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*}
ページ情報
| タイトル | ウォリス積分を含む極限 |
| URL | https://www.nomuramath.com/zotvfo8k/ |
| SNSボタン |
ウォリス積分の定義
\[
\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta
\]
連続関数の和・積・商
ベッセル関数のポアソン積分表示
\[
J_{\nu}(z)=\frac{1}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\left(\frac{z}{2}\right)^{\nu}\int_{-1}^{1}(1-t^{2})^{\nu-\frac{1}{2}}e^{izt}dt
\]
関数の極限
\[
\forall\epsilon>0,\exists\delta>0;\forall x\in\mathbb{R},0<\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-b\right||<\epsilon
\]