对自然数 n≥2, 证明 \bex1π∫π20\sevsin(2n+1)tsint\rdt<2+lnn2.\eex
证明: 仍用 4.3.9 的那两个不等式. 对 \dps∀ 0<x<π2, 有 \beex \bea \frac{1}{\pi}\int_0^\frac{\pi}{2}\sev{\frac{\sin (2n+1)t}{\sin t}}\rd t &=\frac{1}{\pi}\int_0^x+\int_x^\frac{\pi}{2}\sev{\frac{\sin (2n+1)t}{\sin t}}\rd t\\ &\leq \frac{1}{\pi}\int_0^x \frac{(2n+1)|\sin t|}{|\sin t|}\rd t +\frac{1}{\pi}\int_x^\frac{\pi}{2}\frac{1}{\frac{2t}{\pi}}\rd t\\ &<\frac{2n+1}{\pi}x +\frac{1}{2}\ln\frac{\pi}{2x}\\ &\equiv f(x). \eea \eeex 由 \bex f'(x)=\frac{2n+1}{\pi}-\frac{1}{2x}\sedd{\ba{ll} <0,&0<x<\frac{\pi}{2(2n+1)}\\ >0,&\frac{\pi}{2(2n+1)}<x<\frac{\pi}{2} \ea} \eex 即知 \bex1π∫π20\sevsin(2n+1)tsint\rdt≤f\sexπ2(2n+1)=12+ln(2n+1)2<1+ln(en)2=2+lnn2.\eex