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### 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，

$$\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\cos \left(\frac{\pi}{2}-x\right)\right)dx$$

$$\displaystyle=\int^0_{\pi/2}\ln\left(\cos x\right)\left(-dx\right) = \int^{\pi/2}_0\ln\left(\cos x\right)dx$$

$$\displaystyle\int^{\pi/2}_0\ln\left(\sin x\right)dx = \int^{\pi/2}_0\ln\left(\sin \left(\pi-x\right)\right)dx$$

$$\displaystyle=\int^{\pi/2}_{\pi}\ln\left(\sin x\right)\left(-dx\right) = \int^{\pi}_{\pi/2}\ln\left(\sin x\right)dx$$

$$\displaystyle2\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx$$

$$\displaystyle=\int^{\pi/2}_0\ln\left(\sin x\right)dx +\int^{\pi/2}_0\ln\left(\cos x\right)dx$$

$$\displaystyle=\int^{\pi/2}_0\left(\ln\left(\sin x\right)+\ln\left(\cos x\right)\right)dx$$

$$\displaystyle=\int^{\pi/2}_0\ln\left(\frac{\sin\left(2x\right)}{2}\right)dx$$

$$\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(2x\right)\right)dx -\int^{\pi/2}_0\ln2 dx$$

$$\displaystyle=\int^{\pi}_0\ln\left(\sin\left(x\right)\right)\frac{dx}{2} -\int^{\pi/2}_0\ln2 dx$$

$$\displaystyle=\frac{1}{2}\left(\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx+\int^{\pi}_{\pi/2}\ln\left(\sin\left(x\right)\right)dx\right) -\int^{\pi/2}_0\ln2 dx$$

$$\displaystyle=\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx -\int^{\pi/2}_0\ln2 dx$$

$$\displaystyle\int^{\pi/2}_0\ln\left(\sin\left(x\right)\right)dx = -\int^{\pi/2}_0\ln2 dx=-\frac{\pi}{2}\ln2.$$

$\displaystyle \sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\sin\frac{3\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1}=\frac{2k+1}{2^{2k}}$

$$\displaystyle \int^{\pi}_0\ln\left(\sin x\right)dx = \lim_{n \to \infty}\frac{\pi}{n}(\ln\sin\frac{\pi}{n}+\ln\sin\frac{2\pi}{n}+\dots+\ln\sin\frac{(n-1)\pi}{n})$$

$$\displaystyle =\pi\lim_{n \to \infty}\frac{1}{n}\ln(\sin\frac{\pi}{n}\sin\frac{2\pi}{n}\dots\sin\frac{(n-1)\pi}{n})$$

$$\displaystyle =\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1})$$

$$\displaystyle =\pi\lim_{k \to \infty}\frac{1}{2k+1}\ln(\frac{2k+1}{2^{2k}})$$

$$\displaystyle =\pi\lim_{k \to \infty}\frac{\ln(2k+1)-\ln{2^{2k}}}{2k+1}$$

$$\displaystyle =\pi\lim_{k \to \infty}\frac{\ln(2k+1)-2k\ln2}{2k+1}$$

$$\displaystyle =\pi\lim_{k \to \infty}\frac{\frac{ln(2k+1)}{k}-2\ln2}{2+1/k}$$

$$\displaystyle =\pi\frac{0-2\ln2}{2}$$

$$\displaystyle =-\pi\ln2$$

[[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]

$\displaystyle \sin\frac{\pi}{2k+1}\sin\frac{2\pi}{2k+1}\sin\frac{3\pi}{2k+1}\dots\sin\frac{2k\pi}{2k+1}=\frac{2k+1}{2^{2k}}$

... [/quote]

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