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?
This integral can be answered by using Cauchy's residue theorem. Note that the integrand is analytic everywhere except for $z=0$. At $z=0$, the integrand has the residue 1, as can be seen from the following limit definition of the residue of a function, $f$, at point $z_0$: $$ \operatorname{Res}(f,z_0) = \lim_{z\to z_0} (z-z_0)f(z) \implies $$ $$ \operatorname{Res}(\frac{1}{e^z-1},0)=\lim_{z\to 0}\frac{(z-0)}{e^z-1} $$ $$ =\lim_{z\to 0}\frac{z}{1+z+\frac{z^2}{2} + ...-1} $$ $$ =\lim_{z\to 0}\frac{1}{1+\frac{z}{2} + ...} $$ $$ =1 $$ We are justified in using the series expansion of $e^z$ in the third line due to the analyticity of $e^z$. Thus, by the Cauchy residue theorem, the value of the integral is simply $2 \pi i$. In your more concrete (but quite convoluted) approach, the same conclusion is reached by noting that the log function increases by $2 \pi i$ when winding one full $2 \pi$ rotation around the origin in the counter-clockwise direction.