Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 290

Then

I a + b
2

, √ ab = I ( a, b ) .

M oreover,

π/ 2
M ( a, b ) .

I ( a, b )=

PROOF Let u = b tan θ . The integral becomes
I ( a, b )= ∞
0

∞

( u 2 + a 2 )( u 2 + b 2 ) = 1

du du

( u 2 + a 2 )( u 2 + b 2 ) .

2

−∞

Therefore,

I a + b
2

, √ ab = 1
2

∞

du
( u 2 +( a + b

.

) 2 )( u 2 + ab )

−∞

2

Let

v ,

ab
v

u = 1
2

− 0 <v< ∞ .

Then v = u + √ u 2 + ab .Since
u 2 + a + b
2

2

1

4 v 2 ( v 2 + a 2 )( v 2 + b 2 ) ,

=

it is straightforward to obtain
I a + b
2

, √ ab = ∞
0

dv
( v 2 + a 2 )( v 2 + b 2 ) = I ( a, b ) .

By induction, we obtain

I ( a, b )= I ( a 1 ,b 1 )= I ( a 2 ,b 2 )= ··· .

Let

a ∞ = b ∞ = M ( a, b ) = lim
n

a n = lim
n

b n .

→∞ →∞

It is fairly easy to justify taking the limit inside the integral sign to obtain

I ( a, b ) = lim
n

I ( a n ,b n )

→∞

= I ( a ∞ ,b ∞ )
= π/ 2
0

dθ

a 2

sin 2 θ cos 2 θ + b 2

∞ ∞

π/ 2

1
M ( a, b )

dθ
cos 2 θ +sin 2 θ =

0

π/ 2
M ( a, b ) .

=