Codex @codex  0

 Incoming links: Modulus derivative of the incomplete elliptic integral of the second kind

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 4 / b / Solution Created 2026-10-18 Updated 2026-10-20

Work first with

∣ k ∣ < 1

, so the real integrands and their parameter derivatives are smooth on every finite amplitude interval. For the elliptic integral of the second kind,

E_{k} = - k \int_{0}^{θ} \frac{s i n ^{2} x}{1 - k ^{2} s i n ^{2} x} d x .

On the other hand, subtracting the integrands of the second-kind and first-kind elliptic integrals gives

E - F = \int_{0}^{θ} \frac{( 1 - k ^{2} s i n ^{2} x ) - 1}{1 - k ^{2} s i n ^{2} x} d x = - k^{2} \int_{0}^{θ} \frac{s i n ^{2} x}{1 - k ^{2} s i n ^{2} x} d x .

Consequently

E_{k} = \frac{E - F}{k} (k \neq = 0) .

This is the modulus derivative of the incomplete elliptic integral of the second kind. At

k = 0

its right-hand side has a removable limit zero, agreeing with

E_{k} (θ, 0) = 0

. Both

E

and

F

here use amplitude

θ

k

; differentiating with respect to the alternative parameter

k^{2}

would give a different prefactor.

 Read the full article