To appear: Math. Comp LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS

To appear: Math Comp 21 LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS GEORGE BOROS AND VICTOR H MOLL Abstract We present a rational version of the classical Landen transformation for elliptic integrals This is employed to obtain explicit closed-form expressions for a large class of integrals of even rational functions and to develop an algorithm for numerical integration of these functions 1 Introduction We consider the space of even rational functions of degree 2p { E 2p := R(z = P (z p 1 P (z := b z 2(p 1 and Q(z := Q(z = } p a z 2(p with positive real coefficients a, b R + normalized by the condition a = a p = 1, the space E := of normalized even rational functions, and the set of 2p 1 parameters p=1 E 2p P 2p := {a 1,, a p 1 ; b,, b p 1 } We describe an algorithm to determine, as a function of the parameter set P 2p, a closed-form expression of the integral (11 I := R(z dz for a large class of functions R E The function R is called symmetric if its denominator Q satisfies Q(1/z = z 2p Q(z This is equivalent to its coefficients being palindromic, ie a j = a p j for 1 j p The class of symmetric functions plays a crucial role in this algorithm Define E s 2p := {R E 2p den(r is symmetric} (where den(r denotes the denominator of R, the class of rational functions with symmetric denominators of degree 2p, and E s := E s 2p Date: April 25, 21 1991 Mathematics Subject Classification Primary 33 Key words and phrases Rational functions, Landen transformation, Integrals The second author was supported in part by NSF Grant DMS-7567 1 p=1 =

2 GEORGE BOROS AND VICTOR H MOLL For m N define and E m,s 2p E m 2p := {R E (den(r 1/(m+1 is even, symmetric of degree 2p} := E m 2p E s 2p, so a function R E m,s 2p can be written in the form R(z = P (z Q m+1 (z, where P (z is an even polynomial and Q(z is an even symmetric polynomial of degree 2p The method of partial fractions gives (in principle the value of I in terms of the roots of Q Symbolic computations yield either a closed-form answer, an expression in terms of the roots of an associated polynomial, or the integral returned unevaluated The algorithm described here allows only algebraic operations on elementary functions and changes of variables of the same type In particular, we exclude the solution of algebraic equations of degree higher than 2 We say that a rational function R E is computable if its integral can be evaluated by our algorithm The first step in the algorithm is to consider symmetric rational functions In Section 2 we prove a reduction formula, ie a map F p : E s 2p R p that reduces the computability of the integral of the symmetric function R to that of one of degree 1 2 deg(r Here R p is the space of rational functions with denominator of degree p These new functions are not necessarily symmetric, and this is the main limitation of our algorithm The details of F p require the evaluation of some binomial sums which are presented in Appendix A The classical Wallis formula dz (z 2 + 1 m+1 = ( π 2m 2 2m+1 m shows that every R E m 2 is computable The reduction formula now implies that every R E m 4 is computable This is described in Section 3 The computability of R E 4 is also a consequence of the classical theory of hypergeometric functions; the details are given in [2] In Section 4 we compute the integral of every function in E m,s 8, where the reduction method expresses these integrals in terms of functions in E m 4 The algorithm does not, in general, provide a value for the integral of a nonsymmetric function of degree 8 The final piece of the algorithm is described in Section 5: for R E 2p, the symmetrization of its denominator produces a (symmetric rational function in E 4p with the same integral as R The reduction formula now yields a new function in E 2p with the same integral as R We thus obtain a map T 2p : E 2p E 2p such that (12 R(z dz = T 2p (R(z dz In particular, the class of computable rational functions of degree 2p is invariant under forward and bacward iteration of T 2p This map is the rational analog of the original Landen transformation for elliptic integrals described in [4, 13] The map T 2p can also be interpreted as a map on the coefficients Φ 2p : O + 2p O+ 2p where O + 2p = R + p 1 R + p

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 3 The case of Φ 6 is described in detail in Section 6 and is given explicitly by (13 a 1 9 + 5a 1 + 5a 2 + a 1 a 2 (a 1 + a 2 + 2 4/3 a 2 a 1 + a 2 + 6 (a 1 + a 2 + 2 2/3 b b + b 1 + b 2 (a 1 + a 2 + 2 2/3 b 1 b (a 2 + 2 + 2b 1 + b 2 (a 1 + 3 a 1 + a 2 + 2 b + b 2 b 2 (a 1 + a 2 + 2 1/3 Let x := (a 1, a 2 ; b, b 1, b 2 Then Φ 6 : O + 6 O+ 6 is iterated to produce a sequence x n+1 := Φ 6 (x n of points in O + 6 that yield a sequence of rational functions with constant integral We have proved in [3] the existence of L R +, depending upon the initial point x = (a 1, a 2 ; b, b 1, b 2, such that x n (3, 3; L, 2L, L Thus (14 b z 4 + b 1 z 2 + b 2 z 6 + a 1 z 4 + a 2 z 2 + 1 dz = L(x π 2 This establishes a numerical method to compute the integral in (14 Numerical studies on integrals of even degree 2p suggest the existence of a limiting value L = L(x such that the sequence x n := Φ 2p (x n satisfies x n (( ( ( ( ( p p p p 1 p 1,,, ; L, 1 2 p 1 1 L,, ( p 1 p 1 L The integral of the original rational function is thus π 2 L(x The proof of convergence remains open for p 4 Examples are given in Section 7 The most important issues left unresolved in this paper are the convergence of the iteration of the map Φ 2p discussed above and the geometric interpretation of the Landen transformation T 2p Finally, the question of integration of odd functions has not been addressed at all Some history The problem of integration of rational functions R(z = P (z/q(z was considered by J Bernoulli in the 18 th century He completed the original attempt by Leibniz of a general partial fraction decomposition of R(z The main difficulty associated with this procedure is to obtain a complete factorization of Q(z over R Once this is nown the partial fraction decomposition of R(z can be computed The fact is that the primitive of a rational function is always elementary: it consists of a new rational function (its rational part and the logarithm of a second rational function (its transcendental part In his classic monograph [9] G H Hardy states: The solution of the problem (of definite integration in the case of rational functions may therefore be said to be complete; for the difficulty with regard to the explicit solution of algebraical equations is one not of inadequate nowledge but of proved impossibility He goes on to add: It appears from the preceding paragraphs that we can always find the rational part of the integral, and can find the complete integral if we can find the roots of Q(z =

