Contour Integral of Upper Semi Circle of Radius $R$

Question

I have been trying to solve the following question using only what is given without ANY extra theorems. I have been at it for an hour and not getting anywhere. I hope that someone could give me an extra hint to simplify the integral. I have used the parametrization $e^{it}$. Image of question

Answer 1

If you use the parametrization $z=e^{it}, 0\le t\le \pi$, the integral would be$$ \int_{\gamma_R} \frac{e^{iz}}{z}dz=\int_0^\pi \frac{e^{iRe^{it}}}{Re^{it}}Re^{it}idt=i\int_0^\pi e^{iRe^{it}}dt. $$ Divide the RHS integral into three parts,$$ \int_0^\frac{1}{\sqrt{R}} e^{iRe^{it}}dt+\int_\frac{1}{\sqrt{R}}^{\pi-\frac{1}{\sqrt{R}}} e^{iRe^{it}}dt+\int_{\pi-\frac{1}{\sqrt{R}}}^\pi e^{iRe^{it}}dt. $$ Since $\left|e^{iRe^{it}}\right|=\left|e^{-R\sin t+iR\cos t}\right|=e^{-R\sin t}\le 1$, we have $$ \left|\int_0^\frac{1}{\sqrt{R}} e^{iRe^{it}}dt\right|\le \int_0^\frac{1}{\sqrt{R}} \left|e^{iRe^{it}}\right|dt\le \int_0^\frac{1}{\sqrt{R}} dt= \frac{1}{\sqrt{R}}, $$ which tends to $0$ as $R\to \infty.$ The same for $ \int_{\pi-\frac{1}{\sqrt{R}}}^\pi e^{iRe^{it}}dt$.

For the middle part, we have \begin{align} \left|\int_\frac{1}{\sqrt{R}}^{\pi-\frac{1}{\sqrt{R}}} e^{iRe^{it}}dt\right|&\le \int_\frac{1}{\sqrt{R}}^{\pi-\frac{1}{\sqrt{R}}} e^{-R\sin t}dt\le e^{-R\sin \frac{1}{\sqrt R}}\int_\frac{1}{\sqrt{R}}^{\pi-\frac{1}{\sqrt{R}}} dt\\ &= e^{-\sqrt{R}\cdot \sqrt{R}\sin \frac{1}{\sqrt R}}\left(\pi -2\frac{1}{\sqrt{R}}\right)\\ &\le e^{-\sqrt{R}\cdot \frac{1}{2}}\left(\pi -\frac{2}{\sqrt{R}}\right) \end{align} for sufficiently large $R$, and the last term tends to $0$ as $R\to \infty$. Thus we have $$ \lim_{R\to \infty} \int_{\gamma_R} \frac{e^{iz}}{z}dz=0.$$