How to evaluate $\frac{1}{2\pi}\int_0^{2\pi}\log\left\lvert re^{it}-\zeta\right\rvert\,dt$?

Prove that if $\zeta \in \mathbb{C}$ and $r>0$ then $$\frac{1}{2\pi}\int_0^{2\pi}\log\left\lvert re^{it}-\zeta\right\rvert\,dt = \log \left\lvert\zeta\right\rvert\,$$ if $\,r\leq \left\lvert\zeta\right\rvert$, and it is $\,\log r\,$ if $\,r> \left\lvert\zeta\right\rvert$.

My Try:

First I consider the case where $\zeta=0$. Then we have only the case where $r>\left\lvert\zeta\right\rvert$. Then the result is obvious.

Now suppose $\zeta\neq 0$. Then $\left\lvert\zeta\right\rvert>0$. Suppose $0<r'<\left\lvert\zeta\right\rvert$. Let $u\left(z\right)=\log\left\lvert z\right\rvert$ for $z\in \mathbb{C}$. Then $u\left(-\zeta\right)>-\infty$. So $u\left(z\right)$ is harmonic near $-\zeta$. So there is $\rho>0$ $(\rho$ depends on $-\zeta)$ such that $u\left(-\zeta\right)=\log\left\lvert\zeta\right\rvert=\frac{1}{2\pi}\int_0^{2\pi}\log\left\lvert re^{it}-\zeta\right\rvert \, dt\,$ for all $0\leq r<\rho$. Now if $r'<\rho$ then we are done. But what if $r'>\rho$? That is the place where I stuck. Can somebody please help me to solve it? Moreover, I don't know how to handle the case where $r'=\left\lvert\zeta\right\rvert$ for the original problem.

• $\int_0^{2 \pi} \log | r e^{it}-\zeta| dt = -iRe(\int_0^{2 \pi} \log ( r e^{it}-\zeta) \frac{1}{r e^{it}} ir e^{it} dt) = -i Re(\int_{|z|=r} \frac{\log(z-\zeta)}{z} dz)$ with $z = r e^{it}$. use the Cauchy integral formula when $|\zeta| > r$ – reuns May 28 '16 at 3:44
• This is the mean value property for the harmonic function $\log|z-\zeta |$. – user138530 May 28 '16 at 4:02
• @ChristianRemling: That is exactly my try. But I got a problem and I have mentioned it here. – Extremal May 31 '16 at 2:50
• @user1952009: How do we handle the case where $r=|\zeta|$ like this? – Extremal May 31 '16 at 3:54
• why would you be interested in the case $r = |\zeta|$ ? and you can solve it by showing the result is the mean value of $r \to 1^+$ and $r \to 1^-$ – reuns May 31 '16 at 14:14

This is a rare case where Riemann sums can be used to evaluate a definite integral of something other than a polynomial.

For $|\zeta| \neq r$, the integral $\frac1{2\pi}\int_0^{2\pi}\log\left|re^{it}-\zeta\right|\,dt$ is the limit as $N \to \infty$ of $$\frac1N \sum_{n=0}^{N-1} \log \left| r e^{2\pi i n/N} - \zeta \right| = \frac1N \log \Bigl| \, \prod_{n=0}^{N-1} (r e^{2\pi i n/N} - \zeta) \, \Bigr|.$$ Now $e^{2\pi i n/N}$ varies over the $N$-th roots of unity as as $n$ varies over $0, 1, \ldots, N-1$. Hence $$\prod_{n=0}^{N-1} (r e^{2\pi i n/N} - \zeta) = \prod_{w^n = 1} (r w - \zeta) = \pm \left( (rw)^n - \zeta^n \right).$$ Hence our integral is $$\lim_{N \to \infty} \frac1N \log \bigl| (rw)^n - \zeta^n \bigr|.$$ This limit is readily seen to equal $r$ if $|\zeta|<r$ and $|\zeta|$ if $|\zeta|>r$. For $|\zeta|=r$ the integral equals both $r$ and $|\zeta|$, as can be proved either using a limiting argument (as suggested above) or by adapting the Riemann-sum method to that case.

