Evaluate \(\int_0^\infty\frac{\sin x}{x}\,dx\).
Solution
For \(R>1\), let \(\gamma_1\) be the straight line path from \(1/R\) to \(R\), let \(\gamma_2\) be the straight line path from \(R\) to \(R+Ri\), let \(\gamma_3\) be the straight line path from \(R+Ri\) to \(-R+Ri\), let \(\gamma_4\) be the straight line path from \(-R+Ri\) to \(-R\), let \(\gamma_5\) be the straight line path from \(-R\) to \(-1/R\), and let \(\gamma_6\) be the upper semicircle from \(-1/R\) to \(1/R\) given by the parameterization \(\gamma_6(t)=(1/R)e^{it}\) for \(t\) running from \(\pi\) to \(0\). Let \(\gamma\) be the closed loop formed by concatenating these six paths. Cauchy’s theorem implies that \(\int_\gamma\frac{e^{iz}}{z}\,dz=0\).
We have
\[ \abs{\int_{\gamma_2}\frac{e^{iz}}{z}\,dz}\leq\int_0^R\frac{e^{-t}}{R}\,dt=\frac{1-e^{-R}}{R}\to0 \]
as \(R\to\infty\). Similarly \(\int_{\gamma_4}\frac{e^{iz}}{z}\,dz\to0\), and
\[ \abs{\int_{\gamma_3}\frac{e^{iz}}{z}\,dz}\leq\int_{-R}^R\frac{e^{-R}}{R}\,dt=2e^{-R}\to0. \]
On the other hand, \(e^{iz}/z\) differs from \(1/z\) by a holomorphic function, and \(\gamma_6\) is shrinking to a point, so
\[ \begin{aligned} \lim_{R\to\infty}\int_{\gamma_6}\frac{e^{iz}}{z}\,dz&=\lim_{R\to\infty}\int_{\gamma_6}\frac1z\,dz\\ &=\lim_{R\to\infty}\int_\pi^0\frac{1}{(1/R)e^{it}}(1/R)ie^{it}\,dt\\ &=-\pi i. \end{aligned} \]
Thus
\[ \int_{\gamma_1}\frac{e^{iz}}{z}\,dz+\int_{\gamma_5}\frac{e^{iz}}{z}\,dz\to\pi i \]
as \(R\to\infty\). Taking imaginary parts and using the fact that \((\sin z)/z\) is an even function, we find that
\[ 2\int_{1/R}^R\frac{\sin z}{z}\,dz\to\pi \]
as \(R\to\infty\). Since \((\sin z)/z\) is holomorphic, it does not hurt to replace the lower limit \(1/R\) by \(0\), so \(\int_0^\infty\frac{\sin x}{x}\,dx=\pi/2\).