Number Theory: An Introduction to Mathematics

508 XII Elliptic Functions

It follows that

πK(a,c)/ 2 K(a,b)=log( 4 a 1 /c 0 )−

∑∞

n= 1

2 −nlog(an/an+ 1 ). (4)

Finally, to determineE(a,c)we can use the relation

K(a,b)E(a,c)+K(a,c)E(a,b)−a^2 K(a,b)K(a,c)=π/ 2.

By homogeneity we need only establish this relation fora=1. Since

K( 1 ,( 1 −λ)^1 /^2 )=

∫ 1

0

[4x( 1 −x)( 1 −λx)]−^1 /^2 dx,

E( 1 ,( 1 −λ)^1 /^2 )=

∫ 1

0

[( 1 −λx)/ 4 x( 1 −x)]^1 /^2 dx,

it is in fact equivalent to the following relation, due to Legendre, between the complete
elliptic integrals of the first and second kinds:

Proposition 1If

K(λ)=

∫ 1

0

[4x( 1 −x)( 1 −λx)]−^1 /^2 dx, E(λ)=

∫ 1

0

[( 1 −λx)/ 4 x( 1 −x)]^1 /^2 dx,

then

K(λ)E( 1 −λ)+K( 1 −λ)E(λ)−K(λ)K( 1 −λ)=π/ 2 for 0 <λ< 1. (5)

Proof We show first that the derivative of the left side of (5) is zero. Evidently

dE/dλ=−( 1 / 2 )

∫ 1

0

x[4x( 1 −x)( 1 −λx)]−^1 /^2 dx=[E(λ)−K(λ)]/ 2 λ.

Similarly,

dK/dλ=( 1 / 2 )

∫ 1

0

x( 1 −λx)−^1 [4x( 1 −x)( 1 −λx)]−^1 /^2 dx.

Substitutingx=( 1 −u)/( 1 −λu)and writingλ′= 1 −λ, we obtain

dK/dλ=( 1 / 2 λ′)

∫ 1

0

[( 1 −u)/ 4 u( 1 −λu)]^1 /^2 du

=[E(λ)−λ′K(λ)]/ 2 λλ′.

It follows that

d(λλ′dK/dλ)/dλ=K/ 4.

Thusy 1 (λ)=K(λ)is a solution of the second order linear differential equation

d(λλ′dy/dλ)/dλ−y/ 4 = 0. (6)