This page will document useful transformations of symmetric elliptic integrals and relations among them, along with their derivations.These can be applied to simplifications of analytic formulae or numerical evaluations.

Definitions of Integrals
Relations among the Integrals
Relations with Legendre Integrals

Definitions of Integrals

Carlson symmetric forms of elliptic integrals are useful for complex analysis and numerical evaluation due to their simplicity compared to traditional forms. Integrals of the first, second and third kind in symmetric form are

$\begin{array}{c} R_{F} (x, y, z) = \frac{1}{2} \int_{0}^{\infty} \frac{d t}{\sqrt{(t + x) (t + y) (t + z)}} \\ R_{G} (x, y, z) = \frac{1}{4} \int_{0}^{\infty} \frac{t d t}{\sqrt{(t + x) (t + y) (t + z)}} (\frac{x}{t + x} + \frac{y}{t + y} + \frac{z}{t + z}) \\ R_{J} (x, y, z, w) = \frac{3}{2} \int_{0}^{\infty} \frac{d t}{(t + w) \sqrt{(t + x) (t + y) (t + z)}} \end{array}$

There are additionally two useful ‘degenerate’ forms

$\begin{array}{c} R_{C} (x, y) = R_{F} (x, y, y) = \frac{1}{2} \int_{0}^{\infty} \frac{d t}{(t + y) \sqrt{t + x}} \\ R_{D} (x, y, z) = R_{J} (x, y, z, z) = \frac{3}{2} \int_{0}^{\infty} \frac{d t}{(t + z)^{3 / 2} \sqrt{(t + x) (t + y)}} \end{array}$

This last integral is considered to be of second kind as well as degenerate third kind, for a reason to appear in relation to Legendre integrals.

Relations among the Integrals

There several relations among the integrals as defined, based on derivatives of their integrands. Given that

$\frac{d}{d t} \frac{1}{\sqrt{(t + x) (t + y) (t + z)}} = - \frac{1}{2} \frac{(t + y) (t + z) + (t + x) (t + z) + (t + x) (t + y)}{[(t + x) (t + y) (t + z)]^{3 / 2}}$

one can immediately write

$\begin{array}{l} R_{D} (y, z, x) + R_{D} (z, x, y) + R_{D} (x, y, z) \\ = - 3 \int_{0}^{\infty} d t \frac{d}{d t} \frac{1}{\sqrt{(t + x) (t + y) (t + z)}} = \frac{3}{\sqrt{x y z}} \end{array}$

Adding and subtracting t from each single variable x, y and z after taking them inside integrals, one can almost as easily establish

$\begin{array}{l} x R_{D} (y, z, x) + y R_{D} (z, x, y) + z R_{D} (x, y, z) \\ = \frac{3}{2} \int_{0}^{\infty} d t \frac{x (t + y) (t + z) + y (t + x) (t + z) + z (t + x) (t + y)}{[(t + x) (t + y) (t + z)]^{3 / 2}} \\ = 9 R_{F} (x, y, z) + 3 \int_{0}^{\infty} t d t \frac{d}{d t} \frac{1}{\sqrt{(t + x) (t + y) (t + z)}} = 3 R_{F} (x, y, z) \end{array}$

Now consider the derivative

$\begin{array}{l} \frac{d}{d t} \sqrt{\frac{t + y}{(t + x) (t + z)}} = \frac{1}{2} \frac{1}{\sqrt{(t + x) (t + y) (t + z)}} \\ - \frac{1}{2} \frac{t + y}{(t + x)^{3 / 2} \sqrt{(t + y) (t + z)}} - \frac{1}{2} \frac{t + y}{(t + z)^{3 / 2} \sqrt{(t + x) (t + y)}} \end{array}$

Again adding and subtracting t in numerators inside integrals, this allows one to establish

$\begin{array}{l} (x - y) R_{D} (y, z, x) + (z - y) R_{D} (x, y, z) \\ = \frac{3}{2} \int_{0}^{\infty} d t \frac{x + t - t - y}{(t + x)^{3 / 2} \sqrt{(t + y) (t + z)}} + \frac{3}{2} \int_{0}^{\infty} d t \frac{z + t - t - y}{(t + z)^{3 / 2} \sqrt{(t + x) (t + y)}} \\ = 3 R_{F} (x, y, z) + 3 \int_{0}^{\infty} d t \frac{d}{d t} \sqrt{\frac{t + y}{(t + x) (t + z)}} = 3 R_{F} (x, y, z) - 3 \sqrt{\frac{y}{x z}} \end{array}$

Interchanging x and y immediately gives

$(y - x) R_{D} (z, x, y) + (z - x) R_{D} (x, y, z) = 3 R_{F} (x, y, z) - 3 \sqrt{\frac{x}{y z}}$

Since the degenerate integral $R_{D}$ is only symmetric in the first two arguments, these last two relations represent a way to move either of the first two arguments into the second position. This comes in handy in showing that the completely symmetric integral of the second kind is actually a function of two other Carlson integrals:

