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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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  • $\begingroup$ Try with $\int\frac{1}{e^z-1}dz=\int\frac{e^\frac{z}{2}}{e^\frac{z}{2}-e^\frac{-z}{2}}dz$ and write $\sin$, $\cos$ forms. $\endgroup$
    – Nosrati
    Oct 13, 2015 at 6:55
  • $\begingroup$ If you start with $\gamma(t) = e^{2\pi i t}$, then define $u(t) = e^{2\pi i t}$, then $\gamma(t) = u(t)$, and you're right back where you started. You've "undone" your parameterization, and you arrive back at $$\int_\gamma \frac{1}{e^z - 1}\,dz.$$ $\endgroup$ Oct 14, 2015 at 19:53
  • $\begingroup$ Do you have to calculate the integral using the definition, or are you allowed to apply some theoretical result for calculating integrals like this? $\endgroup$
    – mickep
    Oct 14, 2015 at 20:03

2 Answers 2

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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.

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If $I=\int_\gamma \frac{1}{e^z-1}dz$, the residue at $z=0$ is just $1$ thus the value of the integral is $2\pi i$ times $1$, i.e.,

$$I=2\pi i (1)=2\pi i.$$

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