Added later: Here's an amusing route to the $|\zeta|=r$ case. Since then $\zeta = r e^{i\theta}$ for some real $\theta$, we can use the change of variable $t = \theta + \tau$ to reduce the problem to proving that $$\int_0^{2\pi} \log \, \left| e^{i\tau} - 1 \right| \, d\tau = 0$$ (which is the special case $r=\zeta=1$). Call the integral $I$. Then the change of variable $\tau \leftarrow \tau+\pi$ shows that $$I = \int_0^{2\pi} \log \, \left| e^{i\tau} + 1 \right| \, d\tau.$$ But then $$2I = I + I = \int_0^{2\pi} \left( \log \, \left| e^{i\tau} - 1 \right| + \log \, \left| e^{i\tau} + 1 \right| \, \right)\, dt = \int_0^{2\pi} \log \, \left| e^{2i\tau} - 1 \right| \, dt.$$ But that's also $I$, via the further change of variable $\tau \leftarrow \tau/2$: $$\int_0^{2\pi} \log \, \left| e^{2i\tau} - 1 \right| \, dt = \frac12 \int_0^{4\pi} \log \, \left| e^{i\tau} - 1 \right| \, dt = \frac12 \left( \int_0^{2\pi} \cdots \, dt + \int_{2\pi}^{4\pi} \cdots \, dt \right) = \frac12(I+I) = I.$$ Hence $2I=I$, so $I=0$ as claimed.

• But your second variable change makes limits to $\pi$ and $3\pi$ – Extremal Jun 7 '16 at 14:38
• The integrand is $2\pi$-periodic, so $\int_\pi^{3\pi} = \int_0^{2\pi}$. – Noam D. Elkies Jun 7 '16 at 14:57
• I'm a little confused for the original part ( not added later ), but I think I see how to fix. I think the product over the roots of unity should have resulted in $(-r)^N+(-\zeta)^N=\pm(r^N+\zeta^N)$. I think then the rest of the argument is still true, but shows the integral equals $\log(\max(r,|\zeta|))$, not $\max(r,|\zeta|)$ as stated. This then also agrees with the "Added later" section. – Callus - Reinstate Monica Jun 17 '16 at 12:53
• You're right that I omitted a necessary $\log$ from the answer. The formula $\pm \left( (rw)^n - \zeta^n \right)$ should have been $\pm (r^N - \zeta^N)$, but not $\pm(r^N + \zeta^N)$ because the product of $X-w$ over $w$ such that $w^N=1$ comes to $X^N-1$, not $X^N+1$. Will fix, thanks. – Noam D. Elkies Jun 17 '16 at 17:43
• Oops! quite right. Thanks. – Callus - Reinstate Monica Jun 17 '16 at 20:27

When $r\neq|\zeta|$, what you need is the Jensen's formula:

Suppose that $f$ is an analytic function in a region in the complex plane which contains the closed disk $D=\overline{B(0,r)}$ about the origin, $a_1, a_2, \cdots, a_n$ are the zeros of $f$ in the interior of $D$ repeated according to multiplicity, and $f(0)\neq 0$. Jensen's formula states that $${\displaystyle \log |f(0)|=\sum _{k=1}^{n}\log \left({\frac {|a_{k}|}{r}}\right)+{\frac {1}{2\pi }}\int _{0}^{2\pi }\log |f(re^{i\theta })|\,d\theta \tag{*}}$$

You would get the desired identities by considering now $f(z)=z-\zeta$. If $r<|\zeta|$, then the first term on the RHS of $(*)$ would be $0$ since $f$ has no zeros in the interior of $D$. If $r>|\zeta|$, then $f$ has $\zeta$ as its only zero in $D$ and the first term becomes $\log|\zeta|-\log r$.

