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(AMM11924)

\[\int_{0}^{\pi / 2} \frac{\{\tan x\}}{\tan x} {\rm d} x \]

Solution: 記 \(t_n=\arctan n\), 則

\[\begin{aligned} I:&=\int_{0}^{\pi / 2} \frac{\{\tan x\}}{\tan x} {\rm d}x =\int_{0}^{\pi / 2}\left(1-\frac{\lfloor\tan x\rfloor}{\tan x}\right) {\rm d} x=\frac{\pi}{2}-\sum_{n=0}^{\infty} \int_{t_{n}}^{t_{n+1}} \frac{\tan(\arctan n)}{\tan x}{\rm d} x\\ &=\frac{\pi}{2}-\sum_{n=1}^{\infty} n \int_{t_{n}}^{t_{n+1}} \frac{{\rm d} x}{\tan x} =\frac{\pi}{2}-\sum_{n=1}^{\infty} n\left\{\ln \left(\sin t_{n+1}\right)-\ln \left(\sin t_{n}\right)\right\} \end{aligned} \]

注意到當 \(t>0\)

時

\[\sin(\arctan t)=\frac{1}{\sqrt{1+1/t^2}} \]

於是對任意 \(N\geqslant 2\) 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 \(\sin\)

的無窮乘積

\[\sin (\pi z)=\pi z \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right) \]

得到

\[\begin{align*} I&=\frac{\pi}{2}-\frac{1}{2}\ln \left\{\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}}\right)\right\}=\frac{\pi}{2}-\frac{1}{2} \ln \left(\frac{\sin (\pi i)}{\pi i}\right)\\ &=\frac{\pi}{2}-\frac{1}{2} \ln \left(\frac{\sinh (\pi)}{\pi}\right)=\frac{1}{2} \ln \left(\frac{2 \pi}{1-e^{-2 \pi}}\right) \end{align*} \]

\(\square\)