Complex Integration example

Question

Suppose I'm asked to compute the value of the complex integral: $$\int_{C}^{}\frac{\operatorname{Log}(z)}{z}\,dz$$ with $C=[i,1].$

Is it possible to treat the complex integrand like a real one and apply the rules of integration with limits $i$ to $1$, or do I ought to parameterize the given curve and then treat it like a line integral?

Answer 1

When a holomorphic function is defined on an open set containing the smooth curve and when the function also has an antiterivative there, yes you can estimate the integral as the difference of the values of the anti-derivative on the endpoints of your curve. Any proper introductory complex analysis textbook will include this

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, -\ \overbrace{\int_{1}^{\epsilon}{\ln\pars{x} \over x}\,\dd x} ^{\ds{\mbox{over}\,\,\, \pars{1,\epsilon}}}\ -\ \overbrace{\int_{0}^{\pi/2} {\ln\pars{\epsilon} + \ic\theta \over \epsilon\expo{\ic\theta}}\, \epsilon\expo{\ic\theta}\ic\,\dd\theta} ^{\ds{\mbox{over}\,\,\, \epsilon\expo{\ic\pars{0,\pi/2}}}}\ -\ \overbrace{\int_{\epsilon}^{1}{\ln\pars{y} + \ic\pi/2 \over \ic y}\,\ic\,\dd y} ^{\ds{\mbox{over}\,\,\,\pars{\ic\epsilon,\ic}}} \\[5mm] = &\ -\,{1 \over 2}\,\ln^{2}\pars{\epsilon} - \bracks{\ic\ln\pars{\epsilon}\,{\pi \over 2} - \color{#f00}{\pi^{2} \over 8}} - \bracks{-\,{1 \over 2}\,\ln^{2}\pars{\epsilon} -\ic\ln\pars{\epsilon}\,{\pi \over 2}} \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\Large \to}\,\,\, \bbx{\pi^{2} \over 8} \end{align}