PlanetMath: reduction of elliptic integrals to standard form

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (23)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

reduction of elliptic integrals to standard form

(Theorem)

Any integral of the form $\int R(x,\sqrt{P(x)}) \, dx$ , where is a rational function and is a polynomial of degree 3 or 4 can be expressed in terms of elementary functions and the functions , , and $\Pi$ .

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

Make a change of variables . By choosing the coefficients suitably, one can cast P into either Jacobi's normal form or Weierstrass' normal form .

Note that

$\displaystyle R (z, \sqrt{P(z)}) = \frac{A(z) + B(z) \sqrt{P(z)}}{C(z) + D(z) \sqrt{P(z)}}$

for suitable polynomials

. We can rationalize the denominator like so:

$\displaystyle \frac{A(z) + B(z) \sqrt{P(z)}}{C(z) + D(z) \sqrt{P(z)}} \times \frac{C(z) - D(z) \sqrt{P(z)}}{C(z) - D(z) \sqrt{P(z)}} = F(z) + G(z) \sqrt{P(z)}$

The rational functions

and

appearing in the foregoing equation are defined like so:

$\displaystyle F(z)$	$\displaystyle =$	$\displaystyle \frac{A(z) C(z) - B(z) D(z) P(z)}{C^2 (z) - D^2(z) P(z)}$
$\displaystyle G(z)$	$\displaystyle =$	$\displaystyle 2 \frac{B(z) C(z) - A(z) D(z)}{C^2 (z) - D^2(z) P(z)}$

Since $\int F(z) \, dz$ may be expressed in terms of elementary functions, we shall focus our attention on the remaining piece, $\int G(z) \sqrt{P(z)} \, dz$ . Because we may decompose into partial fractions, it suffices to consider the following cases:

$\displaystyle \int z^n \sqrt{P(z)} \, dz$

$\displaystyle \int \frac{\sqrt{P(z)}}{(z - r)^n} \, dz$

Here,

is a non-negative integer and

is a complex number.

At this point, it is convenient to employ the specific form of the polynominal . Let us assume that we have chosen the Jacobi normal form.

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

$\displaystyle \int z^{2m+1} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz$

Then we may make a change of variables

to obtain

$\displaystyle \frac{1}{2} \int y^{2m} \sqrt{(1 - y) (1 - k^2 y)} \, dy ,$

which may be integrated using elementary functions.

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

$\displaystyle d \left( (1 - z^2) (1 - k^2 z^2) \sqrt{(1 - z^2) (1 - k^2 z^2)} \... ...c{9}{2} k^2 z^3 - 3 (1 + k^2) z \right) \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz ,$

we have

$\displaystyle (2 m + 1)$		$\displaystyle \int z^{2m} (1 - z^2) (1 - k^2 z^2) \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz$
$\displaystyle +$		$\displaystyle \int z^{2m+1} \left( \frac{9}{2} k^2 z^3 - 3 (1 + k^2) z \right) \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz$
$\displaystyle =$		$\displaystyle z^{2m+1} (1 - z^2) (1 - k^2 z^2) \sqrt{(1 - z^2) (1 - k^2 z^2)} + C$

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

$\displaystyle \left( 1 + 2 m + \frac{9}{2} k^2 \right)$		$\displaystyle \int z^{2m+4} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz -$
$\displaystyle (4 + 2 m) (1 + k^2)$		$\displaystyle \int z^{2m+2} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz +$
		$\displaystyle \int z^{2m} \sqrt{(1 - z^2) (1 - k^2 z^2)} =$
		$\displaystyle x^{2k+1} (1 - z^2) (1 - k^2 z^2) \sqrt{(1 - z^2) (1 - k^2 z^2)} + C$

By repeated use of this identity, we may express any integral of the form $\int z^{2m} \sqrt{P(z)} \, dz$ as the sum of a linear combination of $\int z^2 \sqrt{P(z)} \, dz$ and $\int \sqrt{P(z)} \, dz$ and the product of a polyomial and $\sqrt{P(z)}$ .

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

$\displaystyle \int \frac{\sqrt{P(z)}}{(z - r)^n} \, dz$

Will finish later -- saving in case of computer crash.

"reduction of elliptic integrals to standard form" is owned by rspuzio.

(view preamble | get metadata)

See Also: expressible in closed form

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: product, linear combination, sum, integration by parts, identities, odd, power, point, complex number, integer, partial fractions, focus, equation, rationalize the denominator, normal form, coefficients, variables, inverse trigonometric functions, square root, radical, factor, roots, functions, elementary functions, terms, degree, polynomial, rational function, integral

This is version 11 of reduction of elliptic integrals to standard form, born on 2006-12-10, modified 2006-12-12.
Object id is 8610, canonical name is ReductionOfEllipticIntegralsToStandardForm.
Accessed 1318 times total.

Classification:

AMS MSC:

33E05 (Special functions :: Other special functions :: Elliptic functions and integrals)

Pending Errata and Addenda

None.

Discussion

forum policy

No messages.

Interact