In mathematics, the Carlson symmetric forms of elliptic integrals are a small canonical set of elliptic integrals to which all others may be reduced. They are a modern alternative to the Legendre forms. The Legendre forms may be expressed in terms of the Carlson forms and vice versa.

The Carlson elliptic integrals are:

$${\displaystyle R_{F}(x,y,z)={\tfrac {1}{2))\int _{0}^{\infty }{\frac {dt}{\sqrt {(t+x)(t+y)(t+z)))))$$

$${\displaystyle R_{J}(x,y,z,p)={\tfrac {3}{2))\int _{0}^{\infty }{\frac {dt}{(t+p){\sqrt {(t+x)(t+y)(t+z)))))}$$

$${\displaystyle R_{C}(x,y)=R_{F}(x,y,y)={\tfrac {1}{2))\int _{0}^{\infty }{\frac {dt}{(t+y){\sqrt {(t+x)))))}$$

$${\displaystyle R_{D}(x,y,z)=R_{J}(x,y,z,z)={\tfrac {3}{2))\int _{0}^{\infty }{\frac {dt}{(t+z)\,{\sqrt {(t+x)(t+y)(t+z)))))}$$

Since ${\displaystyle R_{C))$ and ${\displaystyle R_{D))$ are special cases of ${\displaystyle R_{F))$ and ${\displaystyle R_{J))$, all elliptic integrals can ultimately be evaluated in terms of just ${\displaystyle R_{F))$ and ${\displaystyle R_{J))$.

The term symmetric refers to the fact that in contrast to the Legendre forms, these functions are unchanged by the exchange of certain subsets of their arguments. The value of ${\displaystyle R_{F}(x,y,z)}$ is the same for any permutation of its arguments, and the value of ${\displaystyle R_{J}(x,y,z,p)}$ is the same for any permutation of its first three arguments.

The Carlson elliptic integrals are named after Bille C. Carlson (1924-2013).

Relation to the Legendre forms

Incomplete elliptic integrals

Incomplete elliptic integrals can be calculated easily using Carlson symmetric forms:

${\displaystyle F(\phi ,k)=\sin \phi R_{F}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1\right)}$

${\displaystyle E(\phi ,k)=\sin \phi R_{F}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1\right)-{\tfrac {1}{3))k^{2}\sin ^{3}\phi R_{D}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1\right)}$

${\displaystyle \Pi (\phi ,n,k)=\sin \phi R_{F}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1\right)+{\tfrac {1}{3))n\sin ^{3}\phi R_{J}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1,1-n\sin ^{2}\phi \right)}$

(Note: the above are only valid for ${\displaystyle -{\frac {\pi }{2))\leq \phi \leq {\frac {\pi }{2))}$ and ${\displaystyle 0\leq k^{2}\sin ^{2}\phi \leq 1}$)

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting φ = 1⁄2π:

${\displaystyle K(k)=R_{F}\left(0,1-k^{2},1\right)}$

${\displaystyle E(k)=R_{F}\left(0,1-k^{2},1\right)-{\tfrac {1}{3))k^{2}R_{D}\left(0,1-k^{2},1\right)}$

${\displaystyle \Pi (n,k)=R_{F}\left(0,1-k^{2},1\right)+{\tfrac {1}{3))nR_{J}\left(0,1-k^{2},1,1-n\right)}$

Special cases

When any two, or all three of the arguments of ${\displaystyle R_{F))$ are the same, then a substitution of ${\displaystyle {\sqrt {t+x))=u}$ renders the integrand rational. The integral can then be expressed in terms of elementary transcendental functions.

${\displaystyle R_{C}(x,y)=R_{F}(x,y,y)={\frac {1}{2))\int _{0}^{\infty }{\frac {dt}((\sqrt {t+x))(t+y)))=\int _{\sqrt {x))^{\infty }{\frac {du}{u^{2}-x+y))={\begin{cases}{\frac {\arccos {\sqrt ((x}/{y)))){\sqrt {y-x))},&x<y\\{\frac {1}{\sqrt {y))},&x=y\\{\frac {\operatorname {arcosh} {\sqrt ((x}/{y)))){\sqrt {x-y))},&x>y\\\end{cases))}$

Similarly, when at least two of the first three arguments of ${\displaystyle R_{J))$ are the same,

