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19 Elliptic IntegralsSymmetric Integrals

§19.29 Reduction of General Elliptic Integrals

  1. §19.29(i) Reduction Theorems
  2. §19.29(ii) Reduction to Basic Integrals
  3. §19.29(iii) Examples

§19.29(i) Reduction Theorems

These theorems reduce integrals over a real interval (y,x) of certain integrands containing the square root of a quartic or cubic polynomial to symmetric integrals over (0,) containing the square root of a cubic polynomial (compare §19.16(i)). Let

19.29.1 Xα =aα+bαx,
Yα =aα+bαy,
x>y, 1α5,
19.29.2 dαβ=aαbβaβbα,
dαβ0 if αβ,

and assume that the line segment with endpoints aα+bαx and aα+bαy lies in (,0) for 1α4. If

19.29.3 s(t)=α=14aα+bαt

and α,β,γ,δ is any permutation of the numbers 1,2,3,4, then

19.29.4 yxdts(t)=2RF(U122,U132,U232),


19.29.5 Uαβ =(XαXβYγYδ+YαYβXγXδ)/(xy),
Uαβ =Uβα=Uγδ=Uδγ,
Uαβ2Uαγ2 =dαδdβγ.

There are only three distinct U’s with subscripts 4, and at most one of them can be 0 because the d’s are nonzero. Then

19.29.6 Uαβ =bαbβYγYδ+YαYβbγbδ,
Uαβ =XαXβbγbδ+bαbβXγXδ,
19.29.7 yxaα+bαtaδ+bδtdts(t)=23dαβdαγRD(Uαβ2,Uαγ2,Uαδ2)+2XαYαXδYδUαδ,
19.29.8 yxaα+bαta5+b5tdts(t)=23dαβdαγdαδdα5RJ(U122,U132,U232,Uα52)+2RC(Sα52,Qα52),


19.29.9 Uα52 =Uαβ2dαγdαδdβ5dα5=Uβγ2dαβdαγdδ5dα50,
Sα5 =1xy(XβXγXδXαY52+YβYγYδYαX52),
Qα5 =X5Y5XαYαUα50,
Sα52Qα52 =dβ5dγ5dδ5dα5.

The Cauchy principal value is taken when Uα52 or Qα52 is real and negative. Cubic cases of these formulas are obtained by setting one of the factors in (19.29.3) equal to 1.

The advantages of symmetric integrals for tables of integrals and symbolic integration are illustrated by (19.29.4) and its cubic case, which replace the 8+8+12=28 formulas in Gradshteyn and Ryzhik (2000, 3.147, 3.131, 3.152) after taking x2 as the variable of integration in 3.152. Moreover, the requirement that one limit of integration be a branch point of the integrand is eliminated without doubling the number of standard integrals in the result. (19.29.7) subsumes all 72 formulas in Gradshteyn and Ryzhik (2000, 3.168), and its cubic cases similarly replace the 18+36+18=72 formulas in Gradshteyn and Ryzhik (2000, 3.133, 3.142, and 3.141(1-18)). For example, 3.142(2) is included as

19.29.10 ubat(bt)(tc)3dt=23(ab)(bu)3/2RD+2bc(au)(bu)uc,

where the arguments of the RD function are, in order, (ab)(uc), (bc)(au), (ab)(bc).

§19.29(ii) Reduction to Basic Integrals

(19.2.3) can be written

19.29.11 I(𝐦)=yxα=1h(aα+bαt)1/2j=1n(aj+bjt)mjdt,

where x>y, h=3 or 4, nh, and mj is an integer. Define

19.29.12 𝐦=(m1,,mn)=j=1nmj𝐞j,

where 𝐞j is an n-tuple with 1 in the jth position and 0’s elsewhere. Define also 𝟎=(0,,0) and retain the notation and conditions associated with (19.29.1) and (19.29.2). The integrals in (19.29.4), (19.29.7), and (19.29.8) are I(𝟎), I(𝐞α𝐞δ), and I(𝐞α𝐞5), respectively.

The only cases of I(𝐦) that are integrals of the first kind are the two (h=3 or 4) with 𝐦=𝟎. The only cases that are integrals of the third kind are those in which at least one mj with j>h is a negative integer and those in which h=4 and j=1nmj is a positive integer. All other cases are integrals of the second kind.

I(𝐦) can be reduced to a linear combination of basic integrals and algebraic functions. In the cubic case (h=3) the basic integrals are

19.29.13 I(𝟎);

In the quartic case (h=4) the basic integrals are

19.29.14 I(𝟎);

Basic integrals of type I(𝐞j), 1jh, are not linearly independent, nor are those of type I(𝐞j), 1j4.

