3  Recurrence relations and basic integrals

Theorem 2 Define

A(m)=vm(x)-vm(y),

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vm(t)=  1 s(t)   n Õ j=1  (aj +bj t)mj,       s(t)= h Õ i=1  Ö   ai +bi t    .

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For 1 £ j £ n and h=3 or 4 let E_p(j) be the elementary symmetric function of degree p in

d1j b1  ,¼,  dhj bh  ,

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and define

s0(j) = s0 = h Õ i=1  bi ,       sp(j)=s0 Ep(j) ,

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whence s_h(j)=0 if 1 £ j £ h. Then, for any integer q and for 1 £ j £ n,

h å r=0  æ è q+  r 2 +1 ö ø  sh-r(j) I((q+r)ej) = bjh-1 A æ è (q+1)ej + h å i=1  ei ö ø .

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Choose m=(q+1)e_j +å^h_i=1e_i , whence

vm(t) = s(t)(aj +bj t)q+1,

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v¢m(t) = s¢(t)(aj +bj t)q+1 +(q+1) bj s(t)(aj +bj t)q .

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From

ai +bi t =  bi bj   æ è aj +bj t +  dij bi ö ø ,

it follows that

s2(t) =  s0 bjh   h Õ i=1  æ è aj +bj t +  dij bi ö ø =  s0 bjh   h å r=0  Eh-r(j)(aj +bj t)r =  1 bjh   h å r=0  sh-r(j)(aj +bj t)r (29)

and therefore

s¢(t)=  1 2bjh-1s(t)   h å r=0  rsh-r(j)(aj +bj t)r-1.

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Replacing s by s² /s in the last term of (3.7) and substituting (3.8) and (3.9), we find

v¢m(t)=  1 bjh-1s(t)   h å r=0  æ è  r 2   +q+1 ö ø sh-r(j) (aj +bj t)q+r.

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Integration from y to x proves (3.5).

If 1 £ j £ h one of the quantities listed in (3.3) is 0. Then E_h(j)=s_h(j)=0, and the sum over r in (3.5) effectively starts with r=1. ³ Thus the recurrence relation contains h integrals if 1 £ j £ h or h+1 integrals if h+1 £ j £ n. In the first case I(±qe_j) can be reduced to the h-1 integrals I(-e_j), I(e₀),¼,I((h-3)e_j), but in the second case I((h-2)e_j) is added to the list. Further reduction uses the following theorem in the cases q=1 and 2, cases that could alternatively be obtained less directly from (2.19).

Theorem 3 If q is a positive integer, then

biq I(qej) = q å r=0  ( q r )  bjr djiq-r I(rei).

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Raising both sides of

bi (aj +bj t) = bj (ai +bi t) + dji

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to the power q and making a binomial expansion of the right side, we find

biq (aj +bj t)q = q å r=0  ( q r )  bjr (ai +bi t)r djiq-r .

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Multiplying both sides by Õ^h_k=1 (a_k +b_k t)^-1/2 and integrating from y to x, we obtain (3.11).

If we choose i £ h in the first sum in (2.19), the integrals in that sum and in the first h terms of the second sum can be reduced by (3.5) to integrals of the form I(qe_i),  i £ h, with q=-1,0 if h=3 and q=-1,0,1 if h=4. The remaining terms of the second sum can be reduced by (3.5) to integrals of the form I(qe_j),  j > h, with q=-1,0,1 if h=3 and q=-1,0,1,2 if h=4. Now I(qe_j),  q=1,2, can be expressed by (3.11) in terms of I(re_i) with i £ h and 0 £ r £ q, and these integrals can be treated in the same way as the integrals in the first sum in (2.19). In this way all the integrals on the right side of (2.19) can be reduced to basic integrals of the form I(-e_j),  0 £ j £ n, in the cubic case (h=3) plus I(e_i),  1 £ i £ 4 in the quartic case (h=4).

As the integrand in (1.2) becomes more complicated, so of course does this reduction to basic integrals, but for integrands that occur frequently in practice it is quick and simple. To illustrate this, we show how 136 quartic integrals from [ Gradshteyn and Ryzhik1994] are represented in terms of basic integrals by five formulas. Each entry in Table 1 starts with a list of those integrals that have the form I(m) = I(å_j=1⁵ m_j e_j) displayed in the entry and defined in (1.2). Also displayed is the expression for I(m) in terms of the basic integrals I(e₀),  I(-e₅), and I(±e_i),  i=1,¼,4. Equation (2.19) suffices for I(e₅) and I(e_k -e_i) ; recurrence relations are not needed.

If the basic integrals are to be expressed in terms of Legendre's canonical forms, each of 136 formulas has to be accompanied by inequalities relating the branch points of the integrand and the interval of integration. In the next Section we shall use R-functions to avoid most of this complexity.