504 XII Elliptic Functions
then
tdt/dθ=(a^2 −b^2 )sinθcosθ=[(a^2 −t^2 )(t^2 −b^2 )]^1 /^2and
J=
∫π/ 20φ((a^2 sin^2 θ+b^2 cos^2 θ)^1 /^2 )dθ/(a^2 sin^2 θ+b^2 cos^2 θ)^1 /^2.Now putt 1 =( 1 / 2 )(t+ab/t)and, as before,
a 1 =(a+b)/ 2 , b 1 =(ab)^1 /^2.Then
a 12 −t 12 =(a^2 −t^2 )(t^2 −b^2 )/ 4 t^2 ,
t 12 −b^21 =(t^2 −ab)^2 / 4 t^2 ,
dt 1 /dt=(t^2 −ab)/ 2 t^2.Astincreases frombtob 1 ,t 1 decreases froma 1 tob 1 ,andastfurther increases from
b 1 toa,t 1 increases fromb 1 back toa 1 .Since
t=t 1 ±(t 12 −b^21 )^1 /^2 ,it follows from these observations that
∫a
bφ(t)[(a^2 −t^2 )(t^2 −b^2 )]−^1 /^2 dt=∫a 1b 1ψ(t 1 )[(a 12 −t^21 )(t 12 −b 12 )]−^1 /^2 dt 1 ,where
ψ(t 1 )=( 1 / 2 ){φ[(t 1 +(t 12 −b^21 )^1 /^2 ]+φ[(t 1 −(t 12 −b 12 )^1 /^2 ]}.In particular, if we takeφ(t)=1 and putK(a,b):=∫ab[(a^2 −t^2 )(t^2 −b^2 )]−^1 /^2 dt,we obtain
K(a,b)=K(a 1 ,b 1 ).Hence, by repeating the process,K(a,b)=K(an,bn).But
K(an,bn)=∫π/ 20(a^2 nsin^2 θ+bn^2 cos^2 θ)−^1 /^2 dθ