The French mathematician Adrien Marie Legendre (1752-1833) began the systematic study of elliptic integrals and elliptic functions, and one of his most important discoveries was that all integrals of the form:

∫ R(w, x) dx

where R is a rational function of w and x, and w2 is a cubic or quartic polynomial in x may be expressed in terms of combinations of these three standard forms:

 x            dx
∫  ________________________
 0 sqrt((1 - x2)(1 - k2x2))

 x         x2  dx
∫  ________________________
 0 sqrt((1 - x2)(1 - k2x2))

 x                dx
∫  _________________________________
 0 (1 - nx2)sqrt((1 - x2)(1 - k2x2))

These are known as the (Legendre) elliptic integrals of the first, second, and third kinds respectively.

The actual method for transforming a general elliptic integral into these standard forms is actually quite involved, and requires a lot of fairly tedious algebra, but before automatic computers were widely available for performing numerical quadrature this was the best way to proceed. The method is thus primarily only of historical interest, but it can be useful sometimes because it develops an exact analytical solution, that can provide additional mathematical insight.

(here we go....)

The most general form of elliptic integral may be expressed in the form:

  A + B sqrt(R)
∫ _____________ dx
  C + D sqrt(R)

where A, B, C, and D are rational functions of x and R is a cubic or quartic polynomial (if R is a quintic or higher degree polynomial, then we have not an elliptic, but a hyperelliptic integral).

The integrand may be rewritten as follows:

A + B sqrt(R)    AC - BDR      (AD - CB)R    1
_____________ = ___________ -  __________ ________
C + D sqrt(R)    C2 - D2R       C2 - D2R   sqrt(R)

        Q
= P - ______
      sqrt(R)

where P and Q are rational functions of x. The integral then reduces to:

            Q dx
∫ P dx - ∫ _______
           sqrt(R)

The first integral is simply the integral of a rational function, which anyone who learned undergraduate integral calculus should be able to integrate. The second integral is a bit more troublesome, but we can, in the end, reduce it in terms of the three elliptic integral standard forms above.

Assuming for the moment R to be a quartic polynomial, we can take the expression:

   dx                       dx
_______ = ___________________________________
sqrt(R)   sqrt(A(x - e)(x - f)(x - g)(x - h))

and make the homographic transformation x = (ay + b)/(cy + d), we get:

  dx       (ad-bc) dy
_______ = ____________
sqrt(R)      sqrt(Y)

where Y takes the form:

Y = A(g2y2 + g1t + g0)(h2y2 + h1t + h0)

where

g2 = (a - ce)(a - cf)

g1 = (a - ce)(b - df) + (a - cf)(b - de)

g0 = (b - de)(b - df)

The values for h2, h1, and h0 are obviously the same, except that we substitute g for e and h for f throughout.

Legendre thought to make all of the odd powers of y disappear from the radicand, and that can be done by judicious selection of values for a, b, c, and d. If we take as unknowns the ratios d/b and c/a, the odd powers of y disappear when d/b and c/a simultaneously satisfy:

2 - (d/b + c/a)(e + f) + 2(d/b)(c/a)ef = 0

2 - (d/b + c/a)(g + h) + 2(d/b)(c/a)gh = 0

Particular values for the ratios d/b and c/a may be obtained from these equations, and any values of a, b, c, and d may be chosen that satisfy the ratios thus obtained. These will make the odd powers of y vanish, so Y may then be written in the form:

Y = A(g0 + g2y2)(h0 + h2y2)

By taking p2 = g2/g0 and q2 = h2/h0, and making the substitution y = t/p, we get:

  dx           (1/p)(ad - bc)dt
_______ = _________________________________
sqrt(R)   sqrt(Ag0h0(1 - t2)(1 - (q2/p2)t2))

If we take k2 = q2/p2 and C = (ad - bc)/(p sqrt(Ag0h0)), we finally obtain:

  dx                C dt
_______ = _______________________
sqrt(R)   sqrt((1 - t2)(1 - k2t2))

The value k is called the modulus, and while for theoretical work it may take on any value at all, real or complex, for most applications in geometry and physics the modulus is almost always real and between 0 and 1.

For the cubic case:

  dx                  dx
_______ = ____________________________
sqrt(R)   sqrt(A(x - a)(x - b)(x - c))

The subsititution x = y2 + a suffices to transform the expression into the same standard form. The steps are left as an exercise for the reader.

So, if we make the above subsitutions into the general integral, we get:

   Q dx               f(t) dt
∫ _______ = ∫ _______________________
  sqrt(R)     sqrt((1 - x2)(1 - k2x2))

where f(t) is a rational function of t. f(t) may be written in the following form:

       φ(t2) + tφ1(t2)
f(t) = ________________
       ψ(t2) + tψ1(t2)

where we have collected all of the odd and even powers of t, so the φ's and ψ's are all rational functions. If we multiply the numerator and denominator of this expression by ψ(t2) - tψ1(t2), we can see that f(t) = f0(t2) + zf1(t2), where and f0 and f1 are both rational functions of t.

