Carlson symmetric form

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.

${\displaystyle R_C(x,y)=R_F(x,y,y)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{t+x}(t+y)}={\int }_{\sqrt{x}}^{\mathrm{\infty }}\frac{du}{u^2-x+y}=\{\begin{array}{ll}\frac{\arccos \sqrt{x/y}}{\sqrt{y-x}},& x<y\\ \frac{1}{\sqrt{y}},& x=y\\ \frac{\mathrm{arcosh}\sqrt{x/y}}{\sqrt{x-y}},& x>y\end{array}}$

Similarly, when at least two of the first three arguments of ${\displaystyle R_J}$ are the same,

${\displaystyle R_J(x,y,y,p)=3{\int }_{\sqrt{x}}^{\mathrm{\infty }}\frac{du}{(u^2-x+y)(u^2-x+p)}=\{\begin{array}{ll}\frac{3}{p-y}(R_C(x,y)-R_C(x,p)),& y\ne p\\ \frac{3}{2(y-x)}\left(R_C(x,y)-\frac{1}{y}\sqrt{x}\right),& y=p\ne x\\ \frac{1}{y^{3/2}},& y=p=x\end{array}}$

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

${\displaystyle \begin{array}{cc}{R}_J(x,y,z,p)& =2R_J(x+\lambda ,y+\lambda ,z+\lambda ,p+\lambda )+6R_C(d^2,d^2+(p-x)(p-y)(p-z))\\ & =\frac{1}{4}{R}_J\left(\frac{x+\lambda }{4},\frac{y+\lambda }{4},\frac{z+\lambda }{4},\frac{p+\lambda }{4}\right)+6R_C(d^2,d^2+(p-x)(p-y)(p-z))\end{array}}$

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 ${\displaystyle \mathrm{\Delta }x}$, ${\displaystyle \mathrm{\Delta }y}$ and ${\displaystyle \mathrm{\Delta }z}$ by

${\displaystyle \begin{array}{cc}{R}_F(x,y,z)& =R_F(A(1-\mathrm{\Delta }x),A(1-\mathrm{\Delta }y),A(1-\mathrm{\Delta }z))\\ & =\frac{1}{\sqrt{A}}{R}_F(1-\mathrm{\Delta }x,1-\mathrm{\Delta }y,1-\mathrm{\Delta }z)\end{array}}$

that is ${\displaystyle \mathrm{\Delta }x=1-x/A}$ etc. The differences ${\displaystyle \mathrm{\Delta }x}$, ${\displaystyle \mathrm{\Delta }y}$ and ${\displaystyle \mathrm{\Delta }z}$ 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 ${\displaystyle \mathrm{\Delta }x}$, ${\displaystyle \mathrm{\Delta }y}$ and ${\displaystyle \mathrm{\Delta }z}$. It follows that both the integrand of ${\displaystyle R_F}$ and its integral can be expressed as functions of the elementary symmetric polynomials in ${\displaystyle \mathrm{\Delta }x}$, ${\displaystyle \mathrm{\Delta }y}$ and ${\displaystyle \mathrm{\Delta }z}$ which are

${\displaystyle E_1=\mathrm{\Delta }x+\mathrm{\Delta }y+\mathrm{\Delta }z=0}$

${\displaystyle E_2=\mathrm{\Delta }x\mathrm{\Delta }y+\mathrm{\Delta }y\mathrm{\Delta }z+\mathrm{\Delta }z\mathrm{\Delta }x}$

${\displaystyle E_3=\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }z}$

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

${\displaystyle \begin{array}{cc}{R}_F(x,y,z)& =\frac{1}{2\sqrt{A}}{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{(t+1{)}^3-(t+1{)}^2{E}_1+(t+1)E_2-E_3}}dt\\ & =\frac{1}{2\sqrt{A}}{\int }_0^{\mathrm{\infty }}\left(\frac{1}{(t+1{)}^{\frac{3}{2}}}-\frac{E_2}{2(t+1{)}^{\frac{7}{2}}}+\frac{E_3}{2(t+1{)}^{\frac{9}{2}}}+\frac{3E_2^2}{8(t+1{)}^{\frac{11}{2}}}-\frac{3E_2{E}_3}{4(t+1{)}^{\frac{13}{2}}}+O(E_1)+O({\mathrm{\Delta }}^6)\right)dt\\ & =\frac{1}{\sqrt{A}}\left(1-\frac{1}{10}{E}_2+\frac{1}{14}{E}_3+\frac{1}{24}{E}_2^2-\frac{3}{44}{E}_2{E}_3+O(E_1)+O({\mathrm{\Delta }}^6)\right)\end{array}}$