4 GEORGE BOROS AND VICTOR H MOLL In the middle of the last century Hermite [1] and Ostrogradsy [15] developed algorithms to compute the rational part of the primitive of R(z without factoring Q(z More recently Horowitz [11] rediscovered this method and discussed its complexity The problem of computing the transcendental part of the primitive was finally solved by Lazard and Rioboo [12], Rothstein [17] and Trager [18] For detailed descriptions and proofs of these algorithms the reader is referred to [5] and [6] 2 The reduction formula In this section we present a map F p : E m,s 2p E m p that is the basis of the integration algorithm described in Section 5 The proof is elementary and the binomial sums discussed in the Appendix are employed Let D p (z be the general symmetric polynomial of degree 4p We express the integral of z 2n /Dp m+1 as a linear combination of integrals of z /Ep m+1 where E p is a polynomial of degree 2p whose coefficients are determined by those of D p Theorem 21 Let m, n, p N Define (21 and (22 D p (d 1, d 2,, d p ; z = E p (d 1, d 2,, d p ; z = + p+1 j=1 p i=1 d j p d p+1 (z 2 + z 4p 2 = z 2p + p i+1 2 2i 1 z 2(p i for d i R + Then for n (m + 1p 1, (m+1p n 1 2 m (23 j= j=1 z 2n dz (D p (d 1,, d p ; z m+1 = ( (m + 1p n 1 + j 4 j j + i 1 i ( j + 2i 2 j 1 z 2((m+1p 1 j ( E p (d 1,, d p ; z d j+i, m+1 dz, and for (m + 1p 1 < n < 2p(m + 1 1 we employ the symmetry rule (24 N n,p = N 2p(m+1 1 n,p Proof First observe that (24 follows from the change of variable z 1/z Now consider z 2n dz (25 N n,p (d 1,, d p ; m := ( p = d p+1 (z 4p 2 + z 2 m+1 for n (m + 1p 1 The substitution z = tan θ yields N n,p = π/2 (1 C 2 n C 4(m+1p 2n 2 dθ ( p = d p+1 {(1 C 2 2p C 2 + (1 C 2 C 4p 2 } m+1,

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 5 where C = cos θ Letting ψ = 2 θ and D = cos ψ = 2C 2 1 then gives π (1 D 2 n (1 + D 2(m+1p 2n 1 dψ N n,p = ( p = d p+1 (1 D 2 {(1 D 2p 2 + (1 + D 2p 2 } m+1 Now observe that the integrals of the odd powers of cosine vanish when we expand (1 + D 2(m+1p 2n 1, producing π/2 (1 D 2 n (m+1p n 1 ( 2(m+1p 2n 1 N n,p = 2 m j= D dθ } m+1 D { p = d p+1 (1 D 2 p j= A second double angle substitution ϕ = 2ψ gives π (1 E n (m+1p n 1 N n,p = 2 m j= ( 2(m+1p 2n 1 { p = d p+1 (1 E p j= ( 2p 2 2 p(m+1 n j 1 (1 + E j dϕ ( 2p 2 } m+1, 2 p j (1 + E j where E = cos ϕ = 2D 2 1 The change of variable z = tan(ϕ/2 then yields z 2n (m+1p n 1 ( 2(m+1p 2n 1 N n,p = 2 m j= (1 + z 2 (m+1p n j 1 dz { p ( = d p ( p+1 z 2 2p 2 } m+1 j= (1 + z2 p j (26 Finally, we modify (26 using Lemma A2 and Lemma A4 with N = (m+1p n 1 to produce (23 Note that the previous theorem associates to each rational function of symmetric denominator R 1 (z = a new rational function (m+1p 1 R 2 (z = 2 m such that n= b s z 2s + b s 1 z 2(s 1 + + b ( z 4p + d p z 4p 2 + + 2d 1 z 2p + + 1 m+1 (m+1p n 1 b n j= R 1 (z dz = ( (m + 1p n 1 + j 4 j R 2 (z dz z 2((m+1p 1 j (E p (d 1,, d p ; z m+1 3 The quartic case In this section we describe the computability of rational functions R E m 4 These are functions of the form P (z R(z = (z 4 + 2az 2 + 1 m+1 where P (z is an even polynomial of degree 4m+2 Observe that the normalization a = a 2 = 1 maes the denominator of R automatically symmetric It suffices to evaluate z 2n dz (31 N n,4 (d 1 ; m := (z 4 + 2d 1 z 2 + 1 m+1

