Dawson function

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:

$R_F(x,y,z)=\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dt}{\sqrt{(t+x)(t+y)(t+z)}}$

$R_J(x,y,z,p)=\frac{3}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dt}{(t+p)\sqrt{(t+x)(t+y)(t+z)}}$

$R_C(x,y)=R_F(x,y,y)=\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dt}{(t+y)\sqrt{(t+x)}}$

$R_D(x,y,z)=R_J(x,y,z,z)=\frac{3}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dt}{(t+z)\sqrt{(t+x)(t+y)(t+z)}}$

Since $R_C$ and $R_D$ are special cases of $R_F$ and $R_J$, all elliptic integrals can ultimately be evaluated in terms of just $R_F$ and $R_J$.

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

The Carlson elliptic integrals are named after Bille C. Carlson.

Relation to the Legendre forms

Incomplete elliptic integrals

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

$F(\varphi ,k)=\sin \varphi R_F({\cos }^{}\varphi ,1-k^2{\sin }^{}\varphi ,1)$

$E(\varphi ,k)=\sin \varphi R_F({\cos }^{}\varphi ,1-k^2{\sin }^{}\varphi ,1)-\frac{1}{3}{k}^2{\sin }^{}\varphi R_D({\cos }^{}\varphi ,1-k^2{\sin }^{}\varphi ,1)$

$\mathrm{\Pi }(\varphi ,n,k)=\sin \varphi R_F({\cos }^{}\varphi ,1-k^2{\sin }^{}\varphi ,1)+\frac{1}{3}n{\sin }^{}\varphi R_J({\cos }^{}\varphi ,1-k^2{\sin }^{}\varphi ,1,1-n{\sin }^{}\varphi )$

(Note: the above are only valid for $0\le \varphi \le 2\pi$ and $0\le k^2{\sin }^2\varphi \le 1$)

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting φ = Template:Fracπ:

$K(k)=R_F(0,1-k^2,1)$

$E(k)=R_F(0,1-k^2,1)-\frac{1}{3}{k}^2{R}_D(0,1-k^2,1)$

$\mathrm{\Pi }(n,k)=R_F(0,1-k^2,1)+\frac{1}{3}n{R}_J(0,1-k^2,1,1-n)$

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.

$R_C(x,y)=R_F(x,y,y)=\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{t+x}(t+y)}dt=\underset{\sqrt{x}}{\overset{\mathrm{\infty }}{\int }}\frac{1}{u^2-x+y}du=\{\begin{array}{ll}\frac{\arccos \sqrt{\frac{x}{y}}}{\sqrt{y-x}},& x<y\\ \frac{1}{\sqrt{y}},& x=y\\ \frac{\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\sqrt{\frac{x}{y}}}{\sqrt{x-y}},& x>y\\ \end{array}$

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

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

Properties

Homogeneity

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

$R_F(\kappa x,\kappa y,\kappa z)={\kappa }^{-1/2}{R}_F(x,y,z)$

$R_J(\kappa x,\kappa y,\kappa z,\kappa p)={\kappa }^{-3/2}{R}_J(x,y,z,p)$

Duplication theorem

$R_F(x,y,z)=2R_F(x+\lambda ,y+\lambda ,z+\lambda )=R_F(\frac{x+\lambda }{4},\frac{y+\lambda }{4},\frac{z+\lambda }{4}),$

where $\lambda =\sqrt{xy}+\sqrt{yz}+\sqrt{zx}$.

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

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

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

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

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

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

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

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

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

$\begin{array}{rl}{R}_F(x,y,z)& =\frac{1}{2\sqrt{A}}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{(t+1{)}^3-(t+1{)}^2{E}_1+(t+1)E_2-E_3}}dt\\ & =\frac{1}{2\sqrt{A}}\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{(t+1{)}^{\frac{3}{2}}}-\frac{E_2}{2(t+1{)}^{\frac{7}{2}}}+\frac{E_3}{2(t+1{)}^{\frac{9}{2}}}+\frac{3E_2^2}{8(t+1{)}^{\frac{11}{2}}}-\frac{3E_2{E}_3}{4(t+1{)}^{\frac{13}{2}}}+O(E_1)+O({\mathrm{\Delta }}^6))dt\\ & =\frac{1}{\sqrt{A}}(1-\frac{1}{10}{E}_2+\frac{1}{14}{E}_3+\frac{1}{24}{E}_2^2-\frac{3}{44}{E}_2{E}_3+O(E_1)+O({\mathrm{\Delta }}^6))\end{array}$

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 take to be

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

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

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

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

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

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

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

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

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

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...

$\begin{array}{rl}{R}_J(x,y,z,p)& =\frac{3}{2A^{\frac{3}{2}}}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{(t+1{)}^5-(t+1{)}^4{E}_1+(t+1{)}^3{E}_2-(t+1{)}^2{E}_3+(t+1)E_4-E_5}}dt\\ & =\frac{3}{2A^{\frac{3}{2}}}\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{(t+1{)}^{\frac{5}{2}}}-\frac{E_2}{2(t+1{)}^{\frac{9}{2}}}+\frac{E_3}{2(t+1{)}^{\frac{11}{2}}}+\frac{3E_2^2-4E_4}{8(t+1{)}^{\frac{13}{2}}}+\frac{2E_5-3E_2{E}_3}{4(t+1{)}^{\frac{15}{2}}}+O(E_1)+O({\mathrm{\Delta }}^6))dt\\ & =\frac{1}{A^{\frac{3}{2}}}(1-\frac{3}{14}{E}_2+\frac{1}{6}{E}_3+\frac{9}{88}{E}_2^2-\frac{3}{22}{E}_4-\frac{9}{52}{E}_2{E}_3+\frac{3}{26}{E}_5+O(E_1)+O({\mathrm{\Delta }}^6))\end{array}$

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

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

and

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

where

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

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

${\lambda }_n=\sqrt{x_n{y}_n}+\sqrt{y_n{z}_n}+\sqrt{z_n{x}_n},$

$x_{n+1}=\frac{x_n+{\lambda }_n}{4},y_{n+1}=\frac{y_n+{\lambda }_n}{4},z_{n+1}=\frac{z_n+{\lambda }_n}{4}$

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,

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

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

$R_C(x,y)=R_F(x,y,y).$

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.

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