设 $\dps{f(x)=\int_x^{x+1}\sin t^2\rd t}$, 求证: $x>0$ 时, $\dps{|f(x)|<\frac{1}{x}}$. (北京工业大学)

证明: $$\beex \bea |f(x)|&=\sev{\int_{x^2}^{(x+1)^2}\frac{\sin s}{2\sqrt{s}}\rd s}\quad\sex{t^2=s}\\ &=\sev{\int_{x^2}^{(x+1)^2} \frac{1}{2\sqrt{s}}\rd \cos s}\\ &=\sev{\frac{\cos s}{2\sqrt{s}}|_{x^2}^{(x+1)^2} +\int_{x^2}^{(x+1)^2} \frac{1}{4}s^{-\frac{3}{2}}\cos s\rd s}\\ &< \frac{1}{2x}+\frac{1}{2(x+1)}+\frac{1}{4}\int_{x^2}^{(x+1)^2} s^{-\frac{3}{2}}\rd s\\ &=\frac{1}{x}. \eea \eeex$$