6 GEORGE BOROS AND VICTOR H MOLL where n 2m+1 is required for convergence From (24 we have N n,4 (d 1 ; m = N 2m 1 n,4 (d 1 ; m, so we may assume n m We now employ Theorem 21 to obtain a closed form expression for N n,4 (d 1 ; m Theorem 31 Let m N and assume n m Then (32 N n,4 (d 1 ; m := z 2n dz (z 4 + 2d 1 z 2 + 1 m+1 = m n π 2 3m+3/2 (1 + a ( ( 2m 1 m n + j 2 j (1 + d m+1/2 1 j m j j= For m + 1 n 2m + 1 we have (33 z 2n dz (z 4 + 2d 1 z 2 + 1 m+1 = n m 1 π 2 3m+3/2 (1 + d 1 ( ( 2m 1 m n + j 2 j (1 + d m+1/2 1 j m j j= j= ( j ( 1 m j ( j Proof We apply the result of the Theorem 21 with D 1 (d 1 ; z = z 4 + 2d 1 z 2 + 1 and E 1 (d 1 ; z = (1 + d 1 z 2 + 2, so that z 2n m n dz ( m n + j (z 4 + 2d 1 z 2 + 1 m+1 = 2 m 4 j z 2(m j dz ((1 + d 1 z 2 + 2 m+1 The change of variable u = (1 + d 1 z 2 /2 then yields z 2(m j dz ( (1 + d 1 z 2 + 2 m+1 = 2 (j+3/2 (1 + d 1 m+j 1/2 where we have used ( 2m = π m j u r 1/2 (1 + u s du = π 2 2(s 1 ( 2r r ( j ( m j ( 2(s r 1 s r 1 ( 1 m j u m j 1/2 du (1 + u m+1 1 2 (2m+j+3/2 (1 + d 1 (m j+1/2, ( 1 s 1 r The algorithm also requires a scaled version of N,4 (d 1 ; m Corollary 32 Let b >, c >, a > bc, m N, and n m Define Then for n m, N n,4 (a, b, c; m = π N n,4 (a, b, c; m := z 2n dz (bz 4 + 2az 2 + c m+1 ( c(c/b m n { 8(a + bc} 2m+1 1/2 (34 m n = 2 ( 2m 2 m ( m n + 2 ( 2 ( m 1 ( a + 1, bc

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 7 and for m + 1 n 2m + 1, ( { N n,4 (a, b, c; m = π c(c/b m n 8(a + 2m+1 1/2 bc} (35 n m 1 = 2 ( 2m 2 m ( m n + 2 ( 2 Proof Let n m The substitution u = z(b/c 1/4 yields (36 N n,4 (a, b, c; m = 1 c m n/2+3/4 b n/2+1/4 N n,4 ( m ( a bc ; m 1 ( a + 1 bc so (34 then follows from Theorem 31 From (24 we have N n,4 (a, b, c; m = N 2m+1 n,4 (c, b, a; m for m + 1 n 2m + 1, giving (35 4 The symmetric case of degree 8 In this section we prove the computability of the set E m,s 8 of symmetric rational functions with denominator of degree 8 and establish an explicit formula for the integral N n,8 (a 1, a 2 ; m =, z 2n dz (z 8 + a 2 z 6 + 2a 1 z 4 + a 2 z 2 + 1 m+1 where n 4m + 3 is required for convergence Observe that (24 reduces the discussion to the case n 2m + 1 The expression (22, with p = 2, produces E 2 (a 1, a 2 ; z = (1 + a 1 + a 2 z 4 + 2(a 2 + 4z 2 + 8 Theorem 41 Every function in E m,s 8 is computable More specifically, define c 1 := a 2 + 4, c 2 := 1 + a 1 + a 2, and t,j (m, n; a 1, a 2 := π2 (3m+2++j/2 c (m j/2 2 (c 1 + 8c 2 j m 1/2 ( ( ( ( ( 1 4m n + 2 2m m + j m n m j j j Then for m + 1 n 2m + 1, 1 + a 1 + a 2 > and a 2 + 4 > 8 8(1 + a 1 + a 2, z 2n dz (z 8 + a 2 z 6 + 2a 1 z 4 + a 2 z 2 + 1 m+1 = and for n m, m m =n j= =n t,j (m, n; a 1, a 2 + 2m+1 =n m 1 j= z 2n dz (z 8 + a 2 z 6 + 2a 1 z 4 + a 2 z 2 + 1 m+1 = 2m+1 =m+1 m 1 j= t,j (m, n; a 1, a 2 t,j (m, n; a 1, a 2, Proof The reduction formula yields z 2n dz (z 8 + a 2 z 6 + 2a 1 z 4 + a 2 z 2 + 1 m+1 = 2m+1 ( 4m n + 2 2 3m+2 2 2 z 2 (41 dz n (c 2 z 4 + 2c 1 z 2 + 8 m+1

