计算 $\dps{\int_0^\infty\frac{\sin^2x}{x^2}dx=\frac{\pi}{2}}$

计算 $\dps{\int_0^\infty\frac{\sin^2x}{x^2}dx=\frac{\pi}{2}}$. 由分部积分, $$\bee\label{1}\bea \int_0^\infty\frac{\sin^2x}{x^2}dx &=-\int_0^\infty \sin^2xd\frac{1}{x}\\ &=-\left.\frac{\sin^2x}{x}\right|^\infty_0 +\int_0^\infty \frac{\sin 2x}{x}dx\\ &=\int_0^\infty\frac{\sin 2x}{2x}d(2x)\\ &=\int_0^\infty \frac{\sin t}{t}dt. \eea\eee$$ 往求 $\dps{\int_0^\infty \frac{\sin t}{t}dt}$. 为此, 我们用 Fubini 定理得到 $$\bee\label{2}\bea \int_0^\infty \frac{\sin t}{t}dt &=\int_0^\infty \sin tdt \int_0^\infty e^{-tx}dx\\ &=\int_0^\infty dx\int_0^\infty e^{-tx}\sin tdt. \eea\eee$$ 为求 $\dps{I(x)=\int_0^\infty e^{-tx}\sin tdt}$, 进行两次分部积分, $$\bex\ba{ll} I(x)&=-\int_0^\infty e^{-tx}d\cos t\\ &=-e^{-tx}\cos t|^\infty_0 +\int_0^\infty (-x)e^{-tx}\cos tdt\\ &=1-x\int_0^\infty e^{-tx}d\sin t\\ &=1-x\sez{0+x\int_0^\infty e^{-tx}\sin tdt}\\ &=1-x^2I(x).\ea\eex$$ 于是 $$\bex I(x)=\frac{1}{1+x^2}, \eex$$ 而由 \eqref{1}, \eqref{2} 即知 $$\bex \int_0^\infty\frac{\sin^2x}{x^2}dx =\int_0^\infty I(x)dx =\frac{\pi}{2}. \eex$$