The whole-line principal value of sin x / x is pi, so the half-line integral is pi / 2

On the upper semicircle $z=T{e}^{it}$ one has $\mathrm{\mid }{e}^{iz}\mathrm{\mid }=e^{-T\sin t}$. Convexity of $-\sin t$ on $[0,\pi \mathrm{/}2]$ gives $\sin t\ge 2t\mathrm{/}\pi$ there, and symmetry gives the corresponding bound on the other half. Hence $\mid {\int }_{{\gamma }_T^{+}}\frac{e^{iz}}{z}dz\mid \le {\int }_0^{\pi }{e}^{-T\sin t}dt\le \frac{\pi }{T},$ so the outer arc tends to $0$.

Apply [L1] to the upper contour of radius $T$ with an upper indentation of radius $\epsilon$ at $0$. The contour encloses no pole, while the residue of $e^{iz}\mathrm{/}z$ at $0$ is $1$. Therefore the sum of the two punctured straight integrals, the indentation, and the outer arc is $0$. Letting $\epsilon \downarrow 0$ in this coupled symmetric truncation, then $T\to \mathrm{\infty }$, [L2] and step 1.1 give ${\lim }_{T\to \mathrm{\infty }}{\lim }_{\epsilon \downarrow 0}({\int }_{-T}^{-\epsilon }\frac{e^{ix}}{x}dx+{\int }_{\epsilon }^T\frac{e^{ix}}{x}dx)=i\pi \mathrm{.}$

The imaginary integrand $\sin x\mathrm{/}x$ has the removable value $1$ at $0$ and is locally integrable on the real line, so the imaginary part of step 2.1 is exactly the whole-line principal value in [L3]. Hence $\mathrm{PV}\text{\hspace{0.17em}⁣}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin x}{x}dx=\pi \mathrm{.}$ Since $\sin x\mathrm{/}x$ is even, the symmetric principal value is twice the half-line integral, so ${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}$.