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Consider the integral

$I=\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt$

where $a$ is a complex number and $0. We have

$I=\operatorname{Re}\left(\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt\right)$

Let $\gamma=\partial D(0,r)$. Then

$\begin{align}\int_\gamma\frac{\log(z-a)}{iz}\,dz&=\int_0^{2\pi}\frac{\log\left(re^{it}-a\right)} {ire^{it}}rie^{it}\,dt\\ &=\int_0^{2\pi}\log\left(re^{it}-a\right)\,dt\end{align}$

Thus

$I=\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)$

Now my problem is that $\log(z-a)$ is not holomorphic in $D(0,r)$, so i can't use Cauchy's integral formula to compute $I$. How can I solve this?

  • 2
    Just choose$a$branch of the logarithm that is holomorphic on your disc (not necessarily the principal branch), which is possible since $a \notin D(0,r)$.2012-11-16

2 Answers 2

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Because $0 $\ln(a-z)=\ln{a\left(1-\dfrac{z}{a}\right)}=\ln{a}+\ln{\left(1-\dfrac{z}{a}\right)}=\ln{a}+\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\left(\frac{z}{a} \right)^k }.$ Therefore, Laurent expansion for $\frac{\ln(a-z)}{iz}=-i\frac{\ln(a-z)}{z}$ is $-i\frac{\ln{a}}{z}-i\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\frac{z^{k-1}}{a^k } }.$ Using the residue theorem for integral $\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz$ gives $\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz=2\pi{i}\cdot(-i\ln{a})=2\pi\ln{a}.$ Taking the real part, $\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)=\operatorname{Re}(2\pi\ln{a})=2\pi\ln{|a|}.$

  • 0
    ok, i thank you all for your help. I think there exists a definition of log such that log is holomorphic in every simply connected open set not containing the origin. But i really can't understand this fact in practice. What branch of log i take in order to be sure that it is holomorphic in the disc $D(0,r)$. I'm quite confused about this. Is it sufficient to say: ok, $z-a\neq 0$ hence log is holomorphic?2012-11-16
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{a \in {\mathbb C}}$.

Lets $\ds{z \equiv r\expo{\ic t}\ \imp\ \dd z = r\expo{\ic t}\ic\,\dd t\ \imp\dd t = {\dd z \over \ic z}}$:

\begin{align} I&=\left.\int_{0}^{2\pi}\ln\pars{\verts{re^{\ic t} - a}}\,\dd t \,\right\vert_{\,0\ <\ r\ <\ \verts{a}}\ =\ \Re\int_{0}^{2\pi}\ln\pars{re^{\ic t} - a}\,\dd t \\[5mm] & = \Re\oint_{0\ <\ \verts{z}\ =\ r\ <\ \verts{a}} \ln\pars{z - a}\,{\dd z \over \ic z} = \Re\pars{2\pi\ic\lim_{z\ \to\ 0}\bracks{z\,{\ln\pars{z - a} \over \ic z}}} = 2\pi\,\Re\pars{\ln\pars{-a}} \\[5mm] &= \bbox[10px,border:1px groove navy]{2\pi\ln\pars{\verts{a}}} \end{align}

  • 0
    @user68472 That's the [Residue Theorem](https://en.wikipedia.org/wiki/Residue_theorem). That's the way to evaluate the residue of $\displaystyle{\ln\left(z - a\right) \over \mathrm{i}z}$ at $\displaystyle z = 0$. Thanks.2018-03-16