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

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.

Answer 1

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 $\;\gamma_R\;$