# Integrating $\int_{\Gamma:|z|=1}\frac{e^z-1-z}{z^2}dz$, where $z\in\mathbb{C}$.

I recently came across the exercise of integrating

$$\int_{\Gamma:|z|=1}\frac{e^z-1-z}{z^2}dz,$$

where, naturally, $z\in\mathbb{C}$.

The first thing I thought of was using Cauchy's integral formula

$$f(z_0)=\frac{1}{2\pi i}\int_\Gamma\frac{f(z)}{z-z_0}dz,$$

along with its derivative

$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(\xi)}{(\xi-z)^{n+1}}d\xi.$$

The way I proceeded was by letting $f(\xi)=e^\xi-1-\xi$, implying that $f'(\xi)=e^\xi-1$, and $f'(0)=0$. Hence,

\begin{align} 2\pi i\cdot f'(z)&=\int_\Gamma\frac{f(\xi)}{(\xi-z)^2}d\xi\\ &\Longrightarrow\int_\Gamma\frac{e^\xi-1-\xi}{\xi^2}d\xi=0, \end{align}

whenever $z=0$.

Is this a correct line of thought? The only thing that I cannot wrap my head around is the fact that it seems as though the value of the integral does not necessarily depend on the contour $\Gamma$. Thanks in advance!

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Yes. This is correct. There is no surprise that the integral is independent of the contour. This is so since the integrand is analytic inside any contour except at $0$ where it has a removable singularity.

You could also look at the Laurent series of $f(z) = \dfrac{e^z-1-z}{z^2}$. $$\dfrac{e^z-1-z}{z^2} = \dfrac1{2!} + \dfrac{z}{3!} + \dfrac{z^2}{4!} + \cdots$$ Hence, the residue of $f(z)$ at $z=0$ is $0$.

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Thank you! That makes total sense now: I overlooked the integrand's analyticity property. :) – Josué Jun 26 '12 at 20:15

Another easy way to check that the singularity at $\,z=0\,$ is in fact a removable one is checking the limit exists finitely: $$\lim_{z\to 0}\frac{e^z-1-z}{z^2}\stackrel{L'Hospital}=\lim_{z\to 0}\frac{e^z-1}{2z}\stackrel{L'Hospital}=\lim_{z\to 0}\frac{e^z}{2}=\frac{1}{2}$$

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$\frac{e^z-1-z}{z^2}={1\over z^2}(-1-z+1+z+{z^2\over 2!}+{z^3\over 3!}+...)={1\over z^2}({z^2\over 2!}+{z^3\over 3!}+...)\implies$ Residue of $\frac{e^z-1-z}{z^2}$ at $0$ is $0\implies \int_{\Gamma:|z|=1}\frac{e^z-1-z}{z^2}dz=2\pi i\times 0=0.$ [Related theorem]

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