3
大学数学基礎解説
解説18
117
0
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@sounansya_29 さんがツイートした こちらの問題 の解説です．
以下の式を求めてください．
$\displaystyle\int_0^1\frac{\artanh\sqrt{1-x^2}}{1-x}\dd x$
解答
解説
\begin{align*} &\int_0^1\frac{\artanh\sqrt{1-x^2}}{1-x}\dd x\\ =&\int_0^1\frac{\arsech x}{1-x}\dd x\\ =&\int_0^1\frac{1}{1-x}\r{-\arsech1+\arsech x}\dd x\\ =&\int_0^1\frac{1}{1-x}\r{-\r{\ln\frac{2}{1}-\sum_{n=1}^\infty\r{\frac{\r{2n}!}{2^{2n}\r{n!}^2}}\frac{1^{2n}}{2n}}+\r{\ln\frac{2}{x}-\sum_{n=1}^\infty\r{\frac{\r{2n}!}{2^{2n}\r{n!}^2}}\frac{x^{2n}}{2n}}}\dd x\\ =&\int_0^1\frac{1}{1-x}\r{-\r{\ln\frac{2}{1}-\lim_{b\rightarrow0}\frac{1}{2b}\sum_{n=1}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}}+\r{\ln\frac{2}{x}-\lim_{b\rightarrow0}\frac{1}{2b}\sum_{n=1}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{x^{2n}}{n!}}}\dd x\\ =&\int_0^1\lim_{b\rightarrow0}\frac{1}{2b}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}\frac{x^0-x^{2n}}{1-x}\dd x-\int_0^1\frac{\ln x}{1-x}\dd x\\ =&\lim_{b\rightarrow0}\frac{1}{2b}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}\r{\psi\r{2n+1}-\psi\r{1}}-\int_0^1\ln x\sum_{k=1}^\infty x^{k-1}\dd x\\ =&\lim_{b\rightarrow0}\frac{1}{2b}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}\r{\ln2+\half\r{\psi\r{n+\half}+\psi\r{n+1}}-\psi\r{1}}+\int_0^1\sum_{k=1}^\infty\frac{1}{k^2}\r{-\frac{k}{x}}\r{-\r{k\ln\frac{1}{x}}^{2-1}e^{-k\ln\frac{1}{x}}}\dd x\\ =&\lim_{b\rightarrow0}\frac{1}{4b}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}\r{\frac{1}{\Gamma\r{\half}\r{\half}_n}\lim_{a\rightarrow\half}\pdv{a}\Gamma\r{n+a}-\Gamma\r{1}\r{1}_n\lim_{c\rightarrow1}\pdv{c}\frac{1}{\Gamma\r{n+c}}+2\r{\ln2-\psi\r{1}}}+\s{\sum_{k=1}^\infty\frac{1}{k^2}\Gamma\r{2,k\ln\frac{1}{x}}}_{x=0}^1\\ =&\lim_{b\rightarrow0}\frac{1}{4b}\r{\lim_{a\rightarrow\half}\pdv{a}\frac{\Gamma\r{a}}{\Gamma\r{\half}}\sum_{n=0}^\infty\frac{\r{a}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}-\lim_{c\rightarrow1}\pdv{c}\frac{\Gamma\r{1}}{\Gamma\r{c}}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{c}_n}\frac{1}{n!}+2\r{\ln2-\psi\r{1}}\sum_{n=0}^\infty\frac{\r{\half}_n\r{b}_n}{\r{1}_n}\frac{1}{n!}}+\sum_{k=1}^\infty\frac{1}{k^2}\Gamma\r{2}\\ =&\lim_{b\rightarrow0}\frac{1}{4b}\r{\lim_{a\rightarrow\half}\pdv{a}\frac{\Gamma\r{a}}{\Gamma\r{\half}}\frac{\Gamma\r{1}\Gamma\r{1-a-b}}{\Gamma\r{1-a}\Gamma\r{1-b}}-\lim_{c\rightarrow1}\pdv{c}\frac{\Gamma\r{1}}{\Gamma\r{c}}\frac{\Gamma\r{c}\Gamma\r{c-\half-b}}{\Gamma\r{c-\half}\Gamma\r{c-b}}+\r{2\ln2+2\gamma}\frac{\Gamma\r{1}\Gamma\r{1-\half-b}}{\Gamma\r{1-\half}\Gamma\r{1-b}}}+\zeta\r{2}\Gamma\r{2}\\ =&\lim_{b\rightarrow0}\frac{1}{4b}\r{\frac{\Gamma\r{\half}}{\Gamma\r{\half}}\frac{\Gamma\r{1}\Gamma\r{1-\half-b}}{\Gamma\r{1-\half}\Gamma\r{1-b}}\r{\psi\r{\half}-\psi\r{1-\half-b}+\psi\r{1-\half}}-\frac{\Gamma\r{1}\Gamma\r{1-\half-b}}{\Gamma\r{1-\half}\Gamma\r{1-b}}\r{\psi\r{1-\half-b}-\psi\r{1-\half}-\psi\r{1-b}}+\frac{\Gamma\r{1}\Gamma\r{1-\half-b}}{\Gamma\r{1-\half}\Gamma\r{1-b}}\r{-\psi\r{\half}-\psi\r{1}}}+\frac{\pi^2}{6}\cdot1!\\ =&\lim_{b\rightarrow0}\frac{\Gamma\r{1}\Gamma\r{1-\half-b}}{4\Gamma\r{1-\half}\Gamma\r{1-b}}\r{2\frac{\psi\r{\half-b}-\psi\r{\half}}{-b}-\frac{\psi\r{1-b}-\psi\r{1}}{-b}}+\frac{\pi^2}{6}\\ =&\frac{\Gamma\r{1}\Gamma\r{1-\half-0}}{4\Gamma\r{1-\half}\Gamma\r{1-0}}\r{2\psi^\r{1}\r{\half}-\psi^\r{1}\r{1}}+\frac{\pi^2}{6}\\ =&\frac{\Gamma\r{1}\Gamma\r{\half}}{4\Gamma\r{\half}\Gamma\r{1}}\r{2\frac{\pi^2}{2}-\frac{\pi^2}{6}}+\frac{\pi^2}{6}\\ =&\frac{5\pi^2}{24}+\frac{\pi^2}{6}\\ =&\frac{3\pi^2}{8} \end{align*}
なので，$\displaystyle\int_0^1\frac{\artanh\sqrt{1-x^2}}{1-x}\dd x=\frac{3\pi^2}{8}$です．
投稿日：2021年6月15日
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