$\begin{array}{l} R_{G} (x, y, z) = \frac{1}{4} \int_{0}^{\infty} \frac{d t}{\sqrt{(t + x) (t + y) (t + z)}} \\ \times [\frac{x (t + x - x)}{t + x} + \frac{x (t + y - y)}{t + y} + \frac{z (t + z - z)}{t + z}] \\ = \frac{1}{2} [x R_{F} (x, y, z) + y R_{F} (x, y, z) + z R_{F} (x, y, z)] \\ - \frac{1}{6} [x^{2} R_{D} (y, z, x) + y^{2} R_{D} (z, x, y) + z^{2} R_{D} (x, y, z)] \\ = \frac{1}{2} [x R_{F} (x, y, z) + y R_{F} (x, y, z) + z R_{F} (x, y, z)] \\ - \frac{1}{6} [\frac{x^{2}}{x - y} [3 R_{F} (x, y, z) - (z - y) R_{D} (x, y, z) - 3 \sqrt{\frac{y}{x z}}] \\ - \frac{y^{2}}{x - y} [3 R_{F} (x, y, z) - (z - x) R_{D} (x, y, z) - 3 \sqrt{\frac{x}{y z}}] \\ + z^{2} R_{D} (x, y, z)] \\ R_{G} (x, y, z) = \frac{1}{2} z R_{F} (x, y, z) - \frac{1}{6} (x - z) (y - z) R_{D} (x, y, z) + \frac{1}{2} \sqrt{\frac{x y}{z}} \end{array}$

Relations with Legendre Integrals

The Carlson forms can be converted to Legendre normal form with the change of variable

$u = \frac{sin φ}{\sqrt{t + 1}} t = \frac{{sin}^{2} φ}{u^{2}} - 1 d t = - \frac{2 {sin}^{2} φ}{u^{3}} d u$

The limits of integration are transformed to $sin φ$ and zero, then reversed by the negative sign the in the differential. The integral of the first kind becomes

$\begin{array}{l} R_{F} (x, y, z) = {sin}^{2} φ \int_{0}^{sin φ} \frac{d u}{u^{3} \sqrt{(\frac{{sin}^{2} φ}{u^{2}} - 1 + x) (\frac{{sin}^{2} φ}{u^{2}} - 1 + y) (\frac{{sin}^{2} φ}{u^{2}} - 1 + z)}} \\ = \frac{1}{sin φ} \int_{0}^{sin φ} \frac{d u}{\sqrt{[1 + (x - 1) \frac{u^{2}}{{sin}^{2} φ}] [1 + (y - 1) \frac{u^{2}}{{sin}^{2} φ}] [1 + (z - 1) \frac{u^{2}}{{sin}^{2} φ}]}} \end{array}$

so that with the choices

$x = 1 - {sin}^{2} φ y = 1 - m {sin}^{2} φ z = 1$

one recovers the expression given at the outset of this presentation. This may be written a bit more succinctly as

$R_{F} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1) = \frac{1}{sin φ} F (φ | m)$

The symmetric integral of the third kind has only one additional factor in the denominator, which can be rewritten

$\frac{1}{t + w} = \frac{u^{2}}{{sin}^{2} φ [1 + (w - 1) \frac{u^{2}}{{sin}^{2} φ}]} = \frac{1 - 1 + n u^{2}}{n {sin}^{2} φ [1 + (w - 1) \frac{u^{2}}{{sin}^{2} φ}]}$

so that with the additional choice $w = 1 - n {sin}^{2} φ$ one has

$R_{J} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1, 1 - n {sin}^{2} φ) = \frac{3}{n {sin}^{3} φ} [Π (n; φ | m) - F (φ | m)]$

The first of the two degenerate integrals can be explicitly integrated in terms of inverse circular functions, so need not be considered in the context of conversion to Legendre form. The second degenerate integral again has one additional term in the denominator, which in this case can be rewritten

$\frac{1}{t + w} = \frac{u^{2}}{{sin}^{2} φ [1 + (w - 1) \frac{u^{2}}{{sin}^{2} φ}]} = \frac{1 - 1 + m u^{2}}{m {sin}^{2} φ [1 + (w - 1) \frac{u^{2}}{{sin}^{2} φ}]}$

so that with the additional choice $w = z = 1$ one has

$R_{D} ({cos}^{2} φ, 1 - {sin}^{2} φ, 1) = \frac{3}{m {sin}^{3} φ} [F (φ | m) - E (φ | m)]$

These three expressions can now be inverted to give the Legendre elliptic integrals in terms of the Carlson symmetric integrals:

$\begin{array}{l} F (φ | m) = sin φ R_{F} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1) \\ E (φ | m) = sin φ R_{F} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1) \\ - \frac{m {sin}^{3} φ}{3} R_{D} ({cos}^{2} φ, 1 - {sin}^{2} φ, 1) \\ Π (n; φ | m) = sin φ R_{F} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1) \\ + \frac{n {sin}^{3} φ}{3} R_{J} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1, 1 - n {sin}^{2} φ) \end{array}$

With the same choices of parameters, the completely symmetric Carlson integral of the second kind can now be written using its relation to the other symmetric integrals as

$\begin{array}{l} R_{G} ({cos}^{2} φ, 1 - m {sin}^{2} φ, 1) \\ = \frac{{cos}^{2} φ}{2 sin φ} F (φ | m) + \frac{sin φ}{2} E (φ | m) + \frac{cos φ}{2} \sqrt{1 - m {sin}^{2} φ} \end{array}$

and so is in general a linear combination of Legendre integrals of the first and second kind.

One final relation: the inverse Weierstrass elliptic function is defined by a now familiar integral,

$℘^{- 1} (z) = \frac{1}{2} \int_{z}^{\infty} \frac{d t}{\sqrt{(t - e_{1}) (t - e_{2}) (t - e_{3})}} = R_{F} (z - e_{1}, z - e_{2}, z - e_{3})$

where a linear change of variable of integration is all that is needed to establish this relation.

Contents

Definitions of Integrals

Relations among the Integrals

Relations with Legendre Integrals