Contour integration with logarithms

Question

I'm having trouble calculating the below integral to get the right answer:

$$\frac{1}{2\pi i}\int_\gamma \frac{3}{z-2}\; dz$$ where $\gamma$ is parametrised by $\gamma(t)=3e^{it}, t\in [0,2\pi]$. So using this parametrisation I achieve: $$\frac{1}{2\pi i}\int_\gamma \frac{3}{z-2}\; dz=\frac{1}{2\pi}\int_0^{2\pi}\frac{9e^{it}}{3e^{it}-2}\; dt$$ Now by observation I can see that $$\frac{d}{dt}\left ( -3i \ln\left ( 3e^{it}-2 \right )\right )=\frac{9e^{it}}{3e^{it}-2}$$ and so I have that my integral is equal to $$\frac{1}{2\pi} \left [ -3i\ln(3e^{it}-2)\right ]_0^{2\pi}=0.$$ $\textbf{However}$ if I put this integral into wolfram alpha it gives the answer as $3$ (which I know is the correct answer). I also see that I could have observed that $$\frac{d}{dt}\left ( -3i\ln(2-3e^{it})\right )=\frac{9e^{it}}{3e^{it}-2}$$ and now the integral is equal to $$\frac{1}{2\pi} \left [ -3i\ln(2-3e^{it})\right ]_0^{2\pi}=3$$ Using that $\ln(-1)=i\pi$.

Therefore my question is how can two seeming correct methods give different answers? What is wrong with the first method?

Answer 1

This is something called a branch cut. The natural logarithm is actually multivalued on the complex plane; note that $e^{2\pi i}=e^{0}=1$. So is $\ln{1} = 0$ or $2\pi i$?

The way to resolve this problem is to choose a branch, that is, restrict the argument of the natural log to be between two values, say $-\pi$ and $\pi$. That means if we look at the complex plane, there is what we call a branch cut where the negative real numbers are; that's where the argument jumps by $2\pi$. Your integral from $0$ to $2\pi$ crosses the branch cut; that'll give you the $2\pi i$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Following the @Aurey user answer who is in the right direction and with $\epsilon > 0$:

\begin{align} 0 & = \lim_{\epsilon \to 0^{+}}\pars{\oint_{\gamma_{\epsilon}} {\dd z \over z - 2} + \int_{3}^{2}{\dd x \over x - 2 - \epsilon\ic} + \int_{2}^{3}{\dd x \over x - 2 + \epsilon\ic}} \end{align}

There is a 'small indent' $\pars{~\mbox{see the}\ \gamma_{\epsilon}~}$ of the above $\ds{\oint}$ integral at $\theta = 0^{+}$ and $\theta = \pars{2\pi}^{-}$ which vanish out in the limit $\epsilon \to 0^{+}$. Then,

\begin{align} \imp\color{#f00}{\quad\oint_{\gamma}{\dd z \over z - 2}} & = \lim_{\epsilon \to 0^{+}}\int_{2}^{3}\pars{{1 \over x - 2 - \epsilon\ic} - {1 \over x - 2 + \epsilon\ic}}\,\dd x \\[3mm] &= \braces{\vphantom{\Large A}% \bracks{\vphantom{\large A}\ln\pars{x - 2} + 2\pi\ic} - \bracks{\vphantom{\large A}\ln\pars{x - 2} + 0\ic}}_{\,2}^{\,3} = \color{#f00}{2\pi\ic} \end{align}

Anyway, the result could be find in a straightforward way by using the Residues Theorem.