LANDEN TRANSFORMATIONS AND THE INTEGRATION OF RATIONAL FUNCTIONS. 1. Introduction We consider the space of even rational functions of degree 2p

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1 MATHEMATICS OF COMPUTATION Volume 71, Number 238, Pages S (1) Article electronically published on November 9, 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 := E 2p p=1 of normalized even rational functions, and the set of 2p 1 parameters 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} = Received by the editor April 27, Mathematics Subject Classification Primary 33-XX Key words and phrases Rational functions, Landen transformation, integrals The second author was supported in part by NSF Grant DMS c 21 American Mathematical Society

2 65 GEORGE BOROS AND VICTOR H MOLL (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 For m N define E m 2p := {R E (den(r)) 1/(m+1) is even, symmetric of degree 2p} and E m,s 2p p=1 := E m 2p E s 2p, soafunctionr E m,s 2p canbewrittenintheform P (z) R(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 π 2m (z 2 +1) m+1 = 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 WethusobtainamapT 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

3 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 651 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 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 2 b + 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 )ofpointsino + 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+1 := Φ 2p (x n )satisfies x n (( ) ( ) ( ) ( ) ( ) ( ) ) p p p p 1 p 1 p 1,,, ; L, L,, L 1 2 p 1 1 p 1 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 18th 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) overr Once this is nown the partial fraction decomposition of R(z) canbe computed The fact is that the primitive of a rational function is always elementary: it consists of a new rational function (itsrational 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) =

4 652 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 2j /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) E p (d 1,d 2,,d p ; z) = D p (d 1,d 2,,d p ; z) = p+1 + j=1 p i=1 d j z 2p p d p+1 (z 2 + z 4p 2 ) = p i+1 2 2i 1 z 2(p i) for d i R + Then for n (m +1)p 1, z 2n dz (D p (d 1,,d p ; z)) m+1 (23) (m+1)p n 1 =2 m j=1 j= z 2((m+1)p 1 j) j + i 1 i ( ) (m +1)p n 1+j 4 j 2j ( E p (d 1,,d p ; z)) m+1 dz, ( j +2i 2 j 1 and for (m +1)p 1 <n<2p(m +1) 1 we employ the symmetry rule ) d j+i, (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

5 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 653 for n (m +1)p 1 The substitution z =tanθ yields N n,p = π/2 (1 C 2 ) n C 4(m+1)p 2n 2 dθ ( p = d p+1 {(1 C 2 ) 2p C 2 +(1 C 2 ) C 4p 2 } ) m+1, where C =cosθ Letting ψ =2θ and D =cosψ =2C 2 1thengives π (1 D 2 ) n (1 + D) 2(m+1)p 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+1)p 2n 1, producing N n,p = 2 m π/2 (1 D 2 ) n (m+1)p n 1 j= { p = d p+1 (1 D 2 ) p j= A second double angle substitution ϕ =2ψ gives π (1 E) n (m+1)p n 1 N n,p =2 m j= { p = d p+1 (1 E) p j= ( 2(m+1)p 2n 1 ) 2j D 2j dθ ) } m+1 D 2j ( 2p 2 2j ( 2(m+1)p 2n 1 ) 2j 2 p(m+1) n j 1 (1 + E) j dϕ ( 2p 2 ) } m+1, 2j 2 p j (1 + E) j where E =cosϕ =2D 2 1 The change of variable z =tan(ϕ/2) then yields z 2n (m+1)p n 1 N n,p = 2 m j= (26) { p ( = d p p+1 z 2 j= ( 2(m+1)p 2n 1 ) 2j (1 + z 2 ) (m+1)p n j 1 dz ( 2p 2 ) )} m+1 2j (1 + z2 ) p j Finally, we modify (26) using Lemma A2 and Lemma A4 with N =(m+1)p n 1 to produce (23) Note that the previous theorem associates to each rational function of symmetric denominator R 1 (z) = b s z 2s + b s 1 z 2(s 1) + + b ( z 4p + d p z 4p d 1 z 2p + +1) m+1 a new rational function (m+1)p 1 R 2 (z) =2 m n= (m+1)p n 1 b n j= ( ) (m +1)p n 1+j z 2((m+1)p 1 j) 2j (E p (d 1,,d p ; z)) m+1 4 j such that R 1 (z) dz = R 2 (z) dz

6 654 GEORGE BOROS AND VICTOR H MOLL 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) isanevenpolynomialofdegree4m+2 Observe that the normalization a = a 2 = 1 maes the denominator of R automatically symmetric It suffices to evaluate z 2n (31) dz N n,4 (d 1 ; m) := (z 4 +2d 1 z 2 +1) m+1, 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 z 2n dz (32