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I am interested in evaluating some elliptic integrals, and I have not been able to secure a reference to do exactly what I need. Most of the references I've found seem to focus on reducing more general elliptic integrals to Legendre form, but leave out the part about actually dealing with complete elliptic integrals. In particular, I am interested in the following toy problem, which is to show the following: $$\displaystyle \int_0^\infty \frac{dx}{\sqrt{x(x+2)(x+3)}} = \sqrt{2}K(-1/2),$$ where $K(k)$ is the complete elliptic integral of the first kind. The above result is due to Wolfram Alpha.

I tried the obvious substitution which is $x = \tan \theta$, and after some labour we obtain the integral

$$\displaystyle \int_0^{\pi/2} \frac{2 d \theta}{\sqrt{7 \sin 2\theta + 5 \sin^2 2 \theta + 5 \sin 2 \theta \cos 2 \theta}},$$ which again can be checked to evaluate to $\sqrt{2}K(-1/2)$, although in this case Wolfram only gave the numerical value and not the closed form. Further, this last one does not look like what the 'correct' form should be, which is

$$\displaystyle \int_0^{\pi/2} \frac{\sqrt{2} d \theta}{\sqrt{1 - (1/4)\sin^2 \theta}}.$$

Based on some data I got from playing around with Wolfram, I suspect that the following is true: Suppose that $0 < a < b$. Then

$$\displaystyle \int_0^\infty \frac{dx}{\sqrt{x(x+a)(x+b)}} = \frac{2 K(1 - b/a)}{\sqrt{a}}.$$

Any help would be much appreciated, I apologize if this problem is in fact trivial or well-known.

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  • $\begingroup$ Mathematica yields the last integral instantly, so why "suspect"? Or are you looking for an explicit proof? $\endgroup$ – Suvrit Oct 16 '15 at 4:11
  • $\begingroup$ @Suvrit admittedly, my skill at using mathematica is limited, as I am not familiar with its ability to do symbolic manipulation. But yes, I would like an explicit argument $\endgroup$ – Stanley Yao Xiao Oct 16 '15 at 4:14
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    $\begingroup$ You ma want to look at Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman (Springer, 1971). It contains many explicit formulas of this kind. Moreover, early in the book, they discuss how to reduce general elliptic integrals (with integrands rational in the square root of a polynomial up to degree four) to the standard ones (first, second or third kind). $\endgroup$ – Igor Khavkine Oct 16 '15 at 7:19
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This question has been out for a while, and I think it deserves a thorough answer, even tho I came quite late to this party.

As both the classical Legendre-Jacobi theory and the Carlson theory have been mentioned by other users, I'll treat the OP's integral from both viewpoints.


Legendre-Jacobi

The OP came pretty close to using the correct substitution. One thing that could have been done instead is to recall the Pythagorean identity $1+\tan^2 u=\sec^2 u$, so that the substitution that should have been used is $x=a \tan^2 u$, where $a=2$ or $a=3$. Taking the smaller value of $a$, and after some amount of algebra, we obtain

$$\begin{align*}\int_0^\infty\frac{\mathrm dx}{\sqrt{x(x+2)(x+3)}}&=\int_0^{\pi/2}\frac{2}{\sqrt{3-\sin^2u}}\mathrm du\\ &=\frac2{\sqrt{3}}\int_0^{\pi/2}\frac{\mathrm du}{\sqrt{1-\frac13\sin^2u}}\\&=\frac2{\sqrt{3}}K\left(\frac13\right)\end{align*}$$

where I use the parameter convention for elliptic integrals. (This is the same convention used in Abramowitz and Stegun. Relatedly, see here for an extended discussion on the notational confusion surrounding elliptic integrals.)

Had we chosen the substitution with $a=3$ instead, we would have instead obtained the result $\sqrt{2}K\left(-\frac12\right)$, which is equivalent through the imaginary modulus transformation

$$K(-m)=\frac1{\sqrt{1+m}}K\left(\frac{m}{m+1}\right),\quad m>0$$


Carlson

In Carlson's theory, there is the general hypergeometric function

$$R_{-a}(b_1,\dots,b_k;z_1,\dots,z_k)=\frac1{\mathbf B\left(a,-a+\sum_j b_j\right)}\int_0^\infty u^{-a-1+\sum_j b_j}\prod_j \left(u+z_j\right)^{-b_j}\mathrm du$$

where $\mathbf B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the usual Euler beta function.