The reduction of I(𝐦) is carried out by a relation derived from partial fractions and by use of two recurrence relations. These are given in Carlson (1999, (2.19), (3.5), (3.11)) and simplified in Carlson (2002, (1.10), (1.7), (1.8)) by means of modified definitions. Partial fractions provide a reduction to integrals in which 𝐦 has at most one nonzero component, and these are then reduced to basic integrals by the recurrence relations. A special case of Carlson (1999, (2.19)) is given by

19.29.15 bjI(𝐞l𝐞j)=dljI(𝐞j)+blI(𝟎),

which shows how to express the basic integral I(𝐞j) in terms of symmetric integrals by using (19.29.4) and either (19.29.7) or (19.29.8). The first choice gives a formula that includes the 18+9+18 = 45 formulas in Gradshteyn and Ryzhik (2000, 3.133, 3.156, 3.158), and the second choice includes the 8+8+8+12 = 36 formulas in Gradshteyn and Ryzhik (2000, 3.151, 3.149, 3.137, 3.157) (after setting x2=t in some cases).

If h=3, then the recurrence relation (Carlson (1999, (3.5))) has the special case

19.29.16 bβbγI(𝐞α)=dαβdαγI(𝐞α)+2bα(s(x)aα+bαxs(y)aα+bαy),

where α,β,γ is any permutation of the numbers 1,2,3, and

19.29.17 s(t)=α=13aα+bαt.

(This shows why I(𝐞α) is not needed as a basic integral in the cubic case.) In the quartic case this recurrence relation has an extra term in I(2𝐞α), and hence I(𝐞α), 1α4, is a basic integral. It can be expressed in terms of symmetric integrals by setting a5=1 and b5=0 in (19.29.8).

The other recurrence relation is

19.29.18 bjqI(q𝐞l)=r=0q(qr)blrdljqrI(r𝐞j),

see Carlson (1999, (3.11)). An example that uses (19.29.15)–(19.29.18) is given in §19.34.

For an implementation by James FitzSimons of the method for reducing I(𝐦) to basic integrals and extensive tables of such reductions, see Carlson (1999) and Carlson and FitzSimons (2000).

Another method of reduction is given in Gray (2002). It depends primarily on multivariate recurrence relations that replace one integral by two or more.

§19.29(iii) Examples

The first formula replaces (19.14.4)–(19.14.10). Define Qj(t)=aj+bjt2, j=1,2, and assume both Q’s are positive for 0y<t<x. Then

19.29.19 yxdtQ1(t)Q2(t)=RF(U2+a1b2,U2+a2b1,U2),
19.29.20 yxt2dtQ1(t)Q2(t)=13a1a2RD(U2+a1b2,U2+a2b1,U2)+(xy/U),


19.29.21 yxdtt2Q1(t)Q2(t)=13b1b2RD(U2+a1b2,U2+a2b1,U2)+(xyU)1,


19.29.22 (x2y2)U=xQ1(y)Q2(y)+yQ1(x)Q2(x).

If both square roots in (19.29.22) are 0, then the indeterminacy in the two preceding equations can be removed by using (19.27.8) to evaluate the integral as RG(a1b2,a2b1,0) multiplied either by 2/(b1b2) or by 2/(a1a2) in the cases of (19.29.20) or (19.29.21), respectively. If x=, then U is found by taking the limit. For example,

19.29.23 ydt(t2+a2)(t2b2)=RF(y2+a2,y2b2,y2).

Next, for j=1,2, define Qj(t)=fj+gjt+hjt2, and assume both Q’s are positive for y<t<x. If each has real zeros, then (19.29.4) may be simpler than

19.29.24 yxdtQ1(t)Q2(t)=4RF(U,U+D12+V,U+D12V),


19.29.25 (xy)2U =S1S2,
Sj =(Qj(x)+Qj(y))2hj(xy)2,
Djl =2fjhl+2hjflgjgl,
V =D122D11D22.

(The variables of RF are real and nonnegative unless both Q’s have real zeros and those of Q1 interlace those of Q2.) If Q1(t)=(a1+b1t)(a2+b2t), where both linear factors are positive for y<t<x, and Q2(t)=f2+g2t+h2t2, then (19.29.25) is modified so that

19.29.26 S1 =(X1Y2+Y1X2)2,
Xj =aj+bjx,
Yj =aj+bjy,
D12 =2a1a2h2+2b1b2f2(a1b2+a2b1)g2,
D11 =(a1b2a2b1)2=d122,

with other quantities remaining as in (19.29.25). In the cubic case, in which a2=1, b2=0, (19.29.26) reduces further to

19.29.27 S1 =(X1+Y1)2,
D12 =2a1h2b1g2,
D11 =b12.

For example, because t3a3=(ta)(t2+at+a2), we find that when 0ay<x

19.29.28 yxdtt3a3=4RF(U,U3a+23a,U3a23a),


19.29.29 (xy)2U =(xa+ya)2((ξ+η)2(xy)2),
ξ =x2+ax+a2,
η =y2+ay+a2.

Lastly, define Q(t2)=f+gt2+ht4 and assume Q(t2) is positive and monotonic for y<t<x. Then

19.29.30 yxdtQ(t2)=2RF(U,Ug+2fh,Ug2fh),


19.29.31 (xy)2U=(Q(x2)+Q(y2))2h(x2y2)2.

For example, if 0yx and a40, then

19.29.32 yxdtt4+a4=2RF(U,U+2a2,U2a2),


19.29.33 (xy)2U=(x4+a4+y4+a4)2(x2y2)2.