Solution: Consider the function $f(z)=\frac{e^{iz}}{z}$ and the contour $C$ consisting of

1. A semicircle ${\gamma }_R$ of radius $R$ in the upper half-plane,
2. The real axis from $-R$ to $-ϵ$,
3. A small semicircle ${\gamma }_{ϵ}$ of radius $ϵ$ around the origin in the upper half-plane,
4. The real axis from $ϵ$ to $R$.

By Cauchy's residue theorem, since $f(z)$ is analytic inside and on $C$ except at $z=0$, we have: $${\oint }_{C}f(z)dz=0.$$ Breaking the integral into parts: $${\int }_{-R}^{-ϵ}\frac{e^{ix}}{x}dx+{\int }_{{\gamma }_{ϵ}}\frac{e^{iz}}{z}dz+{\int }_{ϵ}^R\frac{e^{ix}}{x}dx+{\int }_{{\gamma }_R}\frac{e^{iz}}{z}dz=0.$$ As $R\to \mathrm{\infty }$, the integral over ${\gamma }_R$ vanishes by Jordan's lemma. As $ϵ\to 0$, the integral over ${\gamma }_{ϵ}$ approaches $-i\pi$ (half the residue at $z=0$). Taking the imaginary part of the remaining integrals, we get: $$2i{\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=i\pi .$$ Therefore: $${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}.$$