Prove that this function tends to 0

Question

For $r>0$, let be $$I(r)=\int_{\gamma_r}\frac{e^{iz}}{z}dz$$ where $\gamma_r:[0,\pi]\to\mathbb{C}, \gamma_r(t)=re^{it}$. Prove that $\lim_{r\to\infty}I(r)=0$.

I've tried to bound the module of the integral, but what I get is $$|I(r)|\leq \int_{\gamma_r}\left|\frac{e^{iz}}{z}\right|dz=\int_{\gamma_r}\frac{dz}{r}=\frac{\pi r}{r}=\pi.$$

Ideally, this bound would be a function of $\frac{1}{r}$, but obviously it is not. Using the parametrization got me nowhere as well, since $$\int_0^\pi\frac{e^{ire^{it}}}{re^{it}}ire^{it}dt=\int_0^\pi ie^{ire^{it}}dt$$ doesn't look (at least to me) as if it would take me somewhere (and I've tried, but you always end up with some $e^{ie^w}$ inside the integral). Any help or hints would be appreciated.

Answer 1

Hint: When $z=re^{it}$ we have $|e^{iz}|=e^{\Re (ire^{it})}=e^{-r\sin t}$ and $\int_0^{\pi} e^{-r\sin t} dt \to 0$ as $r \to \infty$ by DCT.

Answer 2

It is a particular case of Jordan's lemma. The proof goes essentially as follows. First, you split the integration region into two halves $[0,\pi/2]$ and $[\pi/2,\pi]$. Then you estimate each half separately. For example $(z=Re^{i\varphi},\ \ dz/z=id\varphi)$: \begin{gather} |I|=\Big|\int_{\rm first\ half}e^{iz}\frac{dz}{z}\Big|\leq \int_{\rm first\ half}|e^{iz}|\frac{|dz|}{|z|}= \int_0^{\pi/2}e^{-R\sin\varphi}d\varphi. \end{gather} Next, use the identity $\sin\varphi\geq \frac{2\varphi}{\pi},\ \ \varphi\in[0,\pi/2]$. Hence \begin{gather} |I|\leq \int_0^{\pi/2}e^{-\frac{2R\varphi}{\pi}}d\varphi=\frac{\pi}{2R}(1-e^{-R})\rightarrow0. \end{gather}