Pf: Let $f(z)=\dfrac{e^{iz}}{z}$. We see that it's meromorphic on $\mathbb{C}$ with a simple pole at $z=0$. We evaluate $\displaystyle\int_{\Gamma} f(z)\,dz$ over the indented semi-circle contour $\Gamma=\gamma_R\cup [-R,-\varepsilon]\cup\gamma_{\varepsilon}\cup [\varepsilon, R]$ traversed counterclockwise where $\gamma_{\varepsilon}$ is a half-circle of radius $\varepsilon$ and $\gamma_R$ is a half-circle of radius $R$. Since $\Gamma$ is a closed contour, we have by Cauchy's theorem that\[\int_{\Gamma} f(z)\,dz=0.\]
Now, on $\gamma_{\varepsilon}$, $f(z) \sim \dfrac{e^0}{z}=\dfrac{1}{z}$; thus, $\displaystyle\int_{\gamma_{\varepsilon}}f(z)\,dz \sim-\pi i$ (note that this result is negative since $\gamma_{\varepsilon}$ is actually traversed clockwise). We get equality as $\varepsilon\to 0$.
On $\gamma_R$, let $z=Re^{i\theta}$ for $0\leq \theta\leq \pi$. On $\gamma_R$, we have that
\[|f(z)|= |f(Re^{i\theta})| = \left|\frac{\exp( iRe^{i\theta})}{R e^{i\theta}}\right| = \frac{|\exp(iR(\cos\theta+i\sin\theta))|}{R} = \frac{e^{-R\sin\theta}}{R}\]
and
\[|dz|=|iRe^{i\theta}\,d\theta|=R\,d\theta.\]
Therefore,
\[\left|\int_{\gamma_R}f(z)\,dz\right| \leq\int_0^{\pi} e^{-R\sin\theta}\,d\theta.\]
Now, for any $R$, $\exp(-R\sin\theta)$ is bounded above by $1$. But as $R\to\infty$, $\exp(-R\sin\theta)\to 0$ and hence $\displaystyle\int_{\gamma_R}f(z)\,dz\rightarrow 0$
The remaining two pieces of the integral are
\[\begin{aligned} \int_{-R}^{\varepsilon}\frac{e^{ix}}{x}\,dx+ \int_{\varepsilon}^R \frac{e^{ix}}{x}\,dx &= \int_R^{\varepsilon} \frac{e^{-ix}}{-x}\,d(-x)+\int_{\varepsilon}^R\frac{e^{ix}}{x}\,dx\\ &= \int_{\varepsilon}^R \frac{e^{ix}-e^{-ix}}{x}\,dx\\ &= 2i\int_{\varepsilon}^R\frac{\sin x}{x}\,dx.\end{aligned}\]
Putting all the pieces together, we see that as $\varepsilon\to 0$ and $R\to\infty$, we have
\[\int_{\Gamma}f(z)\,dz=0\implies 2i\int_0^{\infty}\frac{\sin x}{x}\,dx - \pi i = 0\implies \int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}.\]
The justification is now complete. $\hspace{4in}\blacksquare$