ウォリス積分を含む極限
ウォリス積分を含む極限値
(1)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta=\sqrt{\frac{\pi}{2}} \](2)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\cos^{n}\theta d\theta=\sqrt{\frac{\pi}{2}} \](1)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}B\left(\frac{n+1}{2},\frac{1}{2}\right)\\ & =\frac{1}{2}\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\sqrt{n+1}\frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}}\\ & =\frac{\sqrt{\pi}}{2}\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\sqrt{n+1}\frac{2}{n+1}}\\ & =\sqrt{\frac{\pi}{2}}\lim_{n\rightarrow\infty}\sqrt{\frac{1}{\sqrt{1+\frac{1}{n}}}}\\ & =\sqrt{\frac{\pi}{2}} \end{align*}(2)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\cos^{n}\theta d\theta & =\lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\left(\frac{\pi}{2}-\theta\right)d\theta\\ & =\lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}tdt\qquad,\qquad t=-\theta+\frac{\pi}{2}\\ & =\sqrt{\frac{\pi}{2}} \end{align*}
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ウォリス積分の定義
\[
\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta
\]
数列の極限
積分問題
\[
\int_{0}^{\infty}\frac{1}{1+x^{n}}dx
\]
logの2乗の級数表示
\[
\log^{2}(1-x)=2\sum_{k=1}^{\infty}\frac{H_{k}}{k+1}x^{k+1}
\]