Prove the Legendre equation on Elliptic integrals

[alyafey22]

\(\displaystyle K(k)E'(k)+K'(k)E(k)-K(k)K'(k)=\frac{\pi}{2}\)​

Complete Elliptic integral of first kind

\(\displaystyle K(k)= \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}\)

Complete Elliptic integral of second kind

\(\displaystyle E(k)= \int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-k^2x^2}}dx\)

Complementary integral

\(\displaystyle K'(k)=K(k')=K\left(\sqrt{1-k^2} \right)\)BONUS , prove that :

\(\displaystyle K(\sqrt{-1}\, k) = \frac{1}{\sqrt{1+k^2}} K\left(\frac{k}{\sqrt{1+k^2}}\right)\)

\(\displaystyle \frac{K'}{K}\left( \frac{1}{\sqrt{2}}\right)= 1 \, \)​



 

[Aaron Bex]

Let K(k) and K'(k) denote the complete elliptic integral of first kind and its complementary integral respectively.

Since K(k)= \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}} and K'(k)=K\left(\sqrt{1-k^2} \right), we can substitute k = \frac{k}{\sqrt{1+k^2}} into K(k) to get

K(\sqrt{-1}\, k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1+k^2-k^2x^2}}= \frac{1}{\sqrt{1+k^2}} K\left(\frac{k}{\sqrt{1+k^2}}\right)

To prove that \frac{K'}{K}\left( \frac{1}{\sqrt{2}}\right)= 1, we note that since K'(k)=K\left(\sqrt{1-k^2} \right), we have K'\left(\frac{1}{\sqrt{2}}\right)=K\left(\frac{1}{\sqrt{2}}\right).