8 GEORGE BOROS AND VICTOR H MOLL We then use Corollary 32 to evaluate (41 5 A sequence of Landen transformations The transformation theory of elliptic integrals was initiated by Landen in 1771 He proved the invariance of the function (51 under the transformation (52 ie that (53 G(a, b := π/2 a 1 = (a + b/2 d θ a2 cos 2 θ + b 2 sin 2 θ G(a 1, b 1 = G(a, b b 1 = ab, Gauss [7] rediscovered this invariance while numerically calculating the length of a lemniscate An elegant proof of (53 is given by Newman in [14] Here, the substitution x = b tan θ converts 2G(a, b into the integral of [ (a 2 + x 2 (b 2 + x 2 ] 1/2 over R; the change of variable t = (x ab/x/2 then completes the proof The Gauss-Landen transformation can be iterated to produce a double sequence (a n, b n such that a n b n < 2 n It follows that a n and b n converge to a common limit, the so-called arithmetic-geometric mean of a and b, denoted by AGM(a, b Passing to the limit in G(a, b = G(a n, b n produces (54 π 2AGM(a, b = π/2 d θ a2 cos 2 θ + b 2 sin 2 θ The reader is referred to [4] and [13] for details The goal of this section is to produce a map T 2p : E 2p E 2p that preserves the integral, ie (55 R(z dz = T 2p (R(z dz This map is the rational analog of the original Landen transformation (52 Theorem 51 Let R(z = P (z/q(z with (56 p 1 P (z = b j z 2(p 1 j and Q(z = j= p a j z 2(p j j= Define a j = for j > p, b j = for j > p 1, j (57 d p+1 j = a p a j for p 1, (58 d 1 = 1 2 = p = a 2 p, (59 c j = 2p 1 = a j b p 1 j+

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 9 for j 2p 1, and also (51 Let (511 α p (i = for 1 i p 1, and (512 { 2 2i 1 p+1 i =1 1 + p +i 1 i =1 d if i = a + i = b + i = Q(1 2i/p+1/p 2 ( +2i 2 1 d+i if 1 i p α p (i 2 2i Q(1 2(1 i/p [ p 1 i ( ] p 1 + i (c + c 2p 1 2i = for i p 1 Finally, define the polynomials (513 p 1 P + (z = b + i z2(p 1 i and Q + (z = = Then T 2p (R(z := P + (z/q + (z satisfies (55, ie (514 P (z Q(z dz = p a + i z2(p i = P + (z Q + (z dz Proof The first step is to convert the polynomial Q(z to its symmetric form: with Then I := P (z Q(z dz = C(z = P (z z 2p Q(1/z := D(z = Q(z z 2p Q(1/z := I = 2p 1 = 2p 1 = C(z D(z dz c z 2 p d p+1 (z 2 + z 2(2p = c z 2 dz Q s (z Now employ the reduction formula in Section 2 to evaluate j= L := z 2 dz Q s (z Observe that one needs to evaluate L only for p 1 Indeed, the usual symmetry rule yields L = L 2p 1 The reduction formula now gives p 1 ( p 1 + j L = 2 z 2(p 1 j dz p i= α p(iz 2(p i = p 1 α p (p j=1 ( p + j 2 2 λ 2(j 1 2p +1 2 with α p (i as in (51 and λ = [α p (p/α p (] 1/2p z 2(p j dz p i= b+ i z2(p i

1 GEORGE BOROS AND VICTOR H MOLL Note The extension of this transformation to the case of P (z Q m+1 (z dz requires explicit formulae for the coefficients of P (z ( z 2p Q(1/z m+1 and Q m+1 (z ( z 2p Q(1/z m+1 An algorithm for integration Let x = (a, b with a = (a 1,, a p 1, b = (b,, b p 1, and let O + 2p = R + p 1 R + p We then have a map Φ 2p : O + 2p O + 2p x := (a, b x + := (a +, b + where a + i and b + i are given in (511, 512 Iteration of this map, starting at x, produces a sequence x n+1 := Φ 2p (x n of points in O + 2p The rational functions formed with these parameters have integrals that remain constant along this orbit Numerical studies suggest the existence of a number L = L(x R + such that (( ( ( ( ( ( p p p p 1 p 1 p 1 x n,,, ; L, L,, L 1 2 p 1 1 p 1 Thus the integral of the original rational function is π 2 L 6 The sixth degree case We discuss the map T 2p : E 2p E 2p for the case p = 3 The effect of T 6 on the coefficients P 6 = {b, b 1, b 2, a 1, a 2 } is denoted by Φ 6 : O + 6 O+ 6 and is given explicitly by (61 a 1 9 + 5a 1 + 5a 2 + a 1 a 2 (a 1 + a 2 + 2 4/3 a 2 a 1 + a 2 + 6 (a 1 + a 2 + 2 2/3 b b + b 1 + b 2 (a 1 + a 2 + 2 2/3 b 1 b (a 2 + 2 + 2b 1 + b 2 (a 1 + 3 a 1 + a 2 + 2 b + b 2 b 2 (a 1 + a 2 + 2 1/3 using Theorem 51 The map Φ 6 preserves the integral (62 U 6 (a 1, a 2, b ; b 1, b 2 := b z 4 + b 1 z 2 + b 2 z 6 + a 1 z 4 + a 2 x 2 + 1 dz and the convergence of its iterations has been proved in [3], the main result of which is the following theorem Theorem 61 Let x := (a 1, a 2; b, b 1, b 2 R + 5 Define x n+1 := Φ 6 (x n Then U 6 is invariant under Φ 6 Moreover, the sequence {(a n 1, a n 2 } converges to (3, 3 and

