Landen's transformation

Statement

The incomplete elliptic integral of the first kind F is : F(\varphi \setminus \alpha) = F(\varphi, \sin \alpha) = \int_0^\varphi \frac{d \theta}{\sqrt{1-(\sin \theta \sin \alpha)^2}}, where \alpha is the modular angle. Landen's transformation states that if \alpha_0, \alpha_1, \varphi_0, \varphi_1 are such that (1 + \sin\alpha_1)(1 + \cos\alpha_0) = 2 and \tan(\varphi_1 - \varphi_0) = \cos\alpha_0 \tan \varphi_0, then :\begin{align} F(\varphi_0 \setminus \alpha_0) &= (1 + \cos\alpha_0)^{-1} F(\varphi_1 \setminus \alpha_1) \ &= \tfrac{1}{2}(1 + \sin\alpha_1) F(\varphi_1 \setminus \alpha_1). \end{align} Landen's transformation can similarly be expressed in terms of the elliptic modulus k = \sin\alpha and its complement k' = \cos\alpha.

Complete elliptic integral

In Gauss's formulation, the value of the integral :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 and b are replaced by their arithmetic and geometric means respectively, that is :a_1 = \frac{a + b}{2},\qquad b_1 = \sqrt{a b}, :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=\frac{1}{a}K\left(\frac{\sqrt{a^2 - b^2}}{a}\right),
:I_1=\frac{2}{a+b}K\left(\frac{a-b}{a+b}\right).
From Landen's transformation we conclude :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.

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 \theta = \arctan (x/b), d \theta = (\cos^{2}(\theta)/b) d x giving

: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 + \sqrt{t^{2} + a b} gives the desired result

:\begin{align}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 + a b)}} , dt \ & = \int _0^\infty\frac{1}{\sqrt{\left( t^2 + \left( \frac{a + b}{2}\right)^2\right) \left(t^2 + \left(\sqrt{a b}\right)^2\right)}} , dt \end{align}

This latter step is facilitated by writing the radical as

:\sqrt{(x^2 + a^2) (x^2 + b^2)} = 2x \sqrt{t^2 + \left( \frac{a + b}{2}\right)^2}

and the infinitesimal as

: dx = \frac{x}{\sqrt{t^2 + a b}} , dt

so that the factor of x is recognized and cancelled between the two factors.

To prove \left(x^2+a^2\right)\left(x^2+b^2\right) = 4,x^2\left( t^2+\left( \frac{a+b}{2} \right)^2 \right) we write x=t+r with r=\sqrt{t^2+ab}. We also write \bar{x}=t-r. Then x^2\bar{x}^2 = a^2b^2 since x\bar{x} = t^2-r^2 = -ab, and x^2+\bar{x}^2 = 2,\left(t^2+r^2\right) = 4,t^2+2ab. Then \begin{aligned}\left(x^2+a^2\right)\left(x^2+b^2\right) &= x^2\left(x^2+a^2+b^2+\frac{a^2b^2}{x^2} \right)\ &= x^2\left(x^2+\bar{x}^2+a^2+b^2\right)\ &= x^2\left(4,t^2+2ab+a^2+b^2\right)\ &= 4,x^2\left( t^2+\left( \frac{a+b}{2} \right)^2 \right)\end{aligned}

Arithmetic-geometric mean and Legendre's first integral

If the transformation is iterated a number of times, then the parameters a and 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 and b, \operatorname{AGM}(a,b). In the limit, the integrand becomes a constant, so that integration is trivial

: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)

: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, the arithmetic-geometric mean and the complete elliptic integral of the first kind are related by

: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 + \sqrt{a^2 - b^2} ,

:b_{-1} = a - \sqrt{a^2 - b^2} ,

:\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) = \frac{\pi}{2 \operatorname{AGM}(1 + k, 1 - k)} ,

which may be solved for the AGM of a pair of arbitrary arguments;

: \operatorname{AGM}(u,v) = \frac{\pi (u + v)}{4 K\left( \frac{u - v}{v + u}\right)}.

References

• Louis V. King On The Direct Numerical Calculation Of Elliptic Functions And Integrals (Cambridge University Press, 1924)

References

1. (1876). "Arithmetisch geometrisches Mittel, Werke, Bd. 3". Königlichen Gesell. Wiss., Göttingen.
2. {{AS ref