Landen's transformation

The incomplete elliptic integral of the first kind F is

${\displaystyle F(\varphi \setminus \alpha )=F(\varphi ,\sin \alpha )=\int _{0}^{\varphi }{\frac {d\theta }{\sqrt {1-(\sin \theta \sin \alpha )^{2}}}},}$

where ${\displaystyle \alpha }$ is the modular angle. Landen's transformation states that if ${\displaystyle \alpha _{0}}$, ${\displaystyle \alpha _{1}}$, ${\displaystyle \varphi _{0}}$, ${\displaystyle \varphi _{1}}$ are such that ${\displaystyle (1+\sin \alpha _{1})(1+\cos \alpha _{0})=2}$ and ${\displaystyle \tan(\varphi _{1}-\varphi _{0})=\cos \alpha _{0}\tan \varphi _{0}}$, then^[2]

${\displaystyle {\begin{aligned}F(\varphi _{0}\setminus \alpha _{0})&=(1+\cos \alpha _{0})^{-1}F(\varphi _{1}\setminus \alpha _{1})\\&={\tfrac {1}{2}}(1+\sin \alpha _{1})F(\varphi _{1}\setminus \alpha _{1}).\end{aligned}}}$

Landen's transformation can similarly be expressed in terms of the elliptic modulus ${\displaystyle k=\sin \alpha }$ and its complement ${\displaystyle k'=\cos \alpha }$.

In Gauss's formulation, the value of the integral

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

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

${\displaystyle a_{1}={\frac {a+b}{2}},\qquad b_{1}={\sqrt {ab}},}$

${\displaystyle I_{1}=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {a_{1}^{2}\cos ^{2}(\theta )+b_{1}^{2}\sin ^{2}(\theta )}}}\,d\theta .}$

Therefore,

${\displaystyle I={\frac {1}{a}}K\left({\frac {\sqrt {a^{2}-b^{2}}}{a}}\right),}$

${\displaystyle I_{1}={\frac {2}{a+b}}K\left({\frac {a-b}{a+b}}\right).}$

From Landen's transformation we conclude

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

and ${\displaystyle I_{1}=I}$.

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

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

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

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

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

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

the relationship may be written as

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

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

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

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