Number Theory: An Introduction to Mathematics

530 XII Elliptic Functions

Hence, by (49),

u′′=[1+( 1 −λ(τ))^1 /^2 ]u.

By Proposition 5 also,

sn(u′′; 2 τ)=−iθ 00 ( 0 ; 2 τ)θ 10 (v;τ)θ 11 (v;τ)/θ 10 ( 0 ; 2 τ)θ 00 (v;τ)θ 01 (v;τ).

On the other hand,

sn(u;τ)cn(u;τ)/dn(u;τ)=−iθ 002 ( 0 ;τ)θ 10 (v;τ)θ 11 (v;τ)/D,

whereD=θ 102 ( 0 ;τ)θ 00 (v;τ)θ 01 (v;τ).

Since 2θ 00 ( 0 ; 2 τ)θ 10 ( 0 ; 2 τ)=θ 102 ( 0 ;τ), it follows that

sn(u′′; 2 τ)= 2 θ 002 ( 0 ; 2 τ)sn(u;τ)cn(u;τ)/θ 002 ( 0 ;τ)dn(u;τ).

Since 2θ 002 ( 0 ; 2 τ)/θ 002 ( 0 ;τ)=u′′/u, this proves the first assertion of the proposition.
The remaining assertions may be proved similarly.

We show finally how the standard elliptic integrals of the second and third kinds,
defined by (24) and (25), may be expressed in terms of theta functions. If we put

Θ(u)=θ 01 (v), (56)

whereu=πθ 002 ( 0 )v,thensince

λS(u)=λsn^2 u=−θ 102 ( 0 )θ 112 (v)/θ 002 ( 0 )θ 012 (v),

we can rewrite (45) in the form

d{Θ′(u)/Θ(u)}/du=−α+ 1 −λS(u),

whereαis independent ofuand the prime on the left denotes differentiation with
respect tou.SinceΘ′( 0 )=0, by integrating we obtain

E(u)=Θ′(u)/Θ(u)+αu.

To determineαwe takeu=K.Sinceθ 01 ′( 1 / 2 )=θ 00 ′( 1 )=θ 00 ′( 0 )=0, we obtain
α=E/K,where

E=E(K)=

∫ K

0

{ 1 −λS(u)}du=

∫ 1

0

( 1 −λx)dx/gλ(x)^1 /^2

is a complete elliptic integral of the second kind. Thus

E(u)=Θ′(u)/Θ(u)+uE/K. (57)

Substituting this expression forE(u)in (27), we further obtain

Π(u,a)=uΘ′(a)/Θ(a)+( 1 / 2 )log{Θ(u−a)/Θ(u+a)}. (58)