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先ず,次の積分を考えた.
$\BA\D\\
\int_0^z \frac{d\theta}{\sqrt{1-k\sin^2\theta}}
\EA$
ここから,$\D \frac{x}{1-x}=a\tan^2\frac{\theta}{2}$と置換積分する.
$\BA\D\\
\int_0^z \frac{d\theta}{\sqrt{1-k\sin^2\theta}}
&=\sqrt{a}\int_0^\frac{a\tan^2\frac{z}{2}}{1+a\tan^2\frac{z}{2}} \frac{dx}{\sqrt{a^2+2a(1-a-2k)x+((1-a)^2+4ka)x^2}\sqrt{x(1-x)}}\\
&=2\sqrt{a}\int_0^\sqrt{\frac{a\tan^2\frac{z}{2}}{1+a\tan^2\frac{z}{2}}} \frac{dx}{\sqrt{a^2+2a(1-a-2k)x^2+((1-a)^2+4ka)x^4}\sqrt{1-x^2}}
\EA$
この置換方法は,
コチラ
に因んだ.$\D a$は割と自由か.
ここで一度,$\D \frac{a\tan^2\frac{z}{2}}{1+a\tan^2\frac{z}{2}}=1$すなわち$\D z=\pi$について考える.
$\BA\D\\
\int_0^\pi \frac{d\theta}{\sqrt{1-k\sin^2\theta}}
=2\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-k\sin^2\theta}}
=2K(\sqrt{k})
\EA$
また
$\BA\D\\
\int_0^\pi \frac{d\theta}{\sqrt{1-k\sin^2\theta}}
=2\sqrt{a}\int_0^1 \frac{dx}{\sqrt{a^2+2a(1-a-2k)x^2+((1-a)^2+4ka)x^4}\sqrt{1-x^2}}
\EA$
でもある.さらに,右辺の$\sqrt{}$の内容を簡単にしたいと思った.
➀$\D \begin{cases} 2a(1-a-2k)=a^2 \\ (1-a)^2+4ka=a^2 \end{cases} \Longleftarrow \D \begin{cases} a=\frac{1}{\sqrt{3}} \\ k=\frac{(-1+\sqrt{3})^2}{8} \end{cases}$の場合
$\BA\D\\
2\sqrt{a}\int_0^1 \frac{dx}{\sqrt{a^2+2a(1-a-2k)x^2+((1-a)^2+4ka)x^4}\sqrt{1-x^2}}
&=\frac{2}{\sqrt{a}}\int_0^1 \frac{dx}{\sqrt{1+x^2+x^4}\sqrt{1-x^2}}\\
&=2\sqrt[4]{3}\int_0^1 \frac{dx}{\sqrt{1-x^6}}
\EA$
よって
$\BA\D\\
\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-\frac{(-1+\sqrt{3})^2}{8}\sin^2\theta}}=\sqrt[4]{3}\int_0^1 \frac{dx}{\sqrt{1-x^6}}~.
\EA$
➁$\D 2a(1-a-2k)=0 \Longleftrightarrow k=\frac{1-a}{2}$の場合
$\BA\D\\
\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-\frac{1-a}{2}\sin^2\theta}}=\sqrt{a}\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{a^2+(1-a^2)\sin^4\theta}}~.
\EA$
➂$\D (1-a)^2+4ka=0 \Longleftrightarrow k=-\frac{(1-a)^2}{4a}$の場合
$\BA\D\\
\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1+\frac{(1-a)^2}{4a}\sin^2\theta}}=\sqrt{a}\int_0^\frac{\pi}{2} \frac{dx}{\sqrt{a^2+(1-a^2)\sin^2\theta}}~.
\EA$
$z$に代入せず,$\D k=-\frac{(1-a)^2}{4a}$の場合を考える.
$\BA\D\\
\int_0^z \frac{d\theta}{\sqrt{1+\frac{(1-a)^2}{4a}\sin^2\theta}}
=2\sqrt{a}\int_0^\sqrt{\frac{a\tan^2\frac{z}{2}}{1+a\tan^2\frac{z}{2}}} \frac{dx}{\sqrt{a^2+(1-a^2)x^2}\sqrt{1-x^2}}
\EA$
$a=\frac{1}{\sqrt{2}}$とすれば
$\BA\D\\
\int_0^z \frac{d\theta}{\sqrt{4\sqrt{2}+(-1+\sqrt{2})^2\sin^2\theta}}
=\int_0^\sqrt{\frac{\tan^2\frac{z}{2}}{\sqrt{2}+\tan^2\frac{z}{2}}} \frac{dx}{\sqrt{1-x^4}}
\EA$
よって
$\BA\D\\
\int_0^\frac{1}{\sqrt{1+\sqrt{2}}} \frac{dx}{\sqrt{1-x^4}}
&=\int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{4\sqrt{2}+(-1+\sqrt{2})^2\sin^2\theta}}\\
&=\frac{1}{2}\int_0^\pi \frac{d\theta}{\sqrt{4\sqrt{2}+(-1+\sqrt{2})^2\sin^2\theta}}\\
&=\frac{1}{2}\int_0^1 \frac{dx}{\sqrt{1-x^4}}~.
\EA$