Statement
Complete elliptic integral In Gauss's formulation, the value of the integral
I = ∫ 0 π 2 1 a 2 cos 2 ( θ ) + b 2 sin 2 ( θ ) d θ {\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta } is unchanged if a {\displaystyle a} and b {\displaystyle b} are replaced by their arithmetic and geometric means respectively, that is
a 1 = a + b 2 , b 1 = a b , {\displaystyle a_{1}={\frac {a+b}{2}},\qquad b_{1}={\sqrt {ab}},} I 1 = ∫ 0 π 2 1 a 1 2 cos 2 ( θ ) + b 1 2 sin 2 ( θ ) d θ . {\displaystyle I_{1}=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a_{1}^{2}\cos ^{2}(\theta )+b_{1}^{2}\sin ^{2}(\theta )}}}\,d\theta .} Therefore,
I = 1 a K ( a 2 − b 2 a ) , {\displaystyle I={\frac {1}{a}}K\left({\frac {\sqrt {a^{2}-b^{2}}}{a}}\right),} I 1 = 2 a + b K ( a − b a + b ) . {\displaystyle I_{1}={\frac {2}{a+b}}K\left({\frac {a-b}{a+b}}\right).} From Landen's transformation we conclude
K ( a 2 − b 2 a ) = 2 a a + b K ( a − b a + b ) {\displaystyle K\left({\frac {\sqrt {a^{2}-b^{2}}}{a}}\right)={\frac {2a}{a+b}}K\left({\frac {a-b}{a+b}}\right)} and I 1 = I {\displaystyle I_{1}=I} .
Proof The transformation may be effected by integration by substitution . It is convenient to first cast the integral in an algebraic form by a substitution of θ = arctan ( x / b ) {\displaystyle \theta =\arctan(x/b)} , d θ = ( cos 2 ( θ ) / b ) d x {\displaystyle d\theta =(\cos ^{2}(\theta )/b)dx} giving
I = ∫ 0 π 2 1 a 2 cos 2 ( θ ) + b 2 sin 2 ( θ ) d θ = ∫ 0 ∞ 1 ( x 2 + a 2 ) ( x 2 + b 2 ) d x {\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta =\int _{0}^{\infty }{\frac {1}{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}\,dx} A further substitution of x = t + t 2 + a b {\displaystyle x=t+{\sqrt {t^{2}+ab}}} gives the desired result
I = ∫ 0 ∞ 1 ( x 2 + a 2 ) ( x 2 + b 2 ) d x = ∫ − ∞ ∞ 1 2 ( t 2 + ( a + b 2 ) 2 ) ( t 2 + a b ) d t = ∫ 0 ∞ 1 ( t 2 + ( a + b 2 ) 2 ) ( t 2 + ( a b ) 2 ) d t {\displaystyle {\begin{aligned}I&=\int _{0}^{\infty }{\frac {1}{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}\,dx\\&=\int _{-\infty }^{\infty }{\frac {1}{2{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)(t^{2}+ab)}}}}\,dt\\&=\int _{0}^{\infty }{\frac {1}{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)\left(t^{2}+\left({\sqrt {ab}}\right)^{2}\right)}}}\,dt\end{aligned}}} This latter step is facilitated by writing the radical as
( x 2 + a 2 ) ( x 2 + b 2 ) = 2 x t 2 + ( a + b 2 ) 2 {\displaystyle {\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}=2x{\sqrt {t^{2}+\left({\frac {a+b}{2}}\right)^{2}}}} and the infinitesimal as
d x = x t 2 + a b d t {\displaystyle dx={\frac {x}{\sqrt {t^{2}+ab}}}\,dt} so that the factor of x {\displaystyle x} is recognized and cancelled between the two factors.
Arithmetic-geometric mean and Legendre's first integral If the transformation is iterated a number of times, then the parameters a {\displaystyle a} and b {\displaystyle b} converge very rapidly to a common value, even if they are initially of different orders of magnitude. The limiting value is called the arithmetic-geometric mean of a {\displaystyle a} and b {\displaystyle b} , AGM ( a , b ) {\displaystyle \operatorname {AGM} (a,b)} . In the limit, the integrand becomes a constant, so that integration is trivial
I = ∫ 0 π 2 1 a 2 cos 2 ( θ ) + b 2 sin 2 ( θ ) d θ = ∫ 0 π 2 1 AGM ( a , b ) d θ = π 2 AGM ( a , b ) {\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta =\int _{0}^{\frac {\pi }{2}}{\frac {1}{\operatorname {AGM} (a,b)}}\,d\theta ={\frac {\pi }{2\operatorname {AGM} (a,b)}}} The integral may also be recognized as a multiple of Legendre's complete elliptic integral of the first kind . Putting b 2 = a 2 ( 1 − k 2 ) {\displaystyle b^{2}=a^{2}(1-k^{2})}
I = 1 a ∫ 0 π 2 1 1 − k 2 sin 2 ( θ ) d θ = 1 a F ( π 2 , k ) = 1 a K ( k ) {\displaystyle I={\frac {1}{a}}\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {1-k^{2}\sin ^{2}(\theta )}}}\,d\theta ={\frac {1}{a}}F\left({\frac {\pi }{2}},k\right)={\frac {1}{a}}K(k)} Hence, for any a {\displaystyle a} , the arithmetic-geometric mean and the complete elliptic integral of the first kind are related by
K ( k ) = π 2 AGM ( 1 , 1 − k 2 ) {\displaystyle K(k)={\frac {\pi }{2\operatorname {AGM} (1,{\sqrt {1-k^{2}}})}}} By performing an inverse transformation (reverse arithmetic-geometric mean iteration), that is
a − 1 = a + a 2 − b 2 {\displaystyle a_{-1}=a+{\sqrt {a^{2}-b^{2}}}\,} b − 1 = a − a 2 − b 2 {\displaystyle b_{-1}=a-{\sqrt {a^{2}-b^{2}}}\,} AGM ( a , b ) = AGM ( a + a 2 − b 2 , a − a 2 − b 2 ) {\displaystyle \operatorname {AGM} (a,b)=\operatorname {AGM} \left(a+{\sqrt {a^{2}-b^{2}}},a-{\sqrt {a^{2}-b^{2}}}\right)\,} the relationship may be written as
K ( k ) = π 2 AGM ( 1 + k , 1 − k ) {\displaystyle K(k)={\frac {\pi }{2\operatorname {AGM} (1+k,1-k)}}\,} which may be solved for the AGM of a pair of arbitrary arguments;
AGM ( u , v ) = π ( u + v ) 4 K ( u − v v + u ) . {\displaystyle \operatorname {AGM} (u,v)={\frac {\pi (u+v)}{4K\left({\frac {u-v}{v+u}}\right)}}.}
References