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 11 {(b n, b n 1, b n 2 } converges to (L, 2L, L, where the limit L is a function of the initial data x Therefore b z 4 + b 1 z 2 + b 2 z 6 + a 1 z 4 + a 2 z 2 + 1 dz = L(x π 2 This iteration is similar to Landen s transformation for elliptic integrals that has been employed in [4] in the efficient calculation of π Numerical data indicate that the convergence of x n is quadratic The proof of convergence is based on the fact that Φ 6 cuts the distance from (a 1, a 2 to (3, 3 by at least half A sequence of algebraic curves The complete characterization of parameters (a 1, a 2 in the first quadrant that yield computable rational functions R(z := b z 4 + b 1 z 2 + b 2 z 6 + a 1 z 4 + a 2 z 2 + 1 of degree 6 remains open The polynomial z 6 + a 1 z 4 + a 2 z 2 + 1 factors when a 1 = a 2 so the diagonal := {(a 1, a 2 R + R + : a 1 = a 2 } produces computable functions In view of the invariance of the class of computable functions under iterations by Φ 6, the curves X n := Φ ( n 6 (, with n Z, are also computable The curve X 1 has equation (9 + 5a 1 + 5a 2 + a 1 a 2 3 = (a 1 + a 2 + 2 2 (a 1 + a 2 + 6 3 and consists of two branches meeting at the cusp (3, 3 In terms of the coordinates x = a 1 3 and y = a 2 3 the leading order term is T 1 (x, y = 1728(x y 2 This curve is rational and can be parametrized by (63 a 1 (t = t 2 (t 5 t 4 + 2t 3 t 2 + t + 1 a 2 (t = t 3 (t 5 + t 4 t 3 + 2t 2 t + 1 The rationality of X n for n 1 and its significance for the integration algorithm remains open The complexity of these curves increases with n For example, the curve X 2 := Φ ( 2 3 ( is of total degree 9 in x = a 1 3 and y = a 2 3 with leading term T 2 (x, y := 2 121 3 35 (x y 18 [ 163(x 4 + y 4 + 668xy(x 2 + y 2 174x 2 y 2] The diagonal can be replaced by a 2-parameter family of computable curves X(c, d that are produced from the factorization of the sextic with a 1 = c + d and a 2 = cd + 1/d All the images Φ ( n 6 X(c, d with n Z are computable curves The question of whether these are all the computable parameters remains open 7 Examples In this section we present a variety of closed-form evaluations of integrals of rational functions Example 1 The integral z 2 (z 4 + 4z 2 + 1 9 dz = 23698523 π 122359464 6 is computed by Mathematica 3 using (32 in 1 seconds The direct calculation too 1227 seconds and 64 extra seconds to simplify the answer

12 GEORGE BOROS AND VICTOR H MOLL Example 2 The integral of any even rational function with denominator a power of an even quartic polynomial can be computed directly by using Corollary 32 For example: z 6 dz (2z 4 + 2z 2 + 3 11 = 11π(14229567 + 4937288 6 443125674 (1 + 6 21/2 Example 3 The case n = in (32 deserves special attention: π (71 N,4 (a; m = 2 m+3/2 (a + 1 P m(a m+1/2 where (72 P m (a = 2 2m m = 2 ( 2m 2 m ( m + The polynomial P m (a has been studied in [1] and [2] Example 4 The case n = m in (32 yields N m,4 (a; m = m (a + 1 z 2m dz (z 4 + 2az 2 + 1 m+1 = π 2 3m+3/2 (1 + a m+1/2 The change of variable z z converts this integral to which is [8] 32579 N m,4 (a; m = 1 2 z m 1/2 dz (z 2 + 2az + 1 m+1, Example 5 A symmetric function of degree 6 The integral I = x 8 (x 6 + 4x 4 + 4x 2 + 1 5 dx = ( 2m m x 8 [(x 2 + 1(x 4 + 3x 2 + 1] 5 dx can be computed by decomposing the integrand into partial fractions as 1 (x 2 + 1 5 1 (x 2 + 1 4 6 (x 2 + 1 3 11 (x 2 + 1 2 31 (x 2 + 1 1 + (x 4 + 3x 2 + 1 5 + 2x 2 (x 4 + 3x 2 + 1 5 4 (x 4 + 3x 2 + 1 4 3x 2 (x 4 + 3x 2 + 1 4 + 12 (x 4 + 3x 2 + 1 3 + 6x 2 (x 4 + 3x 2 + 1 3 32 (x 4 + 3x 2 + 1 2 14x 2 (x 4 + 3x 2 + 1 2 Each of these terms is now computable yielding I = 147326 5 3146875 16 73 + (x 4 + 3x 2 + 1 + 31x 2 (x 4 + 3x 2 + 1 π Example 6 Non-symmetric functions of degree 6 In this case we can use the scheme (61 to produce numerical approximations to the integral For example,

