ガンマ関数を2つ含む定積分でカタラン定数が出てきます
ガンマ関数を2つ含む定積分でカタラン定数が出てきます
次の定積分を求めよ。
\[ \int_{0}^{\frac{1}{2}}\Gamma\left(1-x\right)\Gamma\left(1+x\right)dx=? \]
次の定積分を求めよ。
\[ \int_{0}^{\frac{1}{2}}\Gamma\left(1-x\right)\Gamma\left(1+x\right)dx=? \]
-
\(\Gamma\left(x\right)\)はガンマ関数\begin{align*}
\int_{0}^{\frac{1}{2}}\Gamma\left(1-x\right)\Gamma\left(1+x\right)dx & =\int_{0}^{\frac{1}{2}}x\Gamma\left(1-x\right)\Gamma\left(x\right)dx\\
& =\int_{0}^{\frac{1}{2}}x\frac{\pi}{\sin\left(\pi x\right)}dx\\
& =\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin\left(x\right)}dx\cmt{\pi x\rightarrow x}\\
& =\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{2ix}{e^{ix}-e^{-ix}}dx\\
& =\frac{2i}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{xe^{-ix}}{1-e^{-2ix}}dx\\
& =\frac{2i}{\pi}\int_{0}^{\frac{\pi}{2}}xe^{-ix}\sum_{k=0}^{\infty}e^{-2kix}dx\cmt{\because0<x\rightarrow\left|e^{-2x}\right|<1}\\
& =\frac{2i}{\pi}\sum_{k=0}^{\infty}\int_{0}^{\frac{\pi}{2}}xe^{-\left(2k+1\right)ix}dx\cmt{\text{積分と総和の順序変更}}\\
& =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left[\frac{1}{-\left(2k+1\right)i}xe^{-\left(2k+1\right)ix}-\frac{1}{\left(-\left(2k+1\right)i\right)^{2}}e^{-\left(2k+1\right)ix}\right]_{0}^{\frac{\pi}{2}}\\
& =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left[\frac{i}{2k+1}xe^{-\left(2k+1\right)ix}+\frac{1}{\left(2k+1\right)^{2}}e^{-\left(2k+1\right)ix}\right]_{0}^{\frac{\pi}{2}}\\
& =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left\{ \frac{i}{2k+1}\cdot\frac{\pi}{2}e^{-\left(2k+1\right)i\frac{\pi}{2}}+\frac{1}{\left(2k+1\right)^{2}}\left(e^{-\frac{2k+1}{2}\pi i}-1\right)\right\} \\
& =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left\{ \frac{i}{2k+1}\cdot\frac{\pi}{2}\left(-1\right)^{k+1}i+\frac{1}{\left(2k+1\right)^{2}}\left(\left(-1\right)^{k+1}i-1\right)\right\} \\
& =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}\\
& =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}\\
& =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\zeta\left(2\right)\left(1-\frac{1}{4}\right)\\
& =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\zeta\left(2\right)\cdot\frac{3}{4}\\
& =\frac{\pi}{4}i+\frac{2}{\pi}C-\frac{3i}{2\pi}\cdot\frac{\pi^{2}}{6}\cmt{C\text{はカタラン定数}}\\
& =\frac{\pi}{4}i+\frac{2}{\pi}C-\frac{\pi}{4}i\\
& =\frac{2}{\pi}C
\end{align*}
ページ情報
| タイトル | ガンマ関数を2つ含む定積分でカタラン定数が出てきます |
| URL | https://www.nomuramath.com/b0iqx9rf/ |
| SNSボタン |
気付かないと解けないかも
\[
\int_{0}^{\infty}\frac{1}{\left(1+x\right)\left(a^{2}+\log^{2}x\right)}dx=?
\]
tanの立方根の積分
\[
\int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C
\]
分母に1乗と2乗ルートの積分
\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz=\frac{\sqrt{z^{2}-1}}{\pm z+1}+C
\]
tanの平方根の積分
\[
\int\sqrt{\tan x}dx=\frac{\sqrt{2}}{4}\log\left(\tan x-\sqrt{2\tan x}+1\right)-\frac{\sqrt{2}}{4}\log\left(\tan x+\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{2}\tan^{\bullet}\left(\sqrt{2\tan x}-1\right)+\frac{\sqrt{2}}{2}\tan^{\bullet}\left(\sqrt{2\tan x}+1\right)+C
\]