问题

证明:\[\lim_{x\rightarrow+\infty}\int_{\frac{1}{x}}^{1}\frac{1-2\sin^2 t}{t^2}dt=+\infty.\]

证明: \[ \begin{split} \int_{\frac{1}{x}}^{1}\frac{1-2\sin^2 t}{t^2}dt&=\int_{\frac{1}{x}}^{1}\frac{\cos 2t}{t^2}dt\\ &=-\int_{\frac{1}{x}}^{1}\cos(2t)d\frac{1}{t}\\ &=-\biggl[\frac{\cos 2t}{t}\biggr|_{\frac{1}{x}}^{1}-\int_{\frac{1}{x}}^{1}\frac{1}{t}d\cos 2t\biggr]\\ &=-2\int_{\frac{1}{x}}^{1}\frac{\sin 2t}{t}dt+x\cos\frac{2}{x}-\cos 2\\ &=-2\int_{\frac{1}{x}}^{1}\frac{\sin 2t}{2t}d2t+x\cos\frac{2}{x}-\cos 2\\ &=-2\int_{\frac{2}{x}}^{2}\frac{\sin u}{u}du+x\cos\frac{2}{x}-\cos 2 \end{split} \]

注意到 Dirichlet 积分

\[ \int_{0}^{+\infty}\frac{\sin t}{t}dt=\frac{\pi}{2}, \] 故\[\lim_{x\rightarrow+\infty}\int_{\frac{1}{x}}^{1}\frac{1-2\sin^2 t}{t^2}dt=\lim_{x\rightarrow+\infty}x\cos\frac{2}{x}=+\infty.\]