reduction of elliptic integrals to standard formReductionOfEllipticIntegralsToStandardForm2006-12-10 15:03:542006-12-12 12:20:16Theoremelliptic integrals and Jacobi elliptic functions% this is the default PlanetMath preamble. as your knowledge
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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 in terms of
elementary functions and the functions $E$, $F$, and $\Pi$.
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 $z = (a x + b) / (c x + d)$. By choosing the coefficients $a, b, c, d$ suitably, one can cast P into either Jacobi's normal form $P(z) = (1 - z^2) (1 - k^2 z^2)$ or Weierstrass' normal form $P(z) = z^3 + g_1 z + g_2$.
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:
\begin{eqnarray*} F(z) &=& \frac{A(z) C(z) - B(z) D(z) P(z)}{C^2 (z) - D^2(z) P(z)} \\ G(z) &=& 2 \frac{B(z) C(z) - A(z) D(z)}{C^2 (z) - D^2(z) P(z)} \end{eqnarray*}
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 $G$ into partial fractions, it suffices to consider
the following cases:
\[\int z^n \sqrt{P(z)} \, dz\]
\[\int \frac{\sqrt{P(z)}}{(z - r)^n} \, dz\]
Here, $n$ is a non-negative integer and $r$ is a complex number.
At this point, it is convenient to employ the specific form of the polynominal $P$. 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:
\[\int z^{2m+1} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz\]
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
\begin{eqnarray*} (2 m + 1) &&\int z^{2m} (1 - z^2) (1 - k^2 z^2) \sqrt{(1 - z^2) (1 -
k^2 z^2)} \, dz \\ + &&\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
\end{eqnarray*}
By colecting terms, this identity may be rewritten as follows:
\begin{eqnarray*} \left( 1 + 2 m + \frac{9}{2} k^2 \right) &&\int z^{2m+4} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz - \\ (4 + 2 m) (1 + k^2) &&\int z^{2m+2} \sqrt{(1 - z^2) (1 - k^2 z^2)} \, dz + \\ &&\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
\end{eqnarray*}
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
\[\int \frac{\sqrt{P(z)}}{(z - r)^n} \, dz\]
\emph{ Will finish later --- saving in case of computer crash.}