Compute $\int_{\gamma}{Log(z)\over z}dz$ for $\gamma(t)=e^{it}$, $t\in[0,2\pi]$ 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?
 A: $$
\begin{align}
\int_\gamma\frac{\log(z)}{z}\,\mathrm{d}z
&=\int_\gamma\log(z)\,\mathrm{d}\log(z)\\
&=\left.\frac12\log(z)^2\right]_{\large e^0}^{\large e^{2\pi i}}\\[9pt]
&=-2\pi^2
\end{align}
$$
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.

A: 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 $$
\int_\gamma \frac{\operatorname{Log (z)}}{z}dz=\int_0^{2\pi} \operatorname{Log}(e^{it})idt=-\int_0^{2\pi}tdt
=-2\pi^2.$$
