TITLE: For $R>0$ , let $\gamma_R$ the arc of circumference with center in 0 and radius R (from R to -R), contained in upper half-plane.
QUESTION [0 upvotes]: For $R>0$ , let $\gamma_R$ the arc of circumference with center in 0 and radius R (from R to -R), contained in upper half-plane.
Show:
$\lim_{R \to \infty} \int_{\gamma_R} \frac{e^{iz}}{z^2}dz = 0$
Let $\epsilon >0$. I consider $\gamma_R: [0,\pi] \rightarrow \mathbb{C}$, such that $\gamma_R(t)=Re^{it}$.
Then:
$|\int_{\gamma_R} \frac{e^{iz}}{z^2}dz-0|=|\int_{\gamma_R} \frac{e^{iz}}{z^2}dz|$
I'm not really sure how to calculate this integral.

REPLY [1 votes]: Use the estimation lemma (which is pretty easy to prove in case you haven't done that):
$$\left|\int_{\gamma_R}\frac{e^{iz}}{z^2} dz\right|\le \max\left|\frac{e^{iz}}{z^2}\right|\cdot \ell(\gamma_R)=\frac{1}{R^2}\cdot \pi R\xrightarrow[R\to\infty]{}0$$
with $\;\ell(\gamma_R)=\,$ the length of $\;\gamma_R\;$
{"set_name": "stack_exchange", "score": 0, "question_id": 3799167}