${\displaystyle R_{J}(x,y,y,p)=3\int _{\sqrt {x))^{\infty }{\frac {du}{(u^{2}-x+y)(u^{2}-x+p)))={\begin{cases}{\frac {3}{p-y))(R_{C}(x,y)-R_{C}(x,p)),&y\neq p\\{\frac {3}{2(y-x)))\left(R_{C}(x,y)-{\frac {1}{y)){\sqrt {x))\right),&y=p\neq x\\{\frac {1}{y^((3}/{2)))),&y=p=x\\\end{cases))}$

Properties

Homogeneity

By substituting in the integral definitions ${\displaystyle t=\kappa u}$ for any constant ${\displaystyle \kappa }$, it is found that

${\displaystyle R_{F}\left(\kappa x,\kappa y,\kappa z\right)=\kappa ^{-1/2}R_{F}(x,y,z)}$

${\displaystyle R_{J}\left(\kappa x,\kappa y,\kappa z,\kappa p\right)=\kappa ^{-3/2}R_{J}(x,y,z,p)}$

Duplication theorem

${\displaystyle R_{F}(x,y,z)=2R_{F}(x+\lambda ,y+\lambda ,z+\lambda )=R_{F}\left({\frac {x+\lambda }{4)),{\frac {y+\lambda }{4)),{\frac {z+\lambda }{4))\right),}$

where ${\displaystyle \lambda ={\sqrt {x)){\sqrt {y))+{\sqrt {y)){\sqrt {z))+{\sqrt {z)){\sqrt {x))}$.

${\displaystyle {\begin{aligned}R_{J}(x,y,z,p)&=2R_{J}(x+\lambda ,y+\lambda ,z+\lambda ,p+\lambda )+6R_{C}(d^{2},d^{2}+(p-x)(p-y)(p-z))\\&={\frac {1}{4))R_{J}\left({\frac {x+\lambda }{4)),{\frac {y+\lambda }{4)),{\frac {z+\lambda }{4)),{\frac {p+\lambda }{4))\right)+6R_{C}(d^{2},d^{2}+(p-x)(p-y)(p-z))\end{aligned))}$^[1]

where ${\displaystyle d=({\sqrt {p))+{\sqrt {x)))({\sqrt {p))+{\sqrt {y)))({\sqrt {p))+{\sqrt {z)))}$ and ${\displaystyle \lambda ={\sqrt {x)){\sqrt {y))+{\sqrt {y)){\sqrt {z))+{\sqrt {z)){\sqrt {x))}$

Series Expansion

In obtaining a Taylor series expansion for ${\displaystyle R_{F))$ or ${\displaystyle R_{J))$ it proves convenient to expand about the mean value of the several arguments. So for ${\displaystyle R_{F))$, letting the mean value of the arguments be ${\displaystyle A=(x+y+z)/3}$, and using homogeneity, define ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$ and ${\displaystyle \Delta z}$ by

${\displaystyle {\begin{aligned}R_{F}(x,y,z)&=R_{F}(A(1-\Delta x),A(1-\Delta y),A(1-\Delta z))\\&={\frac {1}{\sqrt {A))}R_{F}(1-\Delta x,1-\Delta y,1-\Delta z)\end{aligned))}$

that is ${\displaystyle \Delta x=1-x/A}$ etc. The differences ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$ and ${\displaystyle \Delta z}$ are defined with this sign (such that they are subtracted), in order to be in agreement with Carlson's papers. Since ${\displaystyle R_{F}(x,y,z)}$ is symmetric under permutation of ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$, it is also symmetric in the quantities ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$ and ${\displaystyle \Delta z}$. It follows that both the integrand of ${\displaystyle R_{F))$ and its integral can be expressed as functions of the elementary symmetric polynomials in ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$ and ${\displaystyle \Delta z}$ which are

${\displaystyle E_{1}=\Delta x+\Delta y+\Delta z=0}$

${\displaystyle E_{2}=\Delta x\Delta y+\Delta y\Delta z+\Delta z\Delta x}$

${\displaystyle E_{3}=\Delta x\Delta y\Delta z}$

Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term...