This multivariate hypergeometric function is a homogenized/symmetrized version of the classical Lauricella $F_D$ function (see e.g. this paper where Carlson introduced his function, tho that reference uses an opposite sign convention for $a$).

With this consideration, the OP's original integral can be expressed either as a three-variable Carlson integral, or a two-variable Carlson integral:

$$\begin{align*}\int_0^\infty\frac{\mathrm dx}{\sqrt{x(x+2)(x+3)}}&=2\,R_{-\frac12}\left(\frac12,\frac12,\frac12;0,2,3\right)\\&=\pi\,R_{-\frac12}\left(\frac12,\frac12;2,3\right)\end{align*}$$

The three-variable (incomplete) case occurs often enough that it is given the notation

$$\begin{align*}R_F(x,y,z)&=\frac12\int_0^\infty\frac{\mathrm du}{\sqrt{(u+x)(u+y)(u+z)}}\\&=R_{-\frac12}\left(\frac12,\frac12,\frac12;x,y,z\right)\end{align*}$$

and the two-variable form corresponds to the so-called "complete case",

$$\begin{align*}R_K(x,y)&=R_{-\frac12}\left(\frac12,\frac12;x,y\right)\\&=\frac2{\pi}R_F(0,x,y)\end{align*}$$

In fact, the two-variable form has the integral representation

$$R_K(x,y)=\frac2{\pi}\int_0^{\pi/2}\frac{\mathrm du}{\sqrt{x\cos^2 u+y\sin^2 u}}$$

which one recognizes to be related to the integral representation of Gauss's arithmetic-geometric mean (AGM):

$$R_K(x,y)=\frac1{\operatorname{agm}(\sqrt{x},\sqrt{y})}$$

Thus, the OP's integral is $\pi/\operatorname{agm}(\sqrt{2},\sqrt{3})$.

Additionally, the two-variable Carlson function is also related to the Gauss hypergeometric function, being a homogeneous version of it:

$$R_{-a}(b_1,b_2;x,y)=y^{-a}{}_2 F_1\left({{a,b_1}\atop{b_1+b_2}}\middle|1-\frac{x}{y}\right)$$

so one has

$$\begin{align*}\pi\,R_{-\frac12}\left(\frac12,\frac12;2,3\right)&=\frac{\pi}{\sqrt{3}}{}_2 F_1\left({{\frac12,\frac12}\atop{1}}\middle|\frac13\right)\\&=\frac2{\sqrt{3}}K\left(\frac13\right)\end{align*}$$

where we have used the hypergeometric representation of the complete elliptic integral of the first kind.

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    $\begingroup$ (Somewhat relatedly, I recently wrote a Mathematica package implementing the Carlson integrals. I have still been working on them on-and-off despite computer difficulties, so the identities remain fresh in my mind.) $\endgroup$ – J. M. is not a mathematician Jul 30 at 9:57
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It seems to be known as symmetric elliptic integrals of Carlson. Look in the NIST book, 19.15 and further. There are a lot of formulas in it. It seems you seek for exactly the formula 19.22.8 on page 505, note that in it $R_F$ is defined by 19.16.1-19.16.4 and AGM is exactly the complete Legendre integral $K$ as you suggested.

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  • $\begingroup$ Thank you for the reference, it does indeed answer this question. However, the reference provided does not contain a proof for formula 19.22.8, which is dissatisfying to me $\endgroup$ – Stanley Yao Xiao Oct 16 '15 at 5:27
  • $\begingroup$ They give the reference to the original book of Carlson with all proofs, please note it:B. C. Carlson (1977b). Special Functions of Applied Mathematics. New York: Academic Press. $\endgroup$ – Sergei Oct 16 '15 at 5:42

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