Number Theory: An Introduction to Mathematics

496 XII Elliptic Functions

where

σ=α^2 cos^2 θ+β^2 sin^2 θ.

If we now put

u^2 =( 1 −r^2 )/( 1 −σr^2 ),

thenr^2 =( 1 −u^2 )/( 1 −σu^2 )and

rdr/du=−( 1 −σ)u/( 1 −σu^2 )^2.

Hence

S=ab

∫π/ 2

0

dθ

∫ 1

0

( 1 −σ)( 1 −σu^2 )−^2 du.

Inverting the order of integration and givingσits value, we obtain

S=ab

∫ 1

0

du

∫π/ 2

0

[( 1 −α^2 )cos^2 θ+( 1 −β^2 )sin^2 θ]

×[( 1 −α^2 u^2 )cos^2 θ+( 1 −β^2 u^2 )sin^2 θ]−^2 dθ.

It is readily verified that

∫π/ 2

0

cos^2 θ(mcos^2 θ+nsin^2 θ)−^2 dθ=π/ 4 m(mn)^1 /^2 ,

∫π/ 2

0

sin^2 θ(mcos^2 θ+nsin^2 θ)−^2 dθ=π/ 4 n(mn)^1 /^2.

Thus we obtain finally

S=(πab/ 4 )

∫ 1

0

[( 1 −α^2 )/( 1 −α^2 u^2 )+( 1 −β^2 )/( 1 −β^2 u^2 )]

×[( 1 −α^2 u^2 )( 1 −β^2 u^2 )]−^1 /^2 du.

By anelliptic integralone understands today any integral of the form

∫

R(x,w)dx,

whereR(x,w)is a rational function ofxandw,andwherew^2 =g(x)is a polynomial
inxof degree 3 or 4 without repeated roots. The elliptic integral is said to becomplete
if it is a definite integral in which the limits of integration are distinct roots ofg(x).
The case of a quartic is easily reduced to that of a cubic. In the preceding examples
we can simply puty=x^2. Thus, for the lemniscate,

s(Θ)= 2 −^1 /^2

∫Y

0

[4y( 1 −y)( 1 −y/ 2 )]−^1 /^2 dy.