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I want to integrate $f(z)=\frac{1-\mathrm{e}^{\mathrm{i}z}}{z^2}$ over the indented semicircle in the upper half-plane positioned on the $x$-axis as pictured below.

curve

The book (Complex Analysis by Stein and Shakarchi) says that to evaluate $$\int_{\gamma^+_R}\frac{1-\mathrm{e}^{\mathrm{i}z}}{z^2}dz$$ as $R\to\infty$, we should notice that $$\left|\frac{1-\mathrm{e}^{\mathrm{i}z}}{z^2}\right|\leq \frac{2}{|z|^2},$$ as this implies that the integral goes to zero.

My question is: What exactly is the reasoning here? Why does the integral go to zero?

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In general, we have the somewhat naive integral inequality

$$\int_\gamma f(z) dz \leq \underbrace{\max(f(z))}_{\text{on }\gamma} \int_\gamma dz = \underbrace{\max(f(z))}_{\text{on }\gamma} \text{ Len}(\gamma).$$

Here, we happen to see that the integrand $\displaystyle \frac{1-e^{iz}}{z^2} < \frac{2}{\lvert z \rvert^2}$. On the outer circle $\lvert z \rvert^2$ is the constant $R$, the radius of the circle. The length of the outer path is a semicircle, or $\pi R$. So in total,

$$\int_{\gamma^+_R}\frac{1-\mathrm{e}^{\mathrm{i}z}}{z^2}dz < \frac{2\pi R}{R^2} \longrightarrow 0$$

as $R \to \infty$.

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