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

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!

• The singularity $z=0$ is removable, not a pole. – user10354138 May 21 at 17:47
• 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? – J. Doe May 21 at 17:52
• 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 $\int_{C_R}=0$. – logarithm May 21 at 17:53

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 $$z \neq 0$$.