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 13 the evaluation of 45z 4 + 25z 2 + 123 z 6 + z 4 + 3z 2 + 1 is shown below: n a n 1 a n 2 b n b n 1 b n 2 1 3 45 25 123 1 415786 144465 126233 632884 883741 2 26562 317262 42267 15615 836896 3 298142 3338 753541 137717 651111 4 299999 3 696338 139925 72771 5 3 3 699589 139914 699555 6 3 3 699572 139914 699572 7 3 3 699572 139914 699572 Thus L 699572 and 45x 4 + 25x 2 + 123 x 6 + x 4 + 3x 2 + 1 dx 699572 π 2 = 19889 Example 7 Symmetric functions of degree 8 These integrals can be evaluated using Theorem 41 For example: dz (14325195794 + 281536729 26 π (z 8 + 5z 6 + 14z 4 + 5z 2 + 1 4 = 14623232 (9 + 2 26 7/2 Example 8 As in the case of degree 6 we can provide numerical approximations to nonsymmetric integrals of degree 8 The iteration (61 is now replaced by a n+1 1 = an 2 (a n 1 + a n 3 + 4a n 1 a n 3 + 1(a n 1 + a n 3 + 8(a n 2 + 2 (a n 1 + an 2 + an 3 + 23/2 a2 n+1 = an 1 a n 3 + 6(a n 1 + a n 3 + 2(a n 2 + 1 a n 1 + an 2 + a3 3 + 2 a n+1 3 = b n+1 = a n 1 + a n 3 + 8 (a n 1 + an 2 + an 3 + 21/2 b n + b n 1 + b n 2 + b n 3 (a n 1 + an 2 + an 3 + 23/4 b n+1 1 = bn 3 (3a n 1 + a n 2 + 6 + b n 2 (a n 1 + 4 + b n 1 (a n 3 + 4 + b n (3a n 3 + a n 2 + 6 (a n 1 + an 2 + an 3 + 25/4 b n+1 2 = bn 3 (a n 1 + 5 + b n 2 + b n 1 + b n (a n 3 + 5 (a n 1 + an 2 + an 3 + 23/4 b n+1 3 = b n + b n 3 (a n 1 + an 2 + an 3 + 21/4 with initial conditions a 1, a 2, a 3, b, b 1, b 2, b 3 Then (73 U 8 (a 1, a 2, a 3, b, b 1, b 2, b 3 := is invariant under these transformations dz b x 6 + b 1 x 4 + b 2 x 2 + b 3 x 8 + a 1 x 6 + a 2 x 4 + a 3 x 2 + 1 dx

14 GEORGE BOROS AND VICTOR H MOLL Note Numerical calculations show that (a n 1, a n 2, a n 3 (4, 6, 4 and that (b n, b n 1, b n 2, b n 3 (1, 3, 3, 1 L for some L depending upon the initial conditions Example 9 A symmetric function of degree 12 We use Theorem 21 to evaluate as (74 I := z 18 dz (z 12 + 14z 1 + 15z 8 + 4z 6 + 15z 4 + 14z 2 + 1 3 I = 25π(25 56 54 31989888 Here p = 3, n = 9, and m = 2, so n > (m + 1p 1 and we need to apply the transformation z 1/z to reduce the value of n Indeed, we have I = and Theorem 21 now yields z 16 dz (z 12 + 14z 1 + 15z 8 + 4z 6 + 15z 4 + 14z 2 + 1 3, I = 2 17 z 16 dz 13172(1 + z 2 3 (1 + 4z 2 + z 4 3 The new integrand is expanded in partial fractions in the variable t = z 2 to produce (74 Example 1 We use Theorem 21 to evaluate where I := z 1 dz Q 2 (z Q(z = z 2 + 6z 18 + 93z 16 24z 14 + 162z 12 + 548z 1 + 162z 8 24z 6 + 93z 4 + 6z 2 + 1 The factorization with Q(z = (1 + z 2 2 T (zt ( z T (z = z 8 2z 7 + 4z 6 + 14z 5 + 6z 4 14z 3 + 4z 2 + 2z + 1 leads to a partial fraction expansion containing the term 72 51z + 1994z 2 2617z 3 + 1228z 4 43z 5 + 34z 6 55z 7 838868(1 2z + 4z 2 + 14z 3 + 6z 4 14z 5 + 4z 6 + 2z 7 + z 8 2, which we were unable to integrate; furthemore, the roots of T (z = cannot be evaluated by radicals The procedure described in Theorem 21, however, shows that I = z 1 (4 + z 2 (z 6 + 36z 4 + 96z 2 + 64 dz 524288(z 2 + 1 2 (z 8 + 3z 6 + 8z 4 + 3z 2 + 1 2,

