Landen's transformation

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 ${\displaystyle \theta =\arctan \left({\frac {x}{b}}\right)}$, ${\displaystyle d\theta =\left({\frac {1}{b}}\cos ^{2}(\theta )\right)dx}$ giving

${\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta =\int _{0}^{\infty }{\frac {1}{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}\,dx}$

A further substitution of ${\displaystyle x=t+{\sqrt {t^{2}+ab}}}$ gives the desired result

${\displaystyle {\begin{aligned}I&=\int _{0}^{\infty }{\frac {1}{\sqrt {(x^{2}+a^{2})(x^{2}+b^{2})}}}\,dx\\&=\int _{-\infty }^{\infty }{\frac {1}{2{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)(t^{2}+ab)}}}}\,dt\\&=\int _{0}^{\infty }{\frac {1}{\sqrt {\left(t^{2}+\left({\frac {a+b}{2}}\right)^{2}\right)\left(t^{2}+\left({\sqrt {ab}}\right)^{2}\right)}}}\,dt\end{aligned}}}$

This latter step is facilitated by writing the radical as

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

and the infinitesimal as

${\displaystyle dx={\frac {x}{\sqrt {t^{2}+ab}}}\,dt}$

so that the factor of ${\displaystyle x}$ is recognized and cancelled between the two factors.

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

${\displaystyle I=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}\,d\theta =\int _{0}^{\frac {\pi }{2}}{\frac {1}{\operatorname {AGM} (a,b)}}\,d\theta ={\frac {\pi }{2\,\operatorname {AGM} (a,b)}}}$

The integral may also be recognized as a multiple of Legendre's complete elliptic integral of the first kind. Putting ${\displaystyle b^{2}=a^{2}(1-k^{2})}$

${\displaystyle I={\frac {1}{a}}\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {1-k^{2}\sin ^{2}(\theta )}}}\,d\theta ={\frac {1}{a}}F\left({\frac {\pi }{2}},k\right)={\frac {1}{a}}K(k)}$

Hence, for any ${\displaystyle a}$, the arithmetic-geometric mean and the complete elliptic integral of the first kind are related by

${\displaystyle K(k)={\frac {\pi a}{2\,\operatorname {AGM} (a,a{\sqrt {1-k^{2}}})}}}$

By performing an inverse transformation (reverse arithmetic-geometric mean iteration), that is

${\displaystyle a_{-1}=a+{\sqrt {a^{2}-b^{2}}}\,}$

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

${\displaystyle \operatorname {AGM} (a,b)=\operatorname {AGM} (a+{\sqrt {a^{2}-b^{2}}},a-{\sqrt {a^{2}-b^{2}}})\,}$

the relationship may be written as

${\displaystyle K(k)={\frac {\pi a}{2\,\operatorname {AGM} (a(1+k),a(1-k))}}\,}$

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

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

The definition adopted here for ${\displaystyle \scriptstyle {K(k)}}$ differs from that used in the arithmetic-geometric mean article, such that ${\displaystyle \scriptstyle {K(k)}}$ here is ${\displaystyle \scriptstyle {K(m^{2})}}$ in that article.