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We are trying to implement transformations to evaluate the incomplete integral of the third kind $\Pi(n;\phi|m)$ for arbitrary inputs, and I can't find any references for how to calculate this function with phase $\phi$ less than zero or greater than $\pi/2$.

Are there periodic identities for this function for various ranges of $m$ and $n$? Plotting it in Mathematica tends to suggest there is, but the DLMF only lists periodicity for elliptic integrals of the first two kinds.

Update: Wolfram has a periodicity equation for $-1\le n\le 1$ although in practise it seems to work for $n<-1$ as well. But I'm still looking for a similar equation for $n>1$.

2nd update: mjqxxxx below gives periodicity for $n>1$, but for $m=1$ things are slightly different. (I'm trying to be complete.) Here's the real part for $\Pi(n>1,\phi|m=1)$:

real EllipticPi for

Which is symmetric around $\phi=\pi/2$ but anti-symmetric around $\phi=0$. But here's the strange one, the imaginary part:

alt text

Symmetric around $\phi=0$ but around $\phi=\pi/2$ the symmetry breaks down. Can the function be mapped from the region $\pi/2<\phi<\pi$ back to $0<\phi<\pi/2$?

2 Answers 2

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One definition given by Wikipedia for the incomplete elliptic integral of the third kind is: $ \Pi(n; \phi \vert m) = \int_0^{\sin\phi} \frac{1}{1-nt^2}\frac{dt}{\sqrt{(1-mt^2)(1-t^2)}}. $ This indicates that the function is periodic with period $2\pi$, is symmetric about $\phi=\pi/2$, and is antisymmetric about $\phi=0$. So the values for $0\le\phi < \pi/2$ define the function for all $\phi$.

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    Ah, I see now—I'd missed the symmetry around $\pi/2$. Thanks!2011-01-23
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(Not a full answer, but only some preliminary comments, since I've no reputation to do so.)

The identity in the Wolfram Functions site is, as mentioned, valid for characteristic $n$ in the interval $[-1,1)$ ; for $n=1$, your elliptic integral of the third kind degenerates into a particular combination of the elliptic integrals of the other two kinds and trigonometric functions:

$\Pi(n;\phi|m)=F(\phi|m)+\frac{\tan\;\phi\sqrt{1-m\sin^2\phi}-E(\phi|m)}{1-m}$

such that you can use the properties of the other elliptic integrals you already know.

That leaves the circular case $n<-1$ and the hyperbolic case $n>1$; in the sequel I assume that the parameter $m$ is in the usual interval $[0,1]$ (if you need to deal with parameters outside the usual range, I'll have to dig further into my references).

For the circular case, the identity

$\Pi(n;\phi|m)=\frac{n(1-m)}{(n-1)(m-n)}\Pi\left(\frac{m-n}{1-n};\phi|m\right)+\frac{m}{m-n}F(\phi|m)+\frac1{\sqrt{(1-n)(1-m/n)}}\arctan\left(\frac12\sqrt{\frac{n(m-n)}{n-1}}\frac{\sin\;2\phi}{\sqrt{1-m\sin^2\phi}}\right)$

can be used to express the circular case of an incomplete elliptic integral of the third kind back into the case where the characteristic is in the interval $(m,1)$

In the hyperbolic case, the identity you will need is

$\Pi(n;\phi|m)=F(\phi|m)-\Pi\left(\frac{m}{n};\phi|m\right)+\frac1{\sqrt{(n-1)(1-m/n)}}\mathrm{artanh}\left(\sqrt{(n-1)(1-m/n)}\frac{\tan\;\phi}{\sqrt{1-m\sin^2\phi}}\right)$

such that only characteristics in the interval $(0,m)$ are evaluated; the function is $2\pi$-periodic in $\phi$ for the hyperbolic case, so that if your amplitude $\phi$ is outside the usual interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, either add/subtract appropriate multiples of $2\pi$ from the amplitude, or add/subtract appropriate multiples of $\Pi(n|m)=K(m)-\Pi\left(\frac{m}{n}|m\right)$ from the right hand side of that identity.

Alternatively, if you can help it, you might do better to just use the Carlson integral of the third kind instead; there is less worry here of having to split into "circular" and "hyperbolic" cases. The relations to interconvert Carlson and Legendre-Jacobi integrals are in the DLMF.