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 15 the integrand of which can be expanded to yield 9 I = 838868 dz (z 2 + 1 2 75 838868 dz z 2 + 1 1921z 6 + 1815z 4 + 4111z 2 + 1462 + 297152 (z 8 + 3z 6 + 8z 4 + 3z 2 + 1 2 dz 91z 6 + 719z 4 + 1259z 2 5764 + 838868(z 8 + 3z 6 + 8z 4 + 3z 2 + 1 dz Every piece is now computable, with the final result I = (648 59 15 π 241591914 Example 11 The symmetric functions of degree 16 have denominator D 4 (d 1, d 2, d 3, d 4 ; z = z 16 + d 4 z 14 + d 3 z 12 + d 2 z 1 + 2d 1 z 8 + d 2 z 6 + d 3 z 4 + d 4 z 2 + 1, the integral of which is computed in terms of E 4 (d 1, d 2, d 3, d 4 ; z = (1 + d 1 + d 2 + d 3 + d 4 z 8 + 2(16 + d 2 + 4d 3 + 9d 4 z 6 This new integral is symmetric provided [ ] [ ] d1 15 (75 = + d 2 112 + 8(2 + d 3 + 6d 4 z 4 + 32(8 + d 4 z 2 + 128 [ 3 4] d 3 + [ ] 8 d 7 4 Introduce the new parameters ( ( 8 8 e j = d j for 2 j 4 and e 1 = d 1 1 5 j 2 4 Then (75 yields (76 [ e1 e 2 ] = [ ] 3 e 4 3 + Thus, if the original denominator has the form [ ] 8 e 7 4 D 4 (z = (z 16 + 1 + d 4 (z 14 + z 2 + d 3 (z 12 + z 4 + (112 4d 3 + 7d 4 (z 1 + z 6 the integral +2(15 + 3d 3 8d 4 z 8, P (z ( D 4 (z m+1 dz is reduced to an integral with symmetric denominator of degree 8 and these are computable We can thus evaluate a 2-parameter family of symmetric integrals of degree 16 For example, tae d 3 = d 4 = 1 to obtain (77 R 1 (z = The main theorem yields (78 z 4 (z 16 + z 14 + z 12 + 115z 1 + 2z 8 + 115z 6 + z 4 + z 2 + 1 2 R 2 (z = 124z4 + 234z 6 + 1792z 8 + 56z 1 + 6z 12 + z 14 2 7 (16z 8 + 36z 6 + 27z 4 + 36z 2 + 16 2

16 GEORGE BOROS AND VICTOR H MOLL so that R 1 (z dz = Letting f[n] := N n,8 [1, n, 27/32, 9/4] we obtain R 2 (z dz R 1 (z dz = 2 15 (f[] + 6f[1] + 1584f[2] + 496f[3] and conclude that z 4 dz (z 16 + z 14 + z 12 + 115z 1 + 2z 8 + 115z 6 + z 4 + z 2 + 1 2 = = (149288517 + 129473 131π 1124663296 54925 + 4798 131 Example 12 We classify the symmetric denominators of degree 32 that yield computable integrals These functions depend on 8 parameters (79 D 8 (d 1,, d 8 ; z = 8 d 9 (z 2 + z 2(16 = and the main theorem expresses the integral in terms of E 8 The conditions for E 8 to be symmetric yield and the symmetric E 8 is d 1 = 3441 + 35d 5 + 64d 6 312d 7 3264d 8 d 2 = 3472 56d 5 11d 6 + 56d 7 + 4565d 8 d 3 = 3472 + 28d 5 + 64d 6 329d 7 224d 8 d 4 = 496 8d 5 19d 6 + 8d 7 + 938d 8 E 8 (d 5, d 6, d 7, d 8 ; z = 32768(1 + z 16 + (13172 + 8192d 8 (z 2 + z 14 + (212992 + 248d 7 + 28672d 8 (z 4 + z 12 + (18224 + 512d 6 + 6144d 7 + 39424d 8 (z 6 + z 1 + d 5 d 6 +(8448 + 128d 5 + 128d 6 + 6912d 7 + 2688d 8 z 8 The symmetry of E 8 now determines d 5, d 6 in terms of d 7, d 8 and we obtain d 1 63475 9166 5464 d 2 18 14392 86645 d 3 d 4 = 31 47936 13664 + 6895 1964 d 7 + 41664 (71 11471 d 8 222 322 2 224 28 189 The function (z 2 + 1 16 is a symmetric polynomial of degree 32 and yields a particular solution to (71 As before let e j = d j ( 16 9 j for 2 j 8 and e 1 = d 1 1 2 ( 16 8

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 17 Then (711 e 1 e 2 e 3 e 4 e 5 e 6 = 9166 5464 14392 86645 6895 1964 e 7 + 41664 11471 322 2 28 189 and as in the case of degree 16 we can compute a 2-parameter family of symmetric integrals of degree 32 e 8 Appendix A Two binomial sums The closed-form evaluation of sums involving binomials coefficients can be obtained by traditional analytical techniques or by using the powerful WZ-method as described in [16] We discuss two sums used to simplify expressions in later sections, presenting one proof in each style Lemma A1 Let, N be positive integers with N Then N ( ( ( 2N + 1 N j 2N (A1 = 4 N j= Proof Multiply the left hand side of (A1 by x and sum over to produce N N ( ( 2N + 1 N j N ( N j 2N + 1 ( N j x = x = j= = j= = N ( 2N + 1 (x + 1 N j j= = (1 + x + 1 2N+1 (1 x + 1 2N+1 2 x + 1 N ( 2N = 4 N x In order to justify the last step we start with the well nown result ( i 1 1 1 4y = ( 2 + i (A2 y 1 4y 2y (see WILF [19], page 54 Letting x = 4y and i = 2N + 1 in (A2 gives (1 x + 1 2N+1 ( 2N + 1 + 2 2 = ( 1 +1 4 (N+1+ x 2N+1+ ; x + 1 = similarly x = 4y and i = 2N 1 yields (1 + x + 1 2N+1 ( 2N + 1 + 2 2 = ( 1 +1 4 (N+1+ x 2N+1+ x + 1 =2N+1 =

