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Modulus derivative of the incomplete elliptic integral of the second kind (∂k​E(ϕ,k)=[E(ϕ,k)−F(ϕ,k)]/k)

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Complex analysis Elliptic integral Elliptic integral of the second kind
2026-10-20  0 By others on same topic  0 Discussions Create my own version
For real ∣k∣<1, define E(ϕ,k)=∫0ϕ​1−k2sin2x​dx and F(ϕ,k)=∫0ϕ​(1−k2sin2x)−1/2dx. Smoothness of the integrands permits differentiation under the integral sign. Subtracting the two integrands gives E−F=−k2∫0ϕ​sin2x/1−k2sin2x​dx, which is k∂k​E. This proves the displayed identity for k=0. At k=0, both sides extend continuously with value zero. The variable k is the modulus; derivatives with respect to the parameter m=k2 instead obey ∂m​E=(E−F)/(2m), with removable value −21​∫0ϕ​sin2xdx at zero.

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  • 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

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