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

${\displaystyle \begin{array}{cc}{R}_J(x,y,z,p)& =R_J(A(1-\mathrm{\Delta }x),A(1-\mathrm{\Delta }y),A(1-\mathrm{\Delta }z),A(1-\mathrm{\Delta }p))\\ & =\frac{1}{A^{\frac{3}{2}}}{R}_J(1-\mathrm{\Delta }x,1-\mathrm{\Delta }y,1-\mathrm{\Delta }z,1-\mathrm{\Delta }p)\end{array}}$

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

${\displaystyle E_1=\mathrm{\Delta }x+\mathrm{\Delta }y+\mathrm{\Delta }z+2\mathrm{\Delta }p=0}$

${\displaystyle E_2=\mathrm{\Delta }x\mathrm{\Delta }y+\mathrm{\Delta }y\mathrm{\Delta }z+2\mathrm{\Delta }z\mathrm{\Delta }p+\mathrm{\Delta }{p}^2+2\mathrm{\Delta }p\mathrm{\Delta }x+\mathrm{\Delta }x\mathrm{\Delta }z+2\mathrm{\Delta }y\mathrm{\Delta }p}$

${\displaystyle E_3=\mathrm{\Delta }z\mathrm{\Delta }{p}^2+\mathrm{\Delta }x\mathrm{\Delta }{p}^2+2\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }p+\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }z+2\mathrm{\Delta }y\mathrm{\Delta }z\mathrm{\Delta }p+\mathrm{\Delta }y\mathrm{\Delta }{p}^2+2\mathrm{\Delta }x\mathrm{\Delta }z\mathrm{\Delta }p}$

${\displaystyle E_4=\mathrm{\Delta }y\mathrm{\Delta }z\mathrm{\Delta }{p}^2+\mathrm{\Delta }x\mathrm{\Delta }z\mathrm{\Delta }{p}^2+\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }{p}^2+2\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }z\mathrm{\Delta }p}$

${\displaystyle E_5=\mathrm{\Delta }x\mathrm{\Delta }y\mathrm{\Delta }z\mathrm{\Delta }{p}^2}$

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...

${\displaystyle \begin{array}{cc}{R}_J(x,y,z,p)& =\frac{3}{2A^{\frac{3}{2}}}{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{(t+1{)}^5-(t+1{)}^4{E}_1+(t+1{)}^3{E}_2-(t+1{)}^2{E}_3+(t+1)E_4-E_5}}dt\\ & =\frac{3}{2A^{\frac{3}{2}}}{\int }_0^{\mathrm{\infty }}\left(\frac{1}{(t+1{)}^{\frac{5}{2}}}-\frac{E_2}{2(t+1{)}^{\frac{9}{2}}}+\frac{E_3}{2(t+1{)}^{\frac{11}{2}}}+\frac{3E_2^2-4E_4}{8(t+1{)}^{\frac{13}{2}}}+\frac{2E_5-3E_2{E}_3}{4(t+1{)}^{\frac{15}{2}}}+O(E_1)+O({\mathrm{\Delta }}^6)\right)dt\\ & =\frac{1}{A^{\frac{3}{2}}}\left(1-\frac{3}{14}{E}_2+\frac{1}{6}{E}_3+\frac{9}{88}{E}_2^2-\frac{3}{22}{E}_4-\frac{9}{52}{E}_2{E}_3+\frac{3}{26}{E}_5+O(E_1)+O({\mathrm{\Delta }}^6)\right)\end{array}}$

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

${\displaystyle \mathrm{p}.\mathrm{v}.R_C(x,-y)=\sqrt{\frac{x}{x+y}}{R}_C(x+y,y),}$

and

${\displaystyle \begin{array}{cc}\mathrm{p}.\mathrm{v}.R_J(x,y,z,-p)& =\frac{(q-y)R_J(x,y,z,q)-3R_F(x,y,z)+3\sqrt{y}{R}_C(xz,-pq)}{y+p}\\ & =\frac{(q-y)R_J(x,y,z,q)-3R_F(x,y,z)+3\sqrt{\frac{xyz}{xz+pq}}{R}_C(xz+pq,pq)}{y+p}\end{array}}$

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,