Limit of contour integrals over semi-circles

Question

For each $r>0$ let $I(r)=\int_{\gamma} \frac{e^{iz}}{z}dz$ where $\gamma(t)=re^{it}, t \in [0,\pi]$. Show that $\lim_{r \to \infty} I(r)=0$

I used $\int_{\gamma} f(z)dz=\int_{0}^{\pi}f[\gamma(t)]\cdot\gamma'(t)dt$ . but I dont get the answer.

Answer 1

Given $\epsilon > 0$ arbitrary, note that $$|I(r)| = \left|\int_0^\pi \frac{\exp(r i e^{it})}{r e^{it}} ri e^{it} dt\right| \leq \int_0^\pi \exp(-r \sin(t)) dt$$ $$= \int_0^\epsilon e^{-r \sin(t)} dt + \int_\epsilon^{\pi - \epsilon} e^{-r \sin(t)} dt + \int_{\pi - \epsilon}^\pi e^{-r \sin(t)} dt \leq 2\epsilon + (\pi - 2 \epsilon)e^{-r \sin(\epsilon)} \leq 3 \epsilon$$ for $r$ sufficiently large.