\phi_{\large a}}}$ where $\ds{\phi_{r}, \phi_{a} \in \left[0,2\pi\right)}$. Note that \begin{align} &\bbox[15px,#ffe]{\ds{{1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{r\expo{\ic\theta} - a}}\,\dd\theta}} = {1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{\verts{r}\expo{\ic\pars{\phi_{\large r} + \theta}} - \verts{a}\expo{\ic\phi}}}\,\dd\theta \\[5mm] & = {1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{\verts{r} \expo{\ic\pars{\theta + \phi_{\large r}- \phi_{\large a}}} - \verts{a}}}\,\dd\theta = {1 \over 2\pi}\,\Re\int_{0}^{2\pi} \ln\pars{\verts{r}\expo{\ic\pars{\theta + \phi_{\large r}- \phi_{\large a}}} - \verts{a}}\,\dd\theta \\[5mm] = &\ {1 \over 2\pi}\,\Re\oint_{\verts{z}\ =\ \verts{r}}\ln\pars{z - \verts{a}} \,{\dd z \over \ic z} = {1 \over 2\pi}\,\Im\oint_{\verts{z}\ =\ \verts{r}} {\ln\pars{z - \verts{a}} \over z}\,\dd z \end{align} I'll consider the branch-cut $$ \ln\pars{z - \verts{a}} = \ln\pars{\verts{\vphantom{\Large A}z - \verts{a}}} + \mrm{arg}\pars{z - \verts{a}}\ic.\qquad -\pi < \mrm{arg}\pars{z - \verts{a}} < \pi\,,\quad z \not= \verts{a} $$

$\ds{\Large\verts{a} < \verts{r}:\ {\large ?}.\quad}$ $\ds{\large Note\ that\ r \not= 0}$. \begin{align} &{1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{r\expo{\ic\theta} - a}}\,\dd\theta \\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} {1 \over 2\pi}\,\Im\left[% -\int_{-\verts{r}}^{\verts{a}} {\ln\pars{\verts{a} - x} + \ic\pi\over x + \ic\epsilon}\,\dd x - \int_{\pi}^{-\pi}{\ln\pars{\epsilon} + \ic\theta \over \verts{a}} \,\epsilon\expo{\ic\theta}\ic\,\dd\theta\right. \\[2mm] &\ \left.\phantom{\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} {1 \over 2\pi}\,\Im\left[\,\right.} -\int_{\verts{a}}^{-\verts{r}}{\ln\pars{\verts{a} - x} - \ic\pi \over x - \ic\epsilon}\,\dd x\right] \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\large \to}\,\,\, & {1 \over 2\pi}\,\Im\left[% -\,\mrm{P.V.}\int_{-\verts{r}}^{\verts{a}} {\ln\pars{\verts{a} - x} + \ic\pi \over x}\,\dd x + \ic\pi\bracks{-\verts{r} < 0 < \verts{a}}\bracks{\ln\pars{\verts{a}} + \ic\pi}\right. + \\[2mm] & \phantom{{1 \over 2\pi}\left[-\,\,\,\right.} \left.\mrm{P.V.}\int_{-\verts{r}}^{\verts{a}} {\ln\pars{\verts{a} - x} - \ic\pi \over x}\,\dd x + \ic\pi\bracks{-\verts{r} < 0 < \verts{a}}\bracks{\ln\pars{\verts{a}} - \ic\pi} \right] \\[5mm] = &\ -\,\mrm{P.V.}\int_{-\verts{r}}^{\verts{a}}{\dd x \over x} + \bracks{a \not= 0}\ln\pars{\verts{a}} \\[5mm] = &\ -\ \underbrace{\mrm{P.V.}\int_{-\verts{r}}^{\verts{r}}{\dd x \over x}} _{\ds{=\ 0}}\ -\ \int_{\verts{r}}^{\verts{a}}{\dd x \over x} + \bracks{ar \not= 0}\ln\pars{\verts{a}} \\[5mm] & = -\bracks{a \not = 0}\ln\pars{\verts{a} \over \verts{r}} + \bracks{a \not = 0}\ln\pars{\verts{a}} = \bracks{a \not= 0}\ln\pars{\verts{r}} \end{align}
$\ds{\Large\verts{a} > \verts{r}:\ {\large ?}.\quad}$ $\ds{\large Note\ that\ a \not= 0}$. \begin{align} &{1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{r\expo{\ic\theta} - a}}\,\dd\theta = \bracks{r \not= 0}{1 \over 2\pi}\,\Im\bracks{2\pi\ic\ln\pars{-a}} = \bracks{r \not= 0}\ln\pars{\verts{a}} \end{align}
Then, $$ \bbox[#ffe,15px,border:1px dotted navy]{\ds{{1 \over 2\pi}\int_{0}^{2\pi} \ln\pars{\verts{r\expo{\ic\theta} - a}}\,\dd\theta = \max\braces{\ln\pars{r},\ln\pars{\verts{a}}}\,,\qquad ar \not= 0}} $$
Felix Marin
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