렘니스케이트(lemniscate) 곡선의 길이와 타원적분

간단한 소개

[/pages/2090560/attachments/944074 lemniscate-1.png]

• 위의 극좌표로 주어진 곡선을 베르누이의 Lemniscate 라 부름.
• 가우스는 이 곡선의 길이와 관련하여 다음과 같은 기록을 일기에 남김. (Pi-unleashed, 99p)

We have gained some very elegant details about the lemniscate, which have exceeded all expectations, and indeed using methods which open up an entirely new field. That the AGM is equal to \(\frac{\pi }{\omega}\) between 1 and \(\sqrt{2}\) we have confirmed up to the 11th decimal digit; if this is proven, then a truly new field of analysis stands before us.

• Put $x=r(\theta)\cos \theta$ and $y=r(\theta)\sin \theta$.
  The arclength $L$ of the lemniscate is given by $4\int^{\frac{\pi}{4}}_{0}\sqrt{r'(\theta)^2+r(\theta)^2}d\theta$. Since $r(\theta)^2=\cos 2\theta$ and $r'(\theta)=-\frac{\sin 2\theta}{r(\theta)}$, $$L=4\int^{\frac{\pi}{4}}_{0}\sqrt{\frac{\sin^2 2\theta}{r(\theta)^2}+\cos 2\theta}d\theta=4\int^{\frac{\pi}{4}}_{0}\sqrt{\frac{\sin^2 2\theta+\cos^2 2\theta}{\cos 2\theta}}d\theta=4\int^{\frac{\pi}{4}}_{0}\frac{d\theta}{\sqrt{\cos 2\theta}}$$
  
  Now put $\cos 2\theta=\cos^2\phi$ and then $d\theta=\frac{\sin\phi \cos\phi}{\sin 2\theta}d\phi$. We get
  $$L=4\int^{\frac{\pi}{2}}_{0}\frac{\sin\phi \cos\phi }{\cos\phi \sqrt{\sin2\theta}} d\phi=4\int^{\frac{\pi}{2}}_{0}\frac{\sin\phi}{\sqrt{\sin2\theta}}d\phi$$
  Since $\sin^2 2\theta=1-\cos^4\phi=\sin^2 \phi(1+\cos^2\phi)$, $L=4\int^{\frac{\pi}{2}}_{0}\frac{d\phi}{\sqrt{1+\cos^2 \phi}}$.
  
  From this last equation, we get both
  $$L=4\int^{\frac{\pi}{2}}_{0}\frac{d\phi}{\sqrt{1+\cos^2 \phi}}=4\int^{\frac{\pi}{2}}_{0}\frac{d\phi}{\sqrt{2}\sqrt{1-\frac{1}{2}\sin^2 \phi}}=2\sqrt{2}K(\frac{1}{\sqrt{2}})$$ and by putting $x=\cos\phi$,
  $$L=4\int^{\frac{\pi}{2}}_{0}\frac{d\phi}{\sqrt{1+\cos^2 \phi}}=4\int^{1}_{0}\frac{dx}{\sqrt{1-x^4}}$$