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Compute $\int_{\gamma}{Log(z)\over z}dz$ for $\gamma(t)=e^{it}$, $0\le t\le (2\pi)$. (Why is it that using "\le" code suddenly creates "2"?). Before you vote to close this question, know that its duplicate has a confirmed answer understood to the OP but unfamiliar to me in its method. And this is a way for me to, through this question, to better understand the nature of integrals and Logarithm in an adjustable and convenient format.

The use of $\text{Log(z)}$ probably refer to the principal logarithm, but it is defined on $(-\pi,\pi]$. If I split the integral, what should be done with the second part? Another exercise looking at $e^{it},t\in[0,\pi]$ stated that $Log(e^{it})$ is simply $it$ in a well-defined manner, but here it is quite confusing. I don't understand why and how to change this contour to another. Can you please contribute some theory regarding that problem?

Meitar
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    The keyword is : holomorphic anti-derivative. – reuns Dec 28 '16 at 19:34
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    Usually, if you just mention why you are asking a previously asked question, and if it is a good reason, no one will vote to close. – The Count Dec 28 '16 at 19:46
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    Yes, it's the principal branch of the logarithm, which is not continues on $(-\infty,0]$, which includes $-\pi$. What you can do is isolate the $-\pi$ with a small $\varepsilon$ and then take the limit, like in this example http://math.stackexchange.com/questions/1989887/cauchy-integral-theorem-and-the-complex-logarithm-function/ – rtybase Dec 28 '16 at 20:14

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$$ \begin{align} \int_\gamma\frac{\log(z)}{z}\,\mathrm{d}z &=\int_\gamma\log(z)\,\mathrm{d}\log(z)\\ &=\left.\frac12\log(z)^2\right]_{\large e^0}^{\large e^{2\pi i}}\\[9pt] &=-2\pi^2 \end{align} $$ Note that $\log(z)$ is $2\pi i$ greater at the end of the path of integration than at the start because of the branch cut.

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robjohn
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  • Is the antiderivative $\frac{1}{2}\log(z)^2$ differentiable at the intersection of the branch cut and the circle? If it is not differentiable, do we need to separate the contour into two parts, then we take different branch cut? My textbook uses this approach. Example 3 on page 143, starting at the paragraph after example 2. https://math.unice.fr/~nivoche/pdf/Brown-Churchill-Complex%20Variables%20and%20Application%208th%20edition.pdf – user398843 Oct 01 '18 at 02:18
  • On each side of the branch cut, it is differentiable, except at $z=0$, but $\lim\limits_{z\to0}\left|z,\log(z)^2\right|=0$ is sufficient. – robjohn Oct 01 '18 at 03:06
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Since for $z=e^{it}, t\in (\pi,2\pi]$ $$ \operatorname{Log}(z)=\log |e^{it}|+i\arg \left(e^{it}\right)=i\arg e^{i(t-2\pi)}=i(t-2\pi), $$ the second part of the integral should be

$$ \int_\pi^{2\pi} \frac{\operatorname{Log}(e^{it})}{e^{it}}ie^{it}dt=-\int_\pi^{2\pi}\arg\left(e^{i(t-2\pi)}\right)dt= -\int_\pi ^{2\pi}(t-2\pi)dt=\frac{1}{2}\pi^2. $$ Note $\arg(e^{it})=t-2\pi, t\in (\pi,2\pi]$.

Of course the first part is simply $$ -\int_0^{\pi} \arg \left(e^{it}\right)dt=-\int_0^\pi tdt=-\frac{\pi^2}{2}. $$ Thus we have $$ \int_\gamma \frac{\operatorname{Log}(z)}{z}dz=\frac{\pi^2}{2}-\frac{\pi^2}{2}=0. $$

Comment:
If we define $\operatorname{Log}(z)=\operatorname{Log}(e^{it})=i\arg (e^{it})=it $ for $z=e^{it},0\le t<2\pi$, then the integral should be $$ \int_\gamma \frac{\operatorname{Log (z)}}{z}dz=\int_0^{2\pi} \operatorname{Log}(e^{it})idt=-\int_0^{2\pi}tdt =-2\pi^2.$$

ts375_zk26
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