# Complex Integration example

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?

• It's not clear what $C=[i,1]$ means. How are you getting from $i$ to $1$? – Adrian Keister Apr 25 '18 at 19:49
• How do you define $\log z$? – José Carlos Santos Apr 25 '18 at 19:49
• C is the line segment from i to 1 and Logz is the complex logarithm. – Jevaut Apr 25 '18 at 19:54
• @Andrew Tzevas: In that case, since your path never crosses a pole and is analytic on a domain containing the path, I think you could simply compute the antiderivative and use the fundamental theorem for line integrals. – Adrian Keister Apr 25 '18 at 20:00
• $z = \mathrm{i} + \left(1 -\mathrm{i}\right)t$. – Felix Marin Apr 25 '18 at 22:25

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