When $r=|\zeta|$, you need a appropriate definition first for your integral $$\int_0^{2\pi}\log|re^{it}-\zeta|\ dt$$ due to the singularity on the boundary of $D$.

If $\,\left\lvert\zeta\right\rvert<r$ then $\left\lvert re^{it}-\zeta\right\rvert = \big\lvert re^{-it}-\overline\zeta\big\rvert = \big\lvert r-\overline\zeta e^{it}\big\rvert$, so we can apply the mean value property for $z\mapsto \log\big\lvert r-\overline{\zeta}z\big\rvert$ and obtain $\displaystyle\frac1{2\pi}\int = \log\big\lvert r-\overline{\zeta}\cdot0\big\rvert = \log r$.

As mentioned in the comments, $\left\lvert \zeta\right\rvert=r$ can be considered a limit case.

• you say that when $|\zeta| = r$ no need to take the mean value of the limit of $|\zeta| < r$ and $|\zeta| > r$ because they are equal, right ? – reuns May 31 '16 at 16:07
• Both sides give the same limit. – user141614 May 31 '16 at 16:08
• right, by absolute/dominated convergence it is always the case when $\int_0^{2 \pi} |f(r-r e^{it})| dt$ converges, the mean value is when a principal value is needed. – reuns May 31 '16 at 16:13
• I am still confused on the case where $r=|\zeta|$. What do you mean by a 'limit case'? what is the meaning of $r\rightarrow 1^+$, $r\rightarrow 1^-$ in that comment? Where does that $1^+,1^-$ come from? – Extremal May 31 '16 at 17:26
• I am confused on this comment; "Both sides give the same limit". How do we know that this integral is continuous on $r$? – Extremal May 31 '16 at 17:35


such that \begin{align} \partiald{\mathrm{f}\pars{\zeta,\zeta^{*}}}{\zeta} & = -\,{1 \over 4\pi}\int_{0}^{2\pi}{\dd t \over r\expo{\ic t} - \zeta} = -\,{1 \over 4\pi\ic}\oint_{\verts{z} = r}{\dd z \over z\pars{z - \zeta}} \\[3mm] = &\ \left\lbrace\begin{array}{lcrcl} \ds{\underbrace{{1 \over 2\zeta} + \pars{- {1 \over 2\zeta}}}_{\ds{=\ 0}}} & \mbox{if} & \ds{\verts{\zeta}} & \ds{<} & \ds{r} \\[3mm] \ds{{1 \over 2\zeta}} & \mbox{if} & \ds{\verts{\zeta}} & \ds{>} & \ds{r} \end{array}\right. \end{align}

Similarly, \begin{align} \partiald{\mathrm{f}\pars{\zeta,\zeta^{*}}}{{\zeta^{*}}} & = \left\lbrace\begin{array}{lcrcl} \ds{0} & \mbox{if} & \ds{\verts{\zeta}} & \ds{<} & \ds{r} \\[3mm] \ds{{1 \over 2\zeta^{*}}} & \mbox{if} & \ds{\verts{\zeta}} & \ds{>} & \ds{r} \end{array}\right. \end{align}

• Obviously, $\ds{\left.\vphantom{\LARGE A} \mathrm{f}\pars{\zeta,\zeta^{*}}\right\vert _{\ \color{#f00}{\verts{\zeta}\ <\ r}} = \mathrm{f}\pars{0,0} = \color{#f00}{\ln\pars{r}}}$.

• In addition, $\ds{\left.\vphantom{\LARGE A}\mathrm{f}\pars{\zeta,\zeta^{*}}\right\vert _{\ \color{#f00}{\verts{z}\ >\ r}} = \half\,\ln\pars{\zeta\zeta^{*}} = \color{#f00}{\ln\pars{\verts{\zeta}}}}$.