Evaluate \[ \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx. \]
Solution
Put \[ J\coloneqq\int_0^\infty\frac{\sin 2x}{x}\,dx. \]
1. One has\[ \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx=2J. \]
The original integrand is even. It is bounded near \(0\), since \(\sin x/x\to1\), and it is at most \(1/x^2\) for large \(x\), so the improper integral converges. For \(0<\varepsilon<R\), integration by parts gives \[ \int_\varepsilon^R\frac{\sin^2x}{x^2}\,dx = \left[-\frac{\sin^2x}{x}\right]_\varepsilon^R +\int_\varepsilon^R\frac{\sin 2x}{x}\,dx. \] The boundary term tends to \(0\) as \(\varepsilon\downarrow0\) and \(R\to\infty\). Hence \[ \int_0^\infty\frac{\sin^2x}{x^2}\,dx=J, \] and evenness gives the claim.
2. For \(a>0\), define\[ J(a)\coloneqq\int_0^\infty e^{-ax}\frac{\sin 2x}{x}\,dx. \]Then\[ J(a)=\frac\pi2-\arctan\frac a2. \]
For fixed \(a>0\), differentiation under the integral sign is justified on a small neighborhood of \(a\) by an integrable exponential majorant. Thus \[ J'(a) =-\int_0^\infty e^{-ax}\sin 2x\,dx =-\frac{2}{a^2+4}. \] Therefore \[ J(a)=C-\arctan\frac a2 \] for a constant \(C\). With the substitution \(u=ax\), \[ \abs{J(a)} \leq \int_0^\infty e^{-u}\frac{\abs{\sin(2u/a)}}{u}\,du \leq\frac2a, \] so \(J(a)\to0\) as \(a\to\infty\). Since \(\arctan(a/2)\to\pi/2\), it follows that \(C=\pi/2\).
3. The undamped Dirichlet integral satisfies\[ J=\lim_{a\downarrow0}J(a)=\frac\pi2. \]
For \(a\geq0\), let \[ \phi_a(x)\coloneqq\frac{e^{-ax}}{x}. \] On \([R,\infty)\) this function is positive and decreasing to \(0\). Integration by parts against the bounded primitive \(-\cos(2x)/2\) of \(\sin(2x)\) gives, uniformly for \(a\geq0\), \[ \abs{\int_R^\infty\phi_a(x)\sin 2x\,dx} \leq\phi_a(R) \leq\frac1R. \] Thus the tails of \(J(a)\), including \(a=0\), are uniformly small as \(R\to\infty\). On every fixed interval \([0,R]\), \[ e^{-ax}\frac{\sin 2x}{x} \longrightarrow \frac{\sin 2x}{x} \] and the integrands are dominated by \(2\). Dominated convergence on \([0,R]\), followed by the uniform tail estimate, gives \[ J=\lim_{a\downarrow0}J(a). \] Step 2 now yields \(J=\pi/2\).
4. Therefore\[ \boxed{\int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx=\pi}. \]
By step 1 the integral equals \(2J\), and step 3 gives \(J=\pi/2\).
5. Q.E.D.
Step 4 gives the requested value.