Evaluate the following integral:
$\int_\gamma \frac{1}{e^z - 1}dz$ where $\gamma:[0,1]\rightarrow \mathbb C$ is a parameterization of the unit circle oriented counter clockwise.
Attempt at the solution:
Use the parameterization $\gamma(t) = e^{2\pi it}$
$$\int_\gamma \frac{1}{e^z - 1}dz = \int_\gamma \frac{e^{-z}}{1-e^{-z}}dz = \int_{0}^{1}\frac{2\pi ie^{2\pi it}e^{e^{2\pi it}}}{1- e^{e^{2\pi it}}}dt$$
let $u(t) = e^{2\pi it}$ so that $du = 2\pi ie^{2\pi it}dt$
$$\int^{e^{2\pi i}}_{1} \frac{e^{-u}}{1-e^{-u}}du$$
let $v(t) = 1-e^{-u}$ so that $dv = e^{-u}du$
$$\int^{1-e^{2\pi i}}_{1-e^{-1}} \frac{1}{v}dv
= \log(1-e^{2\pi i}) - \log(1-e^{-1})$$
I think there is something fishy going on here. Since the $log$ function is not defined the same way as in the case of the real numbers. Can we treat the upper and lower limits in the integral in the same way as with real numbers? Are the change of variables techniques still applicable?
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