・はじめに
・内容
・最後に
どうも、色数です。
最近僕が始めたDaily integral という遊びをまとめてみます。
毎日積分を錬成して自分で潰すという遊びです。
$\displaystyle \int_0^1\frac{1-x}{1-2x(1-x)}dx=\frac{\pi}{2}$
\begin{align}
\int_0^1\frac{1-x}{1-2x(1-x)}dx&=\sum_{n=0}^\infty2^n\int_0^1x^n(1-x)^{n+1}dx\\&=\frac{1}{2}\sum_{n=1}^\infty\frac{2^n(n-1)!n!}{(2n)!}\\&=\frac{1}{2}\sum_{n=1}^\infty\frac{2^n}{n\binom{2n}{n}}\\&=\arcsin\frac{1}{\sqrt2}…(1)\\&=\frac{\pi}{4}
\end{align}
\begin{align}
(1):\arcsin^2x&=\frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\\
\sum_{n=1}^\infty\frac{2^{2n}x^{2n-1}}{n\binom{2n}{n}}&=\frac{d}{dx}(\arcsin ^2x)\\&=\frac{2\arcsin x}{\sqrt{1-x^2}}\\x=\frac{1}{\sqrt2}\text{とすれば}\sum_{n=1}^\infty\frac{2^n}{n\binom{2n}{n}}&=2\arcsin \frac{1}{\sqrt2}=\frac{\pi}{2}
\end{align}
$\displaystyle \int_0^1\frac{\ln t\ln(1-t)}{t}dt=\zeta(3)$
$\displaystyle \int_0^1\frac{\arcsin^2x}{x\sqrt{1-x^2}}dx=2\pi\beta(2)-\frac{7}{2}\zeta(3)$
$\displaystyle \int_0^1\arcsin t\textup{arctann}\;tdt=\pi\ln2-\frac{\pi}{2}$
$\displaystyle \int_0^\frac{\pi}{2}x\ln\cos xdx=-\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\ln2$
$\displaystyle \int_0^1\frac{dx}{\sqrt{1-x(1-x)}}=\ln3$
\begin{align}
\int_0^1\frac{dx}{\sqrt{1-x(1-x)}}&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}}\int_0^1x^n(1-x)^ndx\\&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}}\frac{1}{(2n+1)\binom{2n}{n}}\\&=\sum_{n=0}^\infty\frac{1}{2^{2n}(2n+1)}
\end{align}
$\displaystyle \sum_{n=0}^\infty\frac{x^{2n}}{2^{2n}}=\frac{1}{1-\frac{x^2}{4}}$
$\displaystyle \int_0^1\frac{dx}{1-\frac{x^2}{4}}=\ln3$より得る
$\displaystyle \int_0^1\frac{dx}{\sqrt{1-x^2}\sqrt{1-x^2(1-x^2)}}=K\left(\frac{1}{4}\right)$
$K$は第一種楕円積分
\begin{align} \int_0^1\frac{dx}{\sqrt{1-x^2}\sqrt{1-x^2(1-x^2)}}&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}}\int_0^1\frac{x^{2n}(1-x^2)^n}{\sqrt{1-x^2}}dx\\&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}}\int_0^1x^{2n}(1-x^2)^{n-\frac{1}{2}}dx\\&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n+1}}\int_0^1x^{n-\frac{1}{2}}(1-x)^{n-\frac{1}{2}}dx\\&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n+1}}\frac{\Gamma(n+\frac{1}{2})\Gamma(n+\frac{1}{2})}{\Gamma(2n+1)}\\&=\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n+1}}\frac{4\pi(2n-1)!^2}{2^{4n}(2n)!(n-1)!^2}\\&=\frac{\pi}{2}\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}}\frac{(2n)!}{2^{4n}n!^2}\\&=K\left(\frac{1}{4}\right) \end{align}
積分もできるようになりたいです。