So now, the whole integral reduces to:

         f(t) dt                   f0(t2)
∫ _______________________ = ∫ _______________________
  sqrt((1 - t2)(1 - k2t2))     sqrt((1 - t2)(1 - k2t2))

         t f1(t2) dt
+ ∫ ______________________
    sqrt((1 - t2)(1 - k2t2))

By making the substitution t2 = z, the second integral on the right hand side may be reduced to:

1         f1(z) dz
_ ∫ ______________________
2    sqrt((1 - z)(1 - k2z))

which may be evaluated in terms of elementary functions, specifically inverse trigonometric functions or logarithmic functions.

f0(t2) in general has the form:

                      Aλi
f0(t2) = g(t2) + ∑ ____________
                 i  (t2 + ai)λi

where g(t2) = C0 + C2t2 + ... is a polynomial, and the summation term is produced by partial fraction expansion of the remainder. The first integral can thus be reduced into the form:

    f(t2) dt           t2m dt                 dt
∫ ___________ = ∑ Cm ∫ ________ + ∑ Aλm ∫ ____________
    sqrt(R')    m      sqrt(R')   m       (t2  + am)λm

where R' = (1 - t2)(1 - k2t2).

By repeatedly differentiating and integrating expressions of the form t2k + 1sqrt((1 - t2)(1 - k2t2) it is not difficult to derive reduction formulae that express one integral of the form ∫ t2k dt/sqrt((1 - t2)(1 - k2t2) into sums of integrals involving successively lower values of k, until ultimately we express such an integral in the form of elliptic integrals of the first and second kind.

For the last form, ∫ dt/((t2 + a)m)sqrt((1 - t2)(1 - k2t2))), once again, by repeatedly differentiating and integrating an expression of the form:

t sqrt((1 - t2)(1 - k2t2))
__________________________
        (t2 - a)m

one can obtain reduction formulae that ultimately express such an integral into elliptic integrals of the first, second, and third kind. The value a that results from this is called the parameter.

Once the integral has been thus reduced into elementary integrals and the Legendre standard forms, algorithms involving Landen's transformation and the arithmetic geometric mean are available to calculate a definite integral for the entire expression. Tables giving the values of the integral for various values of the amplitude (the upper limit of the integrand), the modulus, and in the case of integrals of the third kind, the parameter are also available.

But why are these integrals called elliptic integrals in the first place? By making the substitution x = sin φ in the above standard forms, the integrals reduce to:

           φ        dφ
F(k, φ) = ∫  _________________ ;
           0 sqrt(1 - k2sin2φ)

 φ        dφ            φ
∫  ________________  - ∫  sqrt(1 - k2sin2 φ) dφ ;
 0 sqrt(1 - k2sin2 φ)  0

              φ               dφ
Π(n, k, φ) = ∫  ______________________________
              0 (1 + n sin2 φ)sqrt(1 - k2sin2 φ)

It will be noted that the second term in the second form:

           φ
E(k, φ) = ∫  sqrt(1 - k2 sin2 φ)
           0

is actually the formula for the arc length of an ellipse with eccentricity k up to where the complement of the eccentric angle is φ. By the way, the functions E, K, and Π are the Legendre normal forms for the elliptic integrals of the first, second, and third kinds. These formulas are the cornerstones of the Legendre-Jacobi theory of elliptic integrals and functions.

These elliptic integrals, when taken as functions, are generalizations of the inverse trigonometric and logarithmic functions. Carl Jacobi (1804-1851) inverted these elliptic integrals to come up with the Jacobi Elliptic Functions, which are natural generalizations of the trigonometric functions.

Apparently, Karl Weierstrass (1815-1897), not to be outdone, also developed his own elliptic integral standard forms:

         dt
∫ ___________________;
  sqrt(4t3 - g2t - g3)

         t dt
∫ ___________________;
  sqrt(4t3 - g2t - g3)

              dt
∫ ____________________________
  (t - b) sqrt(4t2 - g2t - g3)

These standard forms are the basis for the Hermite-Weierstrass theory of elliptic integrals and functions. Inverting the first integral develops the Weierstrass elliptic functions. By taking the roots of the equation 4t3 - g2t - g3 = 0, call them e1, e2 and e3, and making the substitution t = e3 + (e1 - e3)/z2, the Hermite-Weierstrass normal forms may be transformed into the corresponding Jacobi-Legendre forms.

Sources

Harris Hancock, Lectures on the Theory of Elliptic Functions

Arthur L. Baker, Elliptic Functions: An Elementary Text-Book for Students of Mathematics

E.T. Whittaker and G.N. Watson, A Course of Modern Analysis

Roland Bulirsch, "Numerical Calclulation of Elliptic Integrals and Elliptic Functions", Numer. Math. 7 pp. 78-90, 1963.

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