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I recently came across the exercise of integrating


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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up vote 1 down vote accepted

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. :) – EsX_Raptor 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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