18 GEORGE BOROS AND VICTOR H MOLL Thus (1 + x + 1 2N+1 (1 x + 1 2N+1 2 x + 1 = = 2N ( (2N + 1 2 ( 1 4 N x = N ( 2N 4 N x = Lemma A2 Let N N Then N ( 2N + 1 (1 + z 2 N j = j= N ( N + j 4 j z 2(N j j= Proof The coefficient of z 2 on the left hand side is N ( ( 2N + 1 N j, j= and the corresponding coefficient on the right hand side is ( 2N 4 N The result then follows from Lemma A1 Lemma A3 Let, N N with N Then ( ( 2N N j = 22 1 N N j= F (; j = ( + N 1 N Proof This lemma could be proven in the same style as Lemma A1 Instead we use the WZ-method as explained in [16] Indeed, let ( ( 2N N j N N2 2 1( +N 1 N and define, with the pacage EKHAD, the function G(; j = F (; j, j( 1 2(N + ( j + 1 Then F (; j F ( + 1; j = G(; j + 1 G(; j, and summing over j we see that the sum of F (; j over j is independent of The case = N produces 1 as the common value Lemma A4 Let p N, d 1, d 2,, d p be parameters, and define d p+1 := 1 Then p p ( 2p 2 (A3 d p+1 z 2 (1 + z 2 p j = p+1 j=1 d j = z 2p + j= p 2 2i 1 z 2(p i i=1 p+1 i j=1 j + i 1 i ( j + 2i 2 j 1 d j+i

LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS 19 Proof For fixed i p 1 the coefficient of z 2i on the right hand side of (A3 is [RHS] (2i = 22(p i 1 p+1 ( r + p i 2 (r 1 d r, p i r p + i 1 r=p i+1 and for i = p we have [RHS] (2p = 1 + p j=1 d j Similarly, for the left hand side of (A3, [LHS] (2i = p+1 r=p+1 i d r p i ( 2r 2 j= ( r j 1 r j 1 + i It is easy to chec that the coefficients of z 2p match It suffices to show that for each i such that i p 1 and for each r such that p + 1 i r p + 1 we have p i ( 2r 2 j= ( r 1 j r 1 p + i = 22(p i 1 (r 1 p i This follows from Lemma A1 with = p i and N = r 1 ( r + p i 2 r p + i 1 The suggestions of the referrees and the editor are gratefully acnowledged References [1] BOROS, G - MOLL, V: A criterion for unimodality Elec Jour of Combinatorics, 6 (1999, #R1 [2] BOROS, G - MOLL, V: An integral hidden in Gradshteyn and Ryzhi Jour Comp Appl Math 16, 361-368, 1999 [3] BOROS, G - MOLL, V: A rational Landen transformation Contemporary Mathematics 251, 83-91, 2 [4] BORWEIN, J - BORWEIN, P: Pi and the AGM Canadian Mathematical Society Wiley- Interscience Publication [5] BRONSTEIN, M: Symbolic Integration I Transcendental functions Algorithms and Computation in Mathematics, 1 Springer-Verlag, 1997 [6] GEDDES, K - CZAPOR, SR - LABAHN, G: Algorithms for Computer Algebra Kluwer, Dordrecht The Netherlands, 1992 [7] GAUSS, KF: Arithmetische Geometrisches Mittel, 1799 In Were, 3, 361-432 Konigliche Gesellschaft der Wissenschaft, Gottingen Reprinted by Olms, Hildescheim, 1981 [8] GRADSHTEYN, IS - RYZHIK, IM: Table of Integrals, Series and Products Fifth Edition, ed Alan Jeffrey Academic Press, 1994 [9] HARDY, GH: The Integration of Functions of a Single Variable Cambridge Tracts in Mathematics and Mathematical Physics, 2, Second Edition, Cambridge University Press, 1958 [1] HERMITE, C: Sur l integration des fractions rationelles Nouvelles Annales de Mathematiques ( 2 eme serie 11, 145-148, 1872 [11] HOROWITZ, E: Algorithms for partial fraction decomposition and rational function integration Proc of SYMSAM 71, ACM Press, 441-457, 1971 [12] LAZARD, D - RIOBOO, R: Integration of Rational Functions: Rational Computation of the Logarithmic Part Journal of Symbolic Computation 9, 113-116, 199 [13] MCKEAN, H - MOLL, V: Elliptic Curves: Function Theory, Geometry, Arithmetic Cambridge University Press, 1997 [14] NEWMAN, D: A simplified version of the fast algorithm of Brent and Salamin Math Comp 44, 27-21, 1985 [15] OSTROGRADSKY, MW: De l integration des fractions rationelles Bulletin de la Classe Physico-Mathematiques de l Academie Imperieriale des Sciences de St Petersbourgh, IV, 145-167, 286-3 1845

2 GEORGE BOROS AND VICTOR H MOLL [16] PETKOVSEK, M - WILF, HS - ZEILBERGER, D: A=B A K Peters, Wellesley, Massachusetts 1996 [17] ROTHSTEIN, M: A new algorithm for the integration of Exponential and Logarithmic Functions, Proc of the 1977 MACSYMA Users Conference, NASA Pub, CP-212, 263-274 [18] TRAGER, BM: Algebraic factoring and rational function integration Proc SYMSAC 76, 219-226 [19] WILF, HS: generatingfunctionology Academic Press, 199 Department of Mathematics, Xavier University, New Orleans, Louisiana 7125 E-mail address: gboros@xulamathedu Department of Mathematics, Tulane University, New Orleans, LA 7118 E-mail address: vhm@mathtulaneedu