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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?

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  • $\begingroup$ It's not clear what $C=[i,1]$ means. How are you getting from $i$ to $1$? $\endgroup$ – Adrian Keister Apr 25 '18 at 19:49
  • $\begingroup$ How do you define $\log z$? $\endgroup$ – José Carlos Santos Apr 25 '18 at 19:49
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    $\begingroup$ C is the line segment from i to 1 and Logz is the complex logarithm. $\endgroup$ – Jevaut Apr 25 '18 at 19:54
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    $\begingroup$ @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. $\endgroup$ – Adrian Keister Apr 25 '18 at 20:00
  • $\begingroup$ $z = \mathrm{i} + \left(1 -\mathrm{i}\right)t$. $\endgroup$ – Felix Marin Apr 25 '18 at 22:25
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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

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

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