請問一題，∫㏑(sinx)dx
ln(sinx) 積分 從0積到 90度 =? 我的做法比較繁瑣，不知是否有人可以提供更簡單的方法，我的做法如下：
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\cos \left(\frac{\pi}{2}-x\right)\right)dx \)
\(\displaystyle=\int^0_{\pi/2}\ln\left(\cos x\right)\left(-dx\right) = \int^{\pi/2}_0\ln\left(\cos x\right)dx \)
且
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\sin \left(\pi-x\right)\right)dx \)
\(\displaystyle=\int^{\pi/2}_{\pi}\ln\left(\sin x\right)\left(-dx\right) = \int^{\pi}_{\pi/2}\ln\left(\sin x\right)dx \)
然後，利用上面兩式，可得如下
\(\displaystyle2\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx\)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin x\right)dx +\int^{\pi/2}_0\ln\left(\cos x\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\left(\ln\left(\sin x\right)+\ln\left(\cos x\right)\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\frac{\sin\left(2x\right)}{2}\right)dx \)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(2x\right)\right)dx -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\int^{\pi}_0\ln\left(\sin\left(x\right)\right)\frac{dx}{2} -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\frac{1}{2}\left(\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx+\int^{\pi}_{\pi/2}\ln\left(\sin\left(x\right)\right)dx\right) -\int^{\pi/2}_0\ln2 dx\)
\(\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx -\int^{\pi/2}_0\ln2 dx\)
因此，
\(\displaystyle\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx = -\int^{\pi/2}_0\ln2 dx=-\frac{\pi}{2}\ln2.\)
題目出處：99台中一中教甄 $$\int_0^{\pi/2}\ln \sin x dx$$其實是一個暇積分(improper integral). 在算此積分之前應該要了解他是收斂的. 也可以直接用極限來做:(以 [b]99台中一中教甄題[/b] 為例 )
其中需用到公式:
\[\displaystyle
\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\sin\frac{3\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1}=\frac{2k+1}{2^{2k}}
\]
\(\displaystyle
\int^{\pi}_0\ln\left(\sin x\right)dx =
\lim_{n \to \infty}\frac{\pi}{n}(\ln\sin\frac{\pi}{n}+\ln\sin\frac{2\pi}{n}+\dots+\ln\sin\frac{(n-1)\pi}{n})
\)
\(\displaystyle
=\pi\lim_{n \to \infty}\frac{1}{n}\ln(\sin\frac{\pi}{n}\sin\frac{2\pi}{n}\dots\sin\frac{(n-1)\pi}{n})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\frac{2k+1}{2^{2k}})
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\ln(2k+1)-\ln{2^{2k}}}{2k+1}
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\ln(2k+1)-2k\ln2}{2k+1}
\)
\(\displaystyle
=\pi\lim_{k \to \infty}\frac{\frac{ln(2k+1)}{k}-2\ln2}{2+1/k}
\)
\(\displaystyle
=\pi\frac{0-2\ln2}{2}
\)
\(\displaystyle
=-\pi\ln2
\)
[[i] 本帖最後由 Joy091 於 2011-2-14 11:39 AM 編輯 [/i]] [quote]原帖由 [i]Joy091[/i] 於 2011-2-14 09:42 AM 發表 [url=https://math.pro/db/redirect.php?goto=findpost&pid=2819&ptid=926][img]https://math.pro/db/images/common/back.gif[/img][/url]
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:
\[\displaystyle
\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\sin\frac{3\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1}=\frac{2k+1}{2^{2k}}
\]
... [/quote]
讚!很漂亮的作法,到後面sin(pi/n)*sin(2pi/n)*........*sin[(n-1)pi/n]=n/2^(n-1) 其實就是一個公式
可以直接做,不需再透過n=2k+1來處理 補充趨近於0時，函數收斂
[attach]6109[/attach]
頁:
[1]