2 完全楕円積分の性質


$ \hspace{3em} 0 \leqq k \leqq 1$ として \[ K(k)=\int_{0}^{\cfrac{\pi}{2}}\cfrac{d\theta}{\sqrt{1-k^2\sin^2\theta}}\hspace{26em} \] \[ E(k)=\int_{0}^{\cfrac{\pi}{2}}\sqrt{1-k^2\sin^2\theta}d\theta \hspace{26em}\] $をそれぞれ、kを母数とする第１種、第２種完全楕円積分といいます。$

$\hspace{2em} 定理３\hspace{2em} \cfrac{dK(k)}{dk}=\cfrac{1}{k(1-k^2)}\{E(k)-(1-k^2)K(k)\} $

$（証明）$
\begin{eqnarray*} \cfrac{dK(k)}{dk} &=&\int_{0}^{\cfrac{\pi}{2}}\cfrac{k\sin^2\theta}{(1-k^2\sin^2\theta)^{\frac{3}{2}}}d\theta \hspace{24em}\\ &=&\int_{0}^{\cfrac{\pi}{2}}\cfrac{k\sin\theta}{(1-k^2\sin^2\theta)^{\frac{3}{2}}}\sin\theta d\theta \\ \end{eqnarray*} $\hspace{5em} ここで \quad \cfrac{d}{d\theta}\Bigl(\cfrac{-\cos\theta}{\sqrt{1-k^2\sin^2\theta}}\Bigr)=\cfrac{(1-k^2)\sin\theta}{(1-k^2\sin^2\theta)^{\frac{3}{2}}} 　だから$
\begin{eqnarray*} \cfrac{dK(k)}{dk} &=&\int_{0}^{\cfrac{\pi}{2}}\cfrac{k}{1-k^2}\cfrac{(1-k^2)\sin\theta}{(1-k^2\sin^2\theta)^{\frac{3}{2}}}\sin\theta d\theta \hspace{19em}\\ &=&\cfrac{k}{1-k^2}\int_{0}^{\cfrac{\pi}{2}}\cfrac{d}{d\theta}\Bigl(\cfrac{-\cos\theta}{\sqrt{1-k^2\sin^2\theta}}\Bigr)\sin\theta d\theta \\ &=&\cfrac{k}{1-k^2}\big[-\cfrac{\cos\theta}{\sqrt{1-k^2\sin^2\theta}} \sin\theta \big]_0^{\cfrac{\pi}{2}}+ \cfrac{k}{1-k^2}\int_{0}^{\cfrac{\pi}{2}}\cfrac{\cos\theta}{\sqrt{1-k^2\sin^2\theta}}\cos\theta d\theta \\ &=&\cfrac{k}{1-k^2}\int_{0}^{\cfrac{\pi}{2}}\cfrac{\cos^2\theta}{\sqrt{1-k^2\sin^2\theta}}d\theta \\ &=&\cfrac{1}{k(1-k^2)}\int_{0}^{\cfrac{\pi}{2}}\cfrac{k^2(1-\sin^2\theta)}{\sqrt{1-k^2\sin^2\theta}}d\theta \\ &=&\cfrac{1}{k(1-k^2)}\int_{0}^{\cfrac{\pi}{2}}\cfrac{(1-k^2\sin^2\theta)+k^2-1}{\sqrt{1-k^2\sin^2\theta}}d\theta \\ &=&\cfrac{1}{k(1-k^2)}\Bigl(\int_{0}^{\cfrac{\pi}{2}}\sqrt{1-k^2\sin^2\theta}d\theta-\int_{0}^{\cfrac{\pi}{2}}\cfrac{1-k^2}{\sqrt{1-k^2\sin^2\theta}}d\theta\Bigr) \\ &=&\cfrac{1}{k(1-k^2)}\{E(k)-(1-k^2)K(k)\} \hspace{16em}\\ \end{eqnarray*}

$\hspace{2em} 定理４ \hspace{2em} \cfrac{dE(k)}{dk}=\cfrac{1}{k}\{E(k)-K(k)\}$

（証明）
\begin{eqnarray*} \cfrac{dE}{dk} &=&\int_{0}^{\cfrac{\pi}{2}}\cfrac{-k\sin^2\theta}{\sqrt{1-k^2\sin^2\theta}}d\theta \hspace{24em}\\ &=&\int_{0}^{\cfrac{\pi}{2}}\cfrac{(1-k^2\sin^2\theta)-1}{k\sqrt{1-k^2\sin^2\theta}}d\theta \\ &=&\cfrac{1}{k}\Bigl(\int_{0}^{\cfrac{\pi}{2}}\sqrt{1-k^2\sin^2\theta}d\theta-\int_{0}^{\cfrac{\pi}{2}}\cfrac{d\theta}{\sqrt{1-k^2\sin^2\theta}}\Bigr) \\ &=&\cfrac{1}{k}\{E(k)-K(k)\} \\ \end{eqnarray*}


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