Carlson symmetric form

Relation to the Legendre forms

Incomplete elliptic integrals

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

:\begin{align} F(\phi,k)&=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right), \[5mu]

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), \[5mu]

\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). \end{align}

(Note: the above are only valid for \textstyle -\tfrac\pi2\le\phi\le\frac\pi2 and 0\le k^2\sin^2\phi\le1)

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting \phi=\tfrac{\pi}{2}:

:\begin{align} K(k) &= R_F\left(0,1-k^2,1\right), \[5mu] 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), \[5mu] \Pi(n,k) &= R_F\left(0,1-k^2,1\right)+\tfrac{1}{3}n R_J \left(0,1-k^2,1,1-n\right) \end{align}

Special cases

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

:\begin{align} 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} \[5mu] &= \begin{cases} \dfrac{\arccos \sqrt}{\sqrt{y - x}}, & x \dfrac{1}{\sqrt{y}}, & x = y \[3mu] \dfrac{\operatorname{arcosh} \sqrt}{\sqrt{x - y}}, & x y \end{cases} \end{align}

Similarly, when at least two of the first three arguments of R_J are the same,

:\begin{align} R_{J}(x,y,y,p) &= 3 \int {\sqrt{x}}^{\infty} \frac{du}{(u^{2} - x + y) (u^{2} - x + p)} \[5mu] &= \begin{cases} \dfrac{3}{p - y} (R{C}(x,y) - R_{C}(x,p)), & y \ne p \[3mu] \dfrac{3}{2 (y - x)} \left( R_{C}(x,y) - \dfrac{1}{y} \sqrt{x}\right), & y = p \ne x \[3mu] \dfrac{1}{y^}, &y = p = x \end{cases} \end{align}

Properties

Homogeneity

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

:\begin{align} R_F\left(\kappa x,\kappa y,\kappa z\right) &= \kappa^{-1/2}R_F(x,y,z), \[5mu] R_J\left(\kappa x,\kappa y,\kappa z,\kappa p\right) &= \kappa^{-3/2}R_J(x,y,z,p). \end{align}

Duplication theorem

: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 \lambda=\sqrt{\vphantom{ty} x}\sqrt{\vphantom{ty} y}+\sqrt{\vphantom{ty} y}\sqrt{\vphantom{ty} z}+\sqrt{\vphantom{ty} z}\sqrt{\vphantom{ty} x}.

:\begin{align}R_{J}(x,y,z,p) & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \[5mu] & = \frac{1}{4} R_{J}\left( \frac{x + \lambda}{4},\frac{y + \lambda}{4},\frac{z + \lambda}{4},\frac{p + \lambda}{4}\right) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \end{align}

where d = \bigl(\sqrt{\vphantom{ty} p} + \sqrt{\vphantom{ty} x}\bigr) \bigl(\sqrt{\vphantom{ty} p} + \sqrt{\vphantom{ty} y}\bigr) \bigl(\sqrt{\vphantom{ty} p} + \sqrt{\vphantom{ty} z}\bigr) and \lambda =\sqrt{\vphantom{ty} x}\sqrt{\vphantom{ty} y}+\sqrt{\vphantom{ty} y}\sqrt{\vphantom{ty} z}+\sqrt{\vphantom{ty} z}\sqrt{\vphantom{ty} x}.

Series Expansion

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

:\begin{align}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{align}

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

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

:E_{2} = \Delta x \Delta y + \Delta y \Delta z + \Delta z \Delta x

: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...

:\begin{align}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{3 E_{2}^{2}}{8 (t + 1)^{\frac{11}{2}}} - \frac{3 E_{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{align}

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

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

:A = \frac{x + y + z + 2 p}{5}

and the differences \Delta x, \Delta y \Delta z and \Delta p defined by

:\begin{align}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^{3/2}} R_{J}(1 - \Delta x,1 - \Delta y,1 - \Delta z,1 - \Delta p) \end{align}

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

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

: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

: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

: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

:E_{5} = \Delta x \Delta y \Delta z \Delta p^{2}

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

:\begin{align}R_{J}(x,y,z,p) & = \frac{3}{2 A^{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}{2 A^{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{3 E_{2}^{2} - 4 E_{4}}{8 (t + 1)^{\frac{13}{2}}} + \frac{2 E_{5} - 3 E_{2} E_{3}}{4 (t + 1)^{\frac{15}{2}}} + O(E_{1}) + O(\Delta^{6})\right) dt \ & = \frac{1}{A^{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{align}

As with R_{J}, by expanding about the mean value of the arguments, more than half the terms (those involving 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 R_C, or the fourth argument, p, of 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

:\mathrm{p.v.}; R_C(x, -y) = \sqrt{\frac{x}{x + y}},R_C(x + y, y),

and

:\begin{align}\mathrm{p.v.}; R_{J}(x,y,z,-p) & = \frac{(q - y) R_{J}(x,y,z,q) - 3 R_{F}(x,y,z) + 3 \sqrt{y\vphantom{t}} R_{C}(x z,- p q)}{y + p} \ & = \frac{(q - y) R_{J}(x,y,z,q) - 3 R_{F}(x,y,z) + 3 \sqrt{\dfrac{x y z}{x z + p q}} R_{C}(x z + p q,p q)}{y + p} \end{align} where

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

which must be greater than zero for 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 R_F(x,y,z): first, define x_0=x, y_0=y and z_0=z. Then iterate the series

:\lambda_n=\sqrt{\vphantom{ty} x_n}\sqrt{\vphantom{ty} y_n}+\sqrt{\vphantom{ty} y_n}\sqrt{\vphantom{ty} z_n}+\sqrt{\vphantom{ty} z_n}\sqrt{\vphantom{ty} x_n}, :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 x, y and z are non-negative, all of the series will converge quickly to a given value, say, \mu. Therefore,

:R_F\left(x,y,z\right)=R_F\left(\mu,\mu,\mu\right)=\mu^{-1/2}.

Evaluating R_C(x,y) is much the same due to the relation

:R_C\left(x,y\right)=R_F\left(x,y,y\right).