<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\D \int_0^1 K(x)^2\,dx=\frac{1}{2}\int_0^1 K'(x)^2\,dx=\frac{\pi^4}{32}\sum_{n=0}^\infty (4n+1)\frac{\binom{2n}{n}^6}{2^{12n}}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D
\int_0^1 \frac{K(x)}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,dx=\frac{\Gamma(\frac{1}{4})^4}{24}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,dx
=\frac{\pi^4}{4}\sum_{n=0}^\infty \frac{\binom{2n}{n}^4}{2^{8n}}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)\ln\frac{1}{x}}{1-x^2}\,dx=\frac{7}{4}\zeta(3)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)\ln\frac{1}{x}}{1-x^2}\,dx=\frac{\pi^3}{8}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{xK(x)K'(x)}{1-t^2x^2}\,dx=\frac{\pi}{4}K(t)^2
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)}{1-t^2x^2}\,dx=\frac{\pi}{2}K(t)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)\tanh^{-1}x}{x}\,dx=\pi\beta(2)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)K'(x)}{x}\ln\frac{1}{1-x^2}\,dx=\frac{7}{8}\pi\zeta(3)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)}{1+x}\,dx=\frac{\pi^2}{8}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)}{1+x}\,dx=2\beta(2)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 K'(x)^3\,dx=\frac{10}{3}\int_0^1 K(x)^3\,dx=\frac{5}{2}\int_0^1 K(\sqrt{x})^3\,dx=6\int_0^1 xK(x)^2K'(x)\,dx=3\int_0^1 K(x)^2K'(x)\,dx=2\int_0^1 K(x)K'(x)^2\,dx=\frac{{\Gamma(\frac{1}{4})}^8}{128\pi^2}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)^2}{\sqrt{1-x^2}}\,dx=\int_0^1 \frac{K(x)^2}{\sqrt{1-x^2}}\,dx=2\int_0^1 K(x)K'(x)\,dx=\frac{\pi^3}{4}\sum_{n=0}^\infty \frac{\binom{2n}{n}^4}{2^{8n}}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\frac{93}{8}\zeta(5)=\int_0^1 (2x-1)K(\sqrt{x})^4\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)}{\sqrt{1+x}}\,dx
=\frac{{\Gamma(\frac{1}{8})}^2{\Gamma(\frac{3}{8})}^2}{32\pi}-\frac{1}{2}\sum_{n=0}^\infty \frac{(4n+1)\binom{4n}{2n}}{(2n+1)^3\binom{2n}{n}^2}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)}{\sqrt{1-x}}\,dx
=\frac{{\Gamma(\frac{1}{8})}^2{\Gamma(\frac{3}{8})}^2}{32\pi}+\frac{1}{2}\sum_{n=0}^\infty \frac{(4n+1)\binom{4n}{2n}}{(2n+1)^3\binom{2n}{n}^2}\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)}{\sqrt{1+x}}\,dx=\frac{1}{3}\int_0^1 \frac{K(x)}{\sqrt{1-x}}\,dx=\frac{{\Gamma(\frac{1}{8})}^2{\Gamma(\frac{3}{8})}^2}{48\sqrt{2}\,\pi}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 (2x-1)K(\sqrt{x})^3K'(\sqrt{x})\,dx=\frac{\pi^5}{64}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 (2x-1)K(\sqrt{x})^2K'(\sqrt{x})\,dx=\frac{32\,\pi^6}{\Gamma\L(\frac{1}{4}\R)^8}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{1-x}{1+x}K(x)^2\,dx=\frac{7}{8}\zeta(3)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 K(x)^2\,dx=\int_0^1 \frac{4K(x)^2-K'(x)^2}{1+x}\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)\ln\frac{1}{x}}{1+x}\,dx=\frac{7}{2}\zeta(3)-\pi\beta(2)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)^2}{1+x^2}\,dx-\int_0^1 K(x)^2\,dx=\int_0^1 \frac{K(\sqrt{x})K'(\sqrt{x})}{1+x}\,dx=\frac{\Gamma\left(\frac{1}{4}\right)^4}{64}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{6K(x)^2K'(x)+2K(x)K'(x)^2-K'(x)^3}{1+x^2}\,dx=\frac{3}{256}\frac{\Gamma\left(\frac{1}{4}\right)^8}{\pi^2}
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K(x)\ln^2\frac{1}{x}}{1-x^2}\,dx=4\beta(2)^2
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{K'(x)\ln^2\frac{1}{x}}{1-x^2}\,dx=\pi^2\beta(2)
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
8\int_0^1 \frac{K(x)^2K'(x)^2}{1+x^2}\,dx=\int_0^1 \L(2K(x)^2K'(x)^2+K'(x)^4\R)\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{24K(x)^4+6K'(x)^4}{1+x^2}\,dx=\int_0^1 \L(8K(x)^4+7K'(x)^4\R)\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{2865K'(x)^5+31608K(x)K'(x)^4-28246K(x)^5}{1+x}\,dx=\frac{753}{128}\pi^6
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{xK(x)^2}{\sqrt[4]{1-x^2}}\,dx=4\int_0^1 \left(K(x)K'(x)-K(x)^2\right)\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{xK'(x)^2}{\sqrt[4]{1-x^2}}\,dx=\int_0^1 \left(K(x)K'(x)-\frac{1}{2}K(x)^2\right)\,dx
\EA$</div></p></div><p></p>

<div style="padding: .5em 1em; margin: 2em 0; border: solid 2px #333333; background: #ffffff; box-shadow: 4px 4px 0 #cacaca;"> <p style=" margin: 0; padding: 0; "><b>命題</b>
<div align="center">$\BA\D 
\int_0^1 \frac{xK(x)K'(x)}{\sqrt[4]{1-x^2}}\,dx=\int_0^1 \left(2K(x)^2-K(x)K'(x)\right)\,dx
\EA$</div></p></div><p></p>


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