Complex contour integration of $\frac{e^{iz}}{z}$

Question

While calculating $\int_0^\infty \frac{\sin(x)}{x} dx $ I integrated the complex function $f(z) = \frac{e^{iz}}{z}$ over the contour $C = [-R,R] \cup \gamma_R $. $\gamma_R = R e^{it}$ where $0 \leq t \leq \pi$. I had some trouble to show that $\int_{\gamma_R} f(z) \rightarrow 0$ as $R \rightarrow \infty$. Is it possible to show this without using Jordan's Lemma but instead using the ML estimate?

Answer 1

Hint: $\int_0^{\pi} e^{-R \sin \, t} dt \to 0$ by DCT.

Answer 2

You have \begin{align} \left\lvert\int_{\gamma_R}\frac{e^{iz}}z\,\mathrm dz\right\rvert&= \left\lvert\int_0^{1/2}\frac{e^{iRe^{2\pi it}}}{Re^{2\pi it}}2\pi iR e^{2\pi it}\,\mathrm dt\right\rvert\\ &=\left\lvert2\pi i\int_0^{1/2}e^{-R\sin(2\pi t)+iR\cos(2\pi t)}\,\mathrm dt \right\rvert\\ &\leqslant2\pi\int_0^{1/2}\left\lvert e^{-R\sin(2\pi t)+iR\cos(2\pi t)}\right\rvert \,\mathrm dt\\ &=2\pi\int_0^{1/2}e^{-R\sin(2\pi t)}\,\mathrm dt\\ &=4\pi\int_0^{1/4}e^{-R\sin(2\pi t)}\,\mathrm dt\text{ (since $(\forall t\in[0,\pi]):\sin(t)=\sin(\pi-t)$)}\\ &\leqslant4\pi\int_0^{1/4}e^{-4Rt}\,\mathrm dt\text{ (since $\left(\forall t\in\left[0,\frac14\right]\right): \sin(2\pi t)\geqslant4t$)}\\ &=4\pi\left[\frac{e^{-4Rt}}{-4R}\right]_{t=0}^{t=1/4}\\ &=\pi\frac{1-e^{-R}}R. \end{align} Now, use the fact that$$\lim_{R\rightarrow\infty}\pi\frac{1-e^{-R}}R=0.$$