Show that
\[ \int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2} .\]
Hint: use the fact that this integral equals \(\frac{1}{2 i} \int_{-\infty}^{\infty} \frac{e^{i x}-1}{x} d x\), and integrate around an indented semicircle.
Solution
Let \(f(z) \coloneqq e^{iz}/z\). For \(0<\varepsilon<R\), let \(\gamma_\varepsilon\) be the upper semicircle \(z=\varepsilon e^{it}\) traversed from \(-\varepsilon\) to \(\varepsilon\) (\(t\) from \(\pi\) to \(0\)), and let \(\Gamma_R\) be the upper semicircle \(z=Re^{it}\), \(t\in[0,\pi]\), traversed counterclockwise. The closed contour \(C_{\varepsilon,R}\) consists of \([-R,-\varepsilon]\), \(\gamma_\varepsilon\), \([\varepsilon,R]\), and \(\Gamma_R\).
1. \(\displaystyle\int_0^{\infty} \frac{\sin x}{x}\, dx = \frac{1}{2}\, \Im\left(\operatorname{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx\right)\), where\[\operatorname{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx \coloneqq \lim_{\varepsilon\to0,\,R\to\infty}\left(\int_{-R}^{-\varepsilon}+\int_{\varepsilon}^{R}\right)\frac{e^{ix}}{x}\,dx.\]
1.1. \(\dfrac{e^{ix}}{x} = \dfrac{\cos x}{x} + i\,\dfrac{\sin x}{x}\) for real \(x\neq0\).
This is Euler’s formula \(e^{ix} = \cos x + i \sin x\).
1.2. \(\displaystyle\left(\int_{-R}^{-\varepsilon}+\int_{\varepsilon}^{R}\right)\frac{\cos x}{x}\,dx = 0\).
The function \(\cos x/x\) is odd and the domain of integration is symmetric about \(0\).
1.3. \(\displaystyle\left(\int_{-R}^{-\varepsilon}+\int_{\varepsilon}^{R}\right)\frac{\sin x}{x}\,dx = 2\int_\varepsilon^{R} \frac{\sin x}{x}\, dx\).
The function \(\sin x / x\) is even.
1.4. Q.E.D.
By steps <2>1–<2>3, the imaginary part of the truncated integral of \(e^{ix}/x\) is \(2\int_\varepsilon^R \frac{\sin x}{x}\,dx\); let \(\varepsilon\to0\) and \(R\to\infty\).
2. \(\displaystyle\int_{C_{\varepsilon,R}} f(z)\,dz = 0\).
The only singularity of \(f\) is the simple pole at \(z=0\), which the indentation \(\gamma_\varepsilon\) excludes. Hence \(f\) is holomorphic on a neighborhood of the closed region bounded by \(C_{\varepsilon,R}\), and Cauchy’s theorem applies.
3. \(\displaystyle\int_{\gamma_\varepsilon} f(z)\,dz \to -i\pi\) as \(\varepsilon \to 0\).
With \(z = \varepsilon e^{it}\), \(dz/z = i\,dt\), so \[\int_{\gamma_\varepsilon} f\, dz = i\int_\pi^0 e^{i\varepsilon e^{it}}\, dt.\] The integrand converges uniformly to \(1\) as \(\varepsilon\to0\), so the integral tends to \(i\int_\pi^0 dt=-i\pi\).
4. \(\displaystyle\int_{\Gamma_R} f(z)\,dz \to 0\) as \(R \to \infty\).
With \(z = Re^{it}\), \(dz/z = i\,dt\), so \[\int_{\Gamma_R} f\, dz = i\int_0^\pi e^{iRe^{it}}\, dt, \qquad \abs{e^{iRe^{it}}} = e^{-R\sin t} \le 1.\] For each \(t\in(0,\pi)\), \(e^{-R\sin t}\to0\) as \(R\to\infty\). Dominated convergence, with dominating function \(1\) on \([0,\pi]\), gives \(\int_0^\pi e^{-R\sin t}\,dt\to0\), hence the claim.
5. \(\displaystyle\operatorname{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx = i\pi\).
By step 2, \[0 = \int_{-R}^{-\varepsilon} f + \int_{\gamma_\varepsilon} f + \int_{\varepsilon}^{R} f + \int_{\Gamma_R} f.\] Let \(\varepsilon \to 0\) and \(R \to \infty\). Steps 3 and 4 give \(\operatorname{PV}\!\int_{-\infty}^{\infty} \frac{e^{ix}}{x}\, dx - i\pi = 0\).
6. Q.E.D.
Steps 1 and 5 give \(\int_0^{\infty} \frac{\sin x}{x}\, dx = \frac{1}{2} \Im(i\pi) = \boxed{\frac{\pi}{2}}\).