对自然数 $n\geq 2$, 证明 $$\bex \frac{1}{\pi}\int_0^\frac{\pi}{2}\sev{\frac{\sin (2n+1)t}{\sin t}}\rd t<\frac{2+\ln n}{2}. \eex$$
证明: 仍用 4.3.9 的那两个不等式. 对 $\dps{\forall\ 0<x<\frac{\pi}{2}}$, 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 $$\bex \frac{1}{\pi}\int_0^\frac{\pi}{2}\sev{\frac{\sin (2n+1)t}{\sin t}}\rd t \leq f\sex{\frac{\pi}{2(2n+1)}} =\frac{1}{2}+\frac{\ln (2n+1)}{2} <\frac{1+\ln (en)}{2}=\frac{2+\ln n}{2}. \eex$$
