For \(t \in \mathbb{R}\), compute
\[\lim_{A \to \infty} \int_{-A}^{A} \frac{\sin x}{x} e^{ixt} \, dx.\]
Solution
The requested limit is \[ \boxed{\begin{cases} \pi,& |t|<1,\\ \pi/2,& |t|=1,\\ 0,& |t|>1. \end{cases}} \] We give \(\sin x/x\) its continuous value one at zero.
1. The basic half-line integral is \(\int_0^\infty\sin x/x\,dx=\pi/2\).
For \(B>A>0\), integration by parts gives \[ \int_A^B\frac{\sin x}{x}\,dx =\left[-\frac{\cos x}{x}\right]_A^B -\int_A^B\frac{\cos x}{x^2}\,dx, \] so the absolute value is at most \(2/A\). This proves convergence at infinity by the Cauchy criterion; continuity settles the origin.
For \(0<\varepsilon<R\), integrate \(F(z)=e^{iz}/z\) along \([-R,-\varepsilon]\), a clockwise upper semicircle of radius \(\varepsilon\), then \([\varepsilon,R]\) and the counterclockwise upper semicircle of radius \(R\). There is no pole in the enclosed half-annulus, so Cauchy’s theorem gives total integral zero [@SS03]. The real segments contribute \(2i\int_\varepsilon^R\sin x/x\,dx\). The small arc, parametrized with angle decreasing from \(\pi\) to zero, contributes \(i\int_\pi^0e^{i\varepsilon e^{i\theta}}\,d\theta\to-i\pi\). The large arc has integral of modulus at most \[ \int_0^\pi e^{-R\sin\theta}\,d\theta \leq2\int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta \leq\frac\pi R\longrightarrow0. \] The middle inequality uses symmetry and concavity of sine: \(\sin\theta\geq2\theta/\pi\) on \([0,\pi/2]\). Passing to the limits yields \(2i\int_0^\infty\sin x/x\,dx-i\pi=0\), which gives the stated value.
2. A real frequency contributes its sign, with frequency zero treated separately.
For \(s>0\), substituting \(u=sx\) gives \[ \lim_{A\to\infty}\int_0^A\frac{\sin(sx)}x\,dx =\frac\pi2. \] For \(s<0\), oddness of sine makes the value \(-\pi/2\). For \(s=0\), the integrand vanishes identically, so the value is zero. Thus the value is \((\pi/2)\operatorname{sgn}(s)\), where \(\operatorname{sgn}(0)=0\).
3. The symmetric transform reduces to the two frequencies \(1+t\) and \(1-t\).
At each finite \(A\), the imaginary part of the integrand is odd, so its integral over \([-A,A]\) is zero. Evenness of the real part and the product-to-sum identity give \[ \int_{-A}^A\frac{\sin x}{x}e^{ixt}\,dx =\int_0^A\frac{\sin((1+t)x)+\sin((1-t)x)}x\,dx. \] By step 2 the limit is \[ \frac\pi2\bigl(\operatorname{sgn}(1+t) +\operatorname{sgn}(1-t)\bigr). \] Both signs are positive for \(|t|<1\), exactly one is zero at \(t=\pm1\), and the signs cancel for \(|t|>1\). These are the three displayed values. All cancellations were made at the finite symmetric endpoints before the limit was taken.