This problem is about the integral
\[I = \int_{-\infty}^{\infty} \frac{\sin x}{x} \, dx.\]
Show directly that \(I\) is a convergent improper Riemann integral.
Use a contour integral to evaluate \(I\).
Solution
The improper integral exists and equals \(\boxed{\pi}\).
1. The integral converges as an ordinary improper Riemann integral.
The quotient \(\sin x/x\) extends continuously across zero with value one. For \(B>A>0\), integration by parts gives \[ \int_A^B\frac{\sin x}{x}\,dx =\left[-\frac{\cos x}{x}\right]_A^B -\int_A^B\frac{\cos x}{x^2}\,dx. \] Consequently the absolute value of this tail is at most \[ \frac1A+\frac1B+\int_A^B\frac{dx}{x^2}=\frac2A. \] The Cauchy criterion proves convergence at positive infinity. Evenness gives convergence at negative infinity separately. Thus \(I=2\int_0^\infty\sin x/x\,dx\) is an improper integral, not merely a symmetric principal value.
2. An indented upper semicircle evaluates the half-line integral.
For \(0<\varepsilon<R\), integrate \(F(z)=e^{iz}/z\) over \([-R,-\varepsilon]\), the clockwise upper semicircle \(\gamma_\varepsilon\) from \(-\varepsilon\) to \(\varepsilon\), \([\varepsilon,R]\), and the counterclockwise upper semicircle \(\Gamma_R\) from \(R\) to \(-R\). The enclosed upper half-annulus contains no pole, so Cauchy’s theorem gives total integral zero [@SS03]. The real-segment contribution is \[ \int_{-R}^{-\varepsilon}\frac{e^{ix}}x\,dx +\int_\varepsilon^R\frac{e^{ix}}x\,dx =2i\int_\varepsilon^R\frac{\sin x}x\,dx. \] Parametrizing the small arc by \(z=\varepsilon e^{it}\), with \(t\) decreasing from \(\pi\) to zero, gives \[ \int_{\gamma_\varepsilon}F(z)\,dz =i\int_\pi^0e^{i\varepsilon e^{it}}\,dt \longrightarrow-i\pi, \] since the integrand tends uniformly to one.
For the large arc, the same parametrization with \(R\) gives \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\int_0^\pi e^{-R\sin t}\,dt \leq2\int_0^{\pi/2}e^{-2Rt/\pi}\,dt \leq\frac\pi R\longrightarrow0. \] Here symmetry reduces the estimate to \([0,\pi/2]\), and concavity of sine on that interval gives \(\sin t\geq2t/\pi\). Passing to the limits in the contour identity, using step 1 for the real integral, yields \[ 2i\int_0^\infty\frac{\sin x}x\,dx-i\pi=0. \] The half-line value is \(\pi/2\), and evenness gives \(I=\pi\).