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19 Elliptic IntegralsLegendre’s Integrals

§19.2 Definitions

  1. §19.2(i) General Elliptic Integrals
  2. §19.2(ii) Legendre’s Integrals
  3. §19.2(iii) Bulirsch’s Integrals
  4. §19.2(iv) A Related Function: RC(x,y)

§19.2(i) General Elliptic Integrals

Let s2(t) be a cubic or quartic polynomial in t with simple zeros, and let r(s,t) be a rational function of s and t containing at least one odd power of s. Then

19.2.1 r(s,t)dt

is called an elliptic integral. Because s2 is a polynomial, we have

19.2.2 r(s,t)=(p1+p2s)(p3p4s)s(p3+p4s)(p3p4s)s=ρs+σ,

where pj is a polynomial in t while ρ and σ are rational functions of t. Thus the elliptic part of (19.2.1) is

19.2.3 ρ(t)s(t)dt.

§19.2(ii) Legendre’s Integrals

Assume 1sin2ϕ(,0] and 1k2sin2ϕ(,0], except that one of them may be 0, and 1α2sin2ϕ{0}. Then

19.2.4 F(ϕ,k) =0ϕdθ1k2sin2θ=0sinϕdt1t21k2t2,
19.2.5 E(ϕ,k) =0ϕ1k2sin2θdθ=0sinϕ1k2t21t2dt.
19.2.6 D(ϕ,k) =0ϕsin2θdθ1k2sin2θ=0sinϕt2dt1t21k2t2=(F(ϕ,k)E(ϕ,k))/k2.
19.2.7 Π(ϕ,α2,k)=0ϕdθ1k2sin2θ(1α2sin2θ)=0sinϕdt1t21k2t2(1α2t2).

The paths of integration are the line segments connecting the limits of integration. The integral for E(ϕ,k) is well defined if k2=sin2ϕ=1, and the Cauchy principal value (§1.4(v)) of Π(ϕ,α2,k) is taken if 1α2sin2ϕ vanishes at an interior point of the integration path. Also, if k2 and α2 are real, then Π(ϕ,α2,k) is called a circular or hyperbolic case according as α2(α2k2)(α21) is negative or positive. The circular and hyperbolic cases alternate in the four intervals of the real line separated by the points α2=0,k2,1.

The cases with ϕ=π/2 are the complete integrals:

19.2.8 K(k) =F(π/2,k),
E(k) =E(π/2,k),
D(k) =D(π/2,k)=(K(k)E(k))/k2,
Π(α2,k) =Π(π/2,α2,k).

The principal branch of K(k) and E(k) is |ph(1k2)|π, that is, the branch-cuts are (,1][1,+). The principal values of K(k) and E(k) are even functions.

Legendre’s complementary complete elliptic integrals are defined via

19.2.8_1 K(k) =01dt1t21(1k2)t2,
19.2.8_2 E(k) =011(1k2)t21t2dt,

with a branch point at k=0 and principal branch |phk|π. Let k=1k2. Then

19.2.9 K(k) ={K(k),|phk|12π,K(k)2iK(k),12π<±phk<π,
E(k) ={E(k),|phk|12π,E(k)2i(K(k)E(k)),12π<±phk<π.

For more details on the analytical continuation of these complete elliptic integrals see Lawden (1989, §§8.12–8.14).

§19.2(iii) Bulirsch’s Integrals

Bulirsch’s integrals are linear combinations of Legendre’s integrals that are chosen to facilitate computational application of Bartky’s transformation (Bartky (1938)). Three are defined by

19.2.11 cel(kc,p,a,b)=0π/2acos2θ+bsin2θcos2θ+psin2θdθcos2θ+kc2sin2θ,
19.2.11_5 el1(x,kc)=0arctanx1cos2θ+kc2sin2θdθ,
19.2.12 el2(x,kc,a,b)=0arctanxa+btan2θ(1+tan2θ)(1+kc2tan2θ)dθ.

Here a,b,p are real parameters, and kc and x are real or complex variables, with p0, kc0. If <p<0, then the integral in (19.2.11) is a Cauchy principal value.


19.2.13 kc =k,
p =1α2,
x =tanϕ,

special cases include

19.2.14 K(k) =cel(kc,1,1,1),
E(k) =cel(kc,1,1,kc2),
D(k) =cel(kc,1,0,1),
(E(k)k2K(k))/k2 =cel(kc,1,1,0),
Π(α2,k) =cel(kc,p,1,1),


19.2.15 F(ϕ,k) =el1(x,kc)=el2(x,kc,1,1),
E(ϕ,k) =el2(x,kc,1,kc2),
D(ϕ,k) =el2(x,kc,0,1).

The integrals are complete if x=. If 1<k1/sinϕ, then kc is pure imaginary.

Lastly, corresponding to Legendre’s incomplete integral of the third kind we have

19.2.16 el3(x,kc,p)=0arctanxdθ(cos2θ+psin2θ)cos2θ+kc2sin2θ=Π(arctanx,1p,k),

§19.2(iv) A Related Function: RC(x,y)

Let x(,0) and y{0}. We define

19.2.17 RC(x,y)=120dtt+x(t+y),

where the Cauchy principal value is taken if y<0. Formulas involving Π(ϕ,α2,k) that are customarily different for circular cases, ordinary hyperbolic cases, and (hyperbolic) Cauchy principal values, are united in a single formula by using RC(x,y).

In (19.2.18)–(19.2.22) the inverse trigonometric and hyperbolic functions assume their principal values (§§4.23(ii) and 4.37(ii)). When x and y are positive, RC(x,y) is an inverse circular function if x<y and an inverse hyperbolic function (or logarithm) if x>y:

19.2.18 RC(x,y)=1yxarctanyxx=1yxarccosx/y,
19.2.19 RC(x,y)=1xyarctanhxyx=1xylnx+xyy,

The Cauchy principal value is hyperbolic:

19.2.20 RC(x,y)=xxyRC(xy,y)=1xyarctanhxxy=1xylnx+xyy,

For the special cases of RC(x,x) and RC(0,y) see (19.6.15).

If the line segment with endpoints x and y lies in (,0], then

19.2.21 RC(x,y)=01(v2x+(1v2)y)1/2dv,
19.2.22 RC(x,y)=2π0π/2RC(y,xcos2θ+ysin2θ)dθ.