How do I evaluate the following integral?
$$\oint_{C}\frac{e^z-1}{z}\mathrm dz$$ where $C$ is the unit circle (counter-clockwise).
I have just learned Cauchy-Goursat's Theorem, but I cannot apply it here since there is a singularity at $z=0$. So I try to go brute force and evaluate the contour integral as follows:
$$\int_{0}^{1}\dfrac{e^{e^{it}}-1}{e^{it}}ie^{it}\,\mathrm dt = \int_{0}^{1}({e^{e^{it}}-1})i\,\mathrm dt .$$
But then I'm stuck again as to how to evaluate this.
Thanks!
Best Answer
There is really no singularity at $0$: let $g(z)=1+\frac z {2!}+\frac {z^{2}} {3!}+...$. Then Goursat theorem can be aplied to $g$. But $f$ and $g$ are one and the same functions on the unit circle so the integral is $0$.