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I read an example evalutating $\int_{-\infty}^\infty {\sin x \over x}dx$ and the author defines a closed curve $C_M$ which is chaining $\Gamma_M:Me^{it}, t\in[0,\pi]$ with $[-M,M]$.

He writes:

According to Cauchy's theorem, $\int _{C_M}{e^{iz}-1\over z} dz=0$ since the integrand has no poles.

(Let $g$ be that integrand). I don't understand this equation: Cauchy Theorem demands $f$ to be analytic in an open region $D$ which contains $C_M$. But there's not such a region since $C_M$ goes throug $0$, but $g$ is not analytic in $0$. I would like for an explanation, thanks!

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    $\begingroup$ The singularity $z=0$ is removable, not a pole. $\endgroup$ – user10354138 May 21 at 17:47
  • $\begingroup$ I know 2 theorems by Cauchy in that context: The first demands analiticity of $f$ in an open region $D$ and the other one is the residue theorem. Do you know which one of Cauchy's theorems they mean? $\endgroup$ – J. Doe May 21 at 17:52
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    $\begingroup$ Alternatively, replace $C_R$ by a loop $C_R'$ that is the same as $C_R$, but that very close to $0$ does an arc around $0$ to leave it outside $C_R'$. On $C_R'$ you can apply Cauchy's theorem as you know it, and the difference between $\int_{C_R}$ and $\int_{C_R'}$ is the integral on a very little loop. Observe that $\frac{e^{iz}-1}{z}$ is bounded near $0$, therefore the integral on a very small loop near zero tends to zero with the length of the loop. Therefore $0=\int_{C_R'}$ and tends to $\int_{C_R}$ as the little loop closes. Therefore $\int_{C_R}=0$. $\endgroup$ – logarithm May 21 at 17:53
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We see that $g(z) = \sum_{n=1}^\infty {1 \over n!} \alpha^n z^{n-1}$ is entire and that ${e^{\alpha z} -1 \over z} = g(z)$ for all $z \neq 0$.

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