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试证: $$\bex 0<\int_0^\infty \frac{\sin t}{\ln(1+x+t)}\rd t<\frac{2}{\ln(1+x)}. \eex$$
证明: $$\beex \bea \int_0^\infty \frac{\sin t}{\ln(1+x+t)}\rd t &=\sum_{k=0}^\infty\sez{ \int_{2k\pi}^{2k\pi+\pi} \frac{\sin t}{\ln(1+x+t)}\rd t +\int_{2k\pi+\pi}^{2k\pi+2\pi} \frac{\sin t}{\ln(1+x+t)}\rd t}\\ &=\sum_{k=0}^\infty \sez{\int_0^\pi \frac{\sin s}{\ln(1+x+2k\pi +s)}\rd s -\int_0^\pi\frac{\sin s}{\ln(1+x+2k\pi+\pi+s)}\rd s}\\ &=\sum_{k=0}^\infty \int_0^\pi \sin s\sez{ \frac{1}{\ln(1+x+2k\pi+s)}-\frac{1}{\ln(1+x+2k\pi+\pi+s)}}\rd s\\ &>0. \eea \eeex$$ 另一方面, $$\beex \bea \int_0^\infty \frac{\sin s}{\ln(1+x+s)}\rd s&=\int_0^\pi \sin s\sez{ \frac{1}{\ln(1+x+2k\pi+s)} -\frac{1}{\ln (1+x+2k\pi+\pi+s)}}\rd s\\ &<\int_0^\pi \sin s\sez{ \frac{1}{\ln(1+x+2k\pi)} -\frac{1}{\ln (1+x+2k\pi+\pi)}}\rd s\\ &\quad\sex{f(s)\equiv\frac{1}{\ln(1+x+2k\pi+s)} -\frac{1}{\ln (1+x+2k\pi+\pi+s)}\ \searrow}\\ &<\int_0^\pi \frac{\sin s}{\ln(1+x)}\rd s\\ &=\frac{2}{\ln(1+x)}. \eea \eeex$$
[家里蹲大学数学杂志]第409期与正弦对数有关的一个积分不等式
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原文地址：http://www.cnblogs.com/zhangzujin/p/4667918.html
