也可以直接用極限來做:(以 99台中一中教甄題 為例 )
其中需用到公式:
\[\displaystyle
\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\sin\frac{3\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1}=\frac{2k+1}{2^{2k}}
\]
\(\displaystyle
\int^{\pi}_0\ln\left(\sin x\right)dx =
\lim_{n \to \infty}\frac{\pi}{n}(\ln\sin\frac{\pi}{n}+\ln\sin\frac{2\pi}{n}+\dots+\ln\sin\frac{(n-1)\pi}{n})
\)
\(\displaystyle
=\pi\lim_{n \to \infty}\frac{1}{n}\ln(\sin\frac{\pi}{n}\sin\frac{2\pi}{n}\dots\sin\frac{(n-1)\pi}{n})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\frac{2k+1}{2^{2k}})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\ln(2k+1)-\ln{2^{2k}}}{2k+1}
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\ln(2k+1)-2k\ln2}{2k+1}
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\frac{ln(2k+1)}{k}-2\ln2}{2+1/k}
\)
\(\displaystyle
=\pi\frac{0-2\ln2}{2}
\)
\(\displaystyle
=-\pi\ln2
\)
[ 本帖最後由 Joy091 於 2011-2-14 11:39 AM 編輯 ]
