$\dps{\int_0^\infty\frac{\sin^2x}{x^2}\rd x=\frac{\pi}{2}}$.
证明: 由分部积分, $$\bee\label{1}\bea \int_0^\infty\frac{\sin^2x}{x^2}\rd x &=-\int_0^\infty \sin^2x\rd\frac{1}{x}\\ &=-\left.\frac{\sin^2x}{x}\right|^\infty_0 +\int_0^\infty \frac{\sin 2x}{x}\rd x\\ &=\int_0^\infty\frac{\sin 2x}{2x}\rd(2x)\\ &=\int_0^\infty \frac{\sin t}{t}\rd t. \eea\eee$$往求 $\dps{\int_0^\infty \frac{\sin t}{t}\rd t}$. 为此, 我们用 Fubini 定理得到 $$\bee \label{2} \bea\int_0^\infty \frac{\sin t}{t}\rd t &=\int_0^\infty \sin t\rd t \int_0^\infty e^{-tx}\rd x\\ &=\int_0^\infty \rd x\int_0^\infty e^{-tx}\sin t\rd t. \eea\eee$$为求 $\dps{I(x)=\int_0^\infty e^{-tx}\sin t\rd t}$, 进行两次分部积分, $$\beex\bea I(x)&=-\int_0^\infty e^{-tx}\rd\cos t\\ &=-e^{-tx}\cos t|^\infty_0 +\int_0^\infty (-x)e^{-tx}\cos t\rd t\\ &=1-x\int_0^\infty e^{-tx}\rd\sin t\\ &=1-x\sez{0+x\int_0^\infty e^{-tx}\sin t\rd t}\\ &=1-x^2I(x). \eea\eeex$$ 于是 $$\beex\bea I(x)=\frac{1}{1+x^2}, \eea\eeex$$ 而由 \eqref{1}, \eqref{2} 即知 $$\beex\bea \int_0^\infty\frac{\sin^2x}{x^2}\rd x =\int_0^\infty I(x)\rd x =\frac{\pi}{2}. \eea\eeex$$
