Landen's transformation

Landen's transformation is a mapping of the parameters of an elliptic integral, which shows how the value of the integral, changes when its parameters:amplitude and modular angle changes following some dependency. As a special case, we can see when the transformation does not change the value of the integral.

It was originally due to John Landen, although independently rediscovered by Carl Friedrich Gauss.

For example,the incomplete elliptic integral of the first kind $F$ is

F(\varphi,k) = F(\varphi \,|\, k^2) = F(\sin \varphi ; k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.

If parameters will be φ' and k' then $F(\varphi',k') = \int_0^{\varphi'} \frac {d\theta}{\sqrt{1 - k'^2 \sin^2 \theta}}.$

Landen's transformation shows how one can calculate second integral through first and formula connecting k and k'.

A comprehensive list of transformations included in the tables 21.6-2 and 21.6-3 "Mathematical Handbook for scientists and engineers" by G.A. Korn and T.M. Korn. Accordingly 21.6-3

K(k)=\frac{2}{1+k'} K(\frac{1-k'}{1+k'})

(aa)

Where

k'={\sqrt{1 - k^2}}

K(k) = \int_0^{\frac{\pi}{2}} \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.

Consider an example when the transformation does not change the value of the integral.

Let

I = \int _0^{\frac{\pi}{2}}\frac{1}{\sqrt{a^2 \cos^2(\theta) + b^2 \sin^2(\theta)}} \, d \theta

and $\scriptstyle{a}$ and $\scriptstyle{b}$ are replaced by their arithmetic and geometric means respectively, that is

a_1 = \frac{a + b}{2},\qquad b_1 = \sqrt{a b}.\,

$I_1 = \int _0^{\frac{\pi}{2}}\frac{1}{\sqrt{a_1^2 \cos^2(\theta) + b_1^2 \sin^2(\theta)}} \, d \theta$

Obviously

I=\frac{1}{a}K(\frac{\sqrt{(a^2 - b^2)}}{a})

I_1=\frac{2}{a+b}K(\frac{a-b}{a+b})

Accordingly formula (aa)

K(\frac{\sqrt{(a^2 - b^2)}}{a})=\frac{2a}{a+b}K(\frac{a-b}{a+b})

(aaa)

As follows from the formula (aaa)

I_1=I

The same equation can be proved using a simple mathematical analysis.

The transformation, may be achieved purely by integration by substitution. It is convenient to first cast the integral in an algebraic form by a substitution of $\scriptstyle{\theta = \arctan\left( \frac{x}{b}\right)}$ , $\scriptstyle{d \theta = \left( \frac{1}{b}\cos^{2}(\theta)\right) d x}$ 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 $\scriptstyle{x = t + \sqrt{t^{2} + a b}}$ gives the desired result (in the algebraic form)

\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 $\scriptstyle{x}$ is easily 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 $\scriptstyle{a}$ and $\scriptstyle{b}$ 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 $\scriptstyle{a}$ and $\scriptstyle{b}$ , $\scriptstyle{\operatorname{AGM}(a,b)}$ . 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 $\scriptstyle{b^2 = a^2 (1 - k^2)}$