reduction of elliptic integrals to standard form

Any integralDlmfPlanetmath of the form ∫R⁒(x,P⁒(x))⁒𝑑x, where R is a rational function and P is a polynomialPlanetmathPlanetmath of degree 3 or 4 can be expressed as a linear combination of elementary functionsMathworldPlanetmath and elliptic integralsMathworldPlanetmath of the first, second, and third kinds.

To begin, we will assume that P has no repeated roots. Were this not the case, we could simply pull the repeated factor out of the radicalMathworldPlanetmathPlanetmath and be left with a polynomial of degree of 1 or 2 inside the square rootMathworldPlanetmath and express the integral in terms of inverse trigonometric functionsDlmfMathworldPlanetmath.

Make a change of variables z=(a⁒x+b)/(c⁒x+d). By choosing the coefficients a,b,c,d suitably, one can cast P into either Jacobi’s normal form P⁒(z)=(1-z2)⁒(1-k2⁒z2) or Weierstrass’ normal form P⁒(z)=4⁒z3-g2⁒z-g3.

Note that


for suitable polynomials A,B,C,D. We can rationalize the denominator like so:


The rational functions F and G appearing in the foregoing equation are defined like so:

F⁒(z) = A⁒(z)⁒C⁒(z)-B⁒(z)⁒D⁒(z)⁒P⁒(z)C2⁒(z)-D2⁒(z)⁒P⁒(z)
G⁒(z) = 2⁒B⁒(z)⁒C⁒(z)-A⁒(z)⁒D⁒(z)C2⁒(z)-D2⁒(z)⁒P⁒(z)

Since ∫F⁒(z)⁒𝑑z may be expressed in terms of elementary functions, we shall focus our attention on the remaining piece, ∫G⁒(z)⁒P⁒(z)⁒𝑑z, which we shall write as ∫H⁒(z)/P⁒(z)⁒𝑑z, where H=P⁒G.. Because we may decompose H into partial fractionsPlanetmathPlanetmath, it suffices to consider the following cases, which we shall all An and Bn:


Here, n is a non-negative integer and r is a complex numberMathworldPlanetmathPlanetmath.

We will reduce thes further using integration by parts. Taking antiderivatives, we have:


These identities will allow us to express An’s and Bn’s with large n in terms of ones with smaller n’s.

At this point, it is convenient to employ the specific form of the polynominal P. We will first conside the Weierstrass normal form and then the Jacobi normal form.

Substituting into our identities and collecting terms, we find


Note that there are some cases which can be integrated in elementary terms. Namely, suppose that the power is odd:


Then we may make a change of variables y=z2 to obtain


which may be integrated using elementary functions.

Next, we derive some identities using integration by parts. Since


we have

(2⁒m+1) ∫z2⁒m⁒(1-z2)⁒(1-k2⁒z2)⁒(1-z2)⁒(1-k2⁒z2)⁒𝑑z
+ ∫z2⁒m+1⁒(92⁒k2⁒z3-3⁒(1+k2)⁒z)⁒(1-z2)⁒(1-k2⁒z2)⁒𝑑z
= z2⁒m+1⁒(1-z2)⁒(1-k2⁒z2)⁒(1-z2)⁒(1-k2⁒z2)+C

By colecting terms, this identity may be rewritten as follows:

(1+2⁒m+92⁒k2) ∫z2⁒m+4⁒(1-z2)⁒(1-k2⁒z2)⁒𝑑z-
(4+2⁒m)⁒(1+k2) ∫z2⁒m+2⁒(1-z2)⁒(1-k2⁒z2)⁒𝑑z+

By repeated use of this identity, we may express any integral of the form ∫z2⁒m⁒P⁒(z)⁒𝑑z as the sum of a linear combination of ∫z2⁒P⁒(z)⁒𝑑z and ∫P⁒(z)⁒𝑑z and the product of a polyomial and P⁒(z).

Likewise, we can use integration by parts to simplify integrals of the form


Will finish later β€” saving in case of computer crash.

Title reduction of elliptic integrals to standard form
Canonical name ReductionOfEllipticIntegralsToStandardForm
Date of creation 2014-02-01 18:13:38
Last modified on 2014-02-01 18:13:38
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 30
Author rspuzio (6075)
Entry type Theorem
Classification msc 33E05
Related topic ExpressibleInClosedForm