Legendre's relation

Remarks

Proof idea: differentiation under the integral sign gives $\frac{dE}{dk} = \frac{E - K}{k}$ and $\frac{dK}{dk} = \frac{E - k'^2 K}{k k'^2}$. With these rules, and the same rules for $E'$ and $K'$ via $dk'/dk = -k/k'$, the derivative of the left side in $k$ is $0$, so the left side is constant. To find the constant, let $k \to 0$: then $K, E \to \pi/2$ and $E' \to 1$, while $K'$ grows only like $\log(4/k)$ and $E - K = O(k^2)$, so $(E - K)K' \to 0$ and the left side tends to $\pi/2$. Example: at $k = 1/\sqrt{2}$ one has $k = k'$, so $K = K'$ and $E = E'$, and the relation becomes $2EK - K^2 = \pi/2$. There is a parallel form for the periods $\omega_1, \omega_2$ and quasi-periods $\eta_1, \eta_2$ of the Weierstrass zeta function: $\eta_1\omega_2 - \eta_2\omega_1 = 2\pi i$ when $\operatorname{Im}(\omega_2/\omega_1) > 0$.

Going deeper

For a student

Legendre's relation is the key identity behind the Gauss–Legendre algorithm for $\pi$, also called the Brent–Salamin algorithm. That algorithm combines it with the arithmetic–geometric mean formula for $K$ and a similar formula for $E$. For an elliptic curve it is the same fact as the quasi-period relation for the Weierstrass zeta function, read for a curve with real periods. It is also used to compute $E$ from $K$ at special moduli, for example at $k = 1/\sqrt 2$, where it gives $E(1/\sqrt2) = \frac{\pi/2 + K^2}{2K}$ with $K = K(1/\sqrt2)$, a value tied to the lemniscate constant.