${\displaystyle {\begin{aligned}R_{F}(x,y,z)&={\frac {1}{2{\sqrt {A))))\int _{0}^{\infty }{\frac {1}{\sqrt {(t+1)^{3}-(t+1)^{2}E_{1}+(t+1)E_{2}-E_{3))))dt\\&={\frac {1}{2{\sqrt {A))))\int _{0}^{\infty }\left({\frac {1}{(t+1)^{\frac {3}{2))))-{\frac {E_{2)){2(t+1)^{\frac {7}{2))))+{\frac {E_{3)){2(t+1)^{\frac {9}{2))))+{\frac {3E_{2}^{2)){8(t+1)^{\frac {11}{2))))-{\frac {3E_{2}E_{3)){4(t+1)^{\frac {13}{2))))+O(E_{1})+O(\Delta ^{6})\right)dt\\&={\frac {1}{\sqrt {A))}\left(1-{\frac {1}{10))E_{2}+{\frac {1}{14))E_{3}+{\frac {1}{24))E_{2}^{2}-{\frac {3}{44))E_{2}E_{3}+O(E_{1})+O(\Delta ^{6})\right)\end{aligned))}$

The advantage of expanding about the mean value of the arguments is now apparent; it reduces ${\displaystyle E_{1))$ identically to zero, and so eliminates all terms involving ${\displaystyle E_{1))$ - which otherwise would be the most numerous.

An ascending series for ${\displaystyle R_{J))$ may be found in a similar way. There is a slight difficulty because ${\displaystyle R_{J))$ is not fully symmetric; its dependence on its fourth argument, ${\displaystyle p}$, is different from its dependence on ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$. This is overcome by treating ${\displaystyle R_{J))$ as a fully symmetric function of five arguments, two of which happen to have the same value ${\displaystyle p}$. The mean value of the arguments is therefore taken to be

${\displaystyle A={\frac {x+y+z+2p}{5))}$

and the differences ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$ ${\displaystyle \Delta z}$ and ${\displaystyle \Delta p}$ defined by

${\displaystyle {\begin{aligned}R_{J}(x,y,z,p)&=R_{J}(A(1-\Delta x),A(1-\Delta y),A(1-\Delta z),A(1-\Delta p))\\&={\frac {1}{A^{\frac {3}{2))))R_{J}(1-\Delta x,1-\Delta y,1-\Delta z,1-\Delta p)\end{aligned))}$

The elementary symmetric polynomials in ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$, ${\displaystyle \Delta z}$, ${\displaystyle \Delta p}$ and (again) ${\displaystyle \Delta p}$ are in full

${\displaystyle E_{1}=\Delta x+\Delta y+\Delta z+2\Delta p=0}$

${\displaystyle E_{2}=\Delta x\Delta y+\Delta y\Delta z+2\Delta z\Delta p+\Delta p^{2}+2\Delta p\Delta x+\Delta x\Delta z+2\Delta y\Delta p}$

${\displaystyle E_{3}=\Delta z\Delta p^{2}+\Delta x\Delta p^{2}+2\Delta x\Delta y\Delta p+\Delta x\Delta y\Delta z+2\Delta y\Delta z\Delta p+\Delta y\Delta p^{2}+2\Delta x\Delta z\Delta p}$

${\displaystyle E_{4}=\Delta y\Delta z\Delta p^{2}+\Delta x\Delta z\Delta p^{2}+\Delta x\Delta y\Delta p^{2}+2\Delta x\Delta y\Delta z\Delta p}$

${\displaystyle E_{5}=\Delta x\Delta y\Delta z\Delta p^{2))$

However, it is possible to simplify the formulae for ${\displaystyle E_{2))$, ${\displaystyle E_{3))$ and ${\displaystyle E_{4))$ using the fact that ${\displaystyle E_{1}=0}$. Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term as before...

${\displaystyle {\begin{aligned}R_{J}(x,y,z,p)&={\frac {3}{2A^{\frac {3}{2))))\int _{0}^{\infty }{\frac {1}{\sqrt {(t+1)^{5}-(t+1)^{4}E_{1}+(t+1)^{3}E_{2}-(t+1)^{2}E_{3}+(t+1)E_{4}-E_{5))))dt\\&={\frac {3}{2A^{\frac {3}{2))))\int _{0}^{\infty }\left({\frac {1}{(t+1)^{\frac {5}{2))))-{\frac {E_{2)){2(t+1)^{\frac {9}{2))))+{\frac {E_{3)){2(t+1)^{\frac {11}{2))))+{\frac {3E_{2}^{2}-4E_{4)){8(t+1)^{\frac {13}{2))))+{\frac {2E_{5}-3E_{2}E_{3)){4(t+1)^{\frac {15}{2))))+O(E_{1})+O(\Delta ^{6})\right)dt\\&={\frac {1}{A^{\frac {3}{2))))\left(1-{\frac {3}{14))E_{2}+{\frac {1}{6))E_{3}+{\frac {9}{88))E_{2}^{2}-{\frac {3}{22))E_{4}-{\frac {9}{52))E_{2}E_{3}+{\frac {3}{26))E_{5}+O(E_{1})+O(\Delta ^{6})\right)\end{aligned))}$

As with ${\displaystyle R_{J))$, by expanding about the mean value of the arguments, more than half the terms (those involving ${\displaystyle E_{1))$) are eliminated.

Negative arguments

In general, the arguments x, y, z of Carlson's integrals may not be real and negative, as this would place a branch point on the path of integration, making the integral ambiguous. However, if the second argument of ${\displaystyle R_{C))$, or the fourth argument, p, of ${\displaystyle R_{J))$ is negative, then this results in a simple pole on the path of integration. In these cases the Cauchy principal value (finite part) of the integrals may be of interest; these are

${\displaystyle \mathrm {p.v.} \;R_{C}(x,-y)={\sqrt {\frac {x}{x+y))}\,R_{C}(x+y,y),}$

and

${\displaystyle {\begin{aligned}\mathrm {p.v.} \;R_{J}(x,y,z,-p)&={\frac {(q-y)R_{J}(x,y,z,q)-3R_{F}(x,y,z)+3{\sqrt {y))R_{C}(xz,-pq)}{y+p))\\&={\frac {(q-y)R_{J}(x,y,z,q)-3R_{F}(x,y,z)+3{\sqrt {\frac {xyz}{xz+pq))}R_{C}(xz+pq,pq)}{y+p))\end{aligned))}$

where

${\displaystyle q=y+{\frac {(z-y)(y-x)}{y+p)).}$

which must be greater than zero for ${\displaystyle R_{J}(x,y,z,q)}$ to be evaluated. This may be arranged by permuting x, y and z so that the value of y is between that of x and z.

Numerical evaluation

The duplication theorem can be used for a fast and robust evaluation of the Carlson symmetric form of elliptic integrals and therefore also for the evaluation of Legendre-form of elliptic integrals. Let us calculate ${\displaystyle R_{F}(x,y,z)}$: first, define ${\displaystyle x_{0}=x}$, ${\displaystyle y_{0}=y}$ and ${\displaystyle z_{0}=z}$. Then iterate the series

${\displaystyle \lambda _{n}={\sqrt {x_{n))}{\sqrt {y_{n))}+{\sqrt {y_{n))}{\sqrt {z_{n))}+{\sqrt {z_{n))}{\sqrt {x_{n))},}$

${\displaystyle x_{n+1}={\frac {x_{n}+\lambda _{n)){4)),y_{n+1}={\frac {y_{n}+\lambda _{n)){4)),z_{n+1}={\frac {z_{n}+\lambda _{n)){4))}$

until the desired precision is reached: if ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ are non-negative, all of the series will converge quickly to a given value, say, ${\displaystyle \mu }$. Therefore,

${\displaystyle R_{F}\left(x,y,z\right)=R_{F}\left(\mu ,\mu ,\mu \right)=\mu ^{-1/2}.}$

Evaluating ${\displaystyle R_{C}(x,y)}$ is much the same due to the relation

${\displaystyle R_{C}\left(x,y\right)=R_{F}\left(x,y,y\right).}$

References and External links

• B. C. Carlson, John L. Gustafson 'Asymptotic approximations for symmetric elliptic integrals' 1993 arXiv
• B. C. Carlson 'Numerical Computation of Real Or Complex Elliptic Integrals' 1994 arXiv
• B. C. Carlson 'Elliptic Integrals:Symmetric Integrals' in Chap. 19 of Digital Library of Mathematical Functions. Release date 2010-05-07. National Institute of Standards and Technology.
• 'Profile: Bille C. Carlson' in Digital Library of Mathematical Functions. National Institute of Standards and Technology.
• Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007), "Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN 978-0-521-88068-8, archived from the original on 2011-08-11, retrieved 2011-08-10
• Fortran code from SLATEC for evaluating RF, RJ, RC, RD,