积分级数题集05
作者：互联网
(AMM11924)
\[\int_{0}^{\pi / 2} \frac{\{\tan x\}}{\tan x} {\rm d} x \]
Solution: 记 \(t_n=\arctan n\), 则
\[\begin{aligned} I:&=\int_{0}^{\pi / 2} \frac{\{\tan x\}}{\tan x} {\rm d}x =\int_{0}^{\pi / 2}\left(1-\frac{\lfloor\tan x\rfloor}{\tan x}\right) {\rm d} x=\frac{\pi}{2}-\sum_{n=0}^{\infty} \int_{t_{n}}^{t_{n+1}} \frac{\tan(\arctan n)}{\tan x}{\rm d} x\\ &=\frac{\pi}{2}-\sum_{n=1}^{\infty} n \int_{t_{n}}^{t_{n+1}} \frac{{\rm d} x}{\tan x} =\frac{\pi}{2}-\sum_{n=1}^{\infty} n\left\{\ln \left(\sin t_{n+1}\right)-\ln \left(\sin t_{n}\right)\right\} \end{aligned} \]注意到当 \(t>0\) 时
\[\sin(\arctan t)=\frac{1}{\sqrt{1+1/t^2}} \]于是对任意 \(N\geqslant 2\) 有
\[\begin{aligned} \sum_{n=1}^{N-1} n\left(\ln \left(\sin t_{n+1}\right)-\ln \left(\sin t_{n}\right)\right) &=\sum_{n=1}^{N-1}\left[(n+1) \ln \left(\sin t_{n+1}\right)-n \ln \left(\sin t_{n}\right)\right]-\sum_{n=1}^{N-1} \ln \left(\sin t_{n+1}\right) \\ &=N \ln \left(\sin t_{N}\right)-\sum_{n=0}^{N-1} \ln \left(\sin t_{n+1}\right) \\ &=N \ln \left(\frac{1}{\sqrt{1+1 / N^{2}}}\right)-\ln \left(\prod_{n=1}^{N} \frac{1}{\sqrt{1+1 / n^{2}}}\right)\\ &=\frac12\ln \left\{\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}}\right)\right\}\quad(N\to\infty) \end{aligned} \]利用 \(\sin\) 的无穷乘积
\[\sin (\pi z)=\pi z \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right) \]得到
\[\begin{align*} I&=\frac{\pi}{2}-\frac{1}{2}\ln \left\{\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}}\right)\right\}=\frac{\pi}{2}-\frac{1}{2} \ln \left(\frac{\sin (\pi i)}{\pi i}\right)\\ &=\frac{\pi}{2}-\frac{1}{2} \ln \left(\frac{\sinh (\pi)}{\pi}\right)=\frac{1}{2} \ln \left(\frac{2 \pi}{1-e^{-2 \pi}}\right) \end{align*} \]\(\square\)
标签：right,frac,级数,ln,05,题集,pi,sin,left 来源： https://www.cnblogs.com/deer2021/p/15118191.html