Why does $\int_{C_M}{e^{iz}-1\over z}dz=0$ where $C_M$ goes through $0$?

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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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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 $z \neq 0$.