# 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:

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

Since $\int F(z)\,dz$ may be expressed in terms of elementary functions, we shall focus our attention on the remaining piece, $\int G(z)\sqrt{P(z)}\,dz$, 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:

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

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

 $4(2n+3)A_{n+2}=(2n+1)g_{2}A_{n}+2ng_{3}A_{n-1}+z^{n}\sqrt{4z^{3}-g_{2}x-g_{3}}+C$
 $2n(4r^{3}-g_{2}r-g_{3})B_{n+1}+(2n-1)(12r^{2}-g_{2})B_{n}+24(n-1)rB_{n-1}+4(2n% -3)B_{n-2}+\frac{\sqrt{4z^{3}-g_{2}x-g_{3}}}{(z-r)^{n}}+C=0$

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

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

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

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

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

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

Will finish later — saving in case of computer crash.

Title reduction of elliptic integrals to standard form ReductionOfEllipticIntegralsToStandardForm 2014-02-01 18:13:38 2014-02-01 18:13:38 rspuzio (6075) rspuzio (6075) 30 rspuzio (6075) Theorem msc 33E05 ExpressibleInClosedForm