reduction of elliptic integrals to standard form

Any integral of the form $\int R(x,\sqrt{P(x)})𝑑x$, where $R$ is a rational function and $P$ is a polynomial of degree 3 or 4 can be expressed as a linear combination of elementary functions and elliptic integrals 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 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 $z=(ax+b)/(cx+d)$. By choosing the coefficients $a,b,c,d$ suitably, one can cast P into either Jacobi’s normal form $P(z)=(1-z^2)(1-k^2{z}^2)$ or Weierstrass’ normal form $P(z)=4z^3-g_{2}z-g_3$.

Note that

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

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

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

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

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

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

Since $\int F(z)𝑑z$ may be expressed in terms of elementary functions, we shall focus our attention on the remaining piece, $\int G(z)\sqrt{P(z)}𝑑z$, which we shall write as $\int H(z)/\sqrt{P(z)}𝑑z$, where $H=PG$.. Because we may decompose $H$ into partial fractions, it suffices to consider the following cases, which we shall all $A_n$ and $B_n$:

$$A_n(z)=\int \frac{z^n}{\sqrt{P(z)}}𝑑z$$

$$B_n(z,r)=\int \frac{1}{{(z-r)}^n\sqrt{P(z)}}𝑑z$$

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

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

$$\int \frac{z^{n-1}(z{P}^{\prime }(z)+2nP(z))}{2\sqrt{P(z)}}𝑑z=z^n\sqrt{P(z)}+C$$

$$\int \frac{(z-r)P^{\prime }(z)-2nP(z)}{2{(z-r)}^{n+1}\sqrt{P(z)}}𝑑z=\frac{\sqrt{P(z)}}{{(z-r)}^n}+C$$

These identities will allow us to express $A_n$’s and $B_n$’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

$$4(2n+3)A_{n+2}=(2n+1)g_2{A}_n+2n{g}_3{A}_{n-1}+z^n\sqrt{4z^3-g_{2}x-g_3}+C$$

$$2n(4r^3-g_{2}r-g_3)B_{n+1}+(2n-1)(12r^2-g_2)B_n+24(n-1)r{B}_{n-1}+4(2n-3)B_{n-2}+\frac{\sqrt{4z^3-g_{2}x-g_3}}{{(z-r)}^n}+C=0$$

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

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

Then we may make a change of variables $y=z^2$ to obtain

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

which may be integrated using elementary functions.

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

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

we have

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

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

$=$ $z^{2m+1}(1-z^2)(1-k^2{z}^2)\sqrt{(1-z^2)(1-k^2{z}^2)}+C$

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

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

$(4+2m)(1+k^2)$ ${\displaystyle \int z^{2m+2}\sqrt{(1-z^2)(1-k^2{z}^2)}𝑑z}+$

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

$x^{2k+1}(1-z^2)(1-k^2{z}^2)\sqrt{(1-z^2)(1-k^2{z}^2)}+C$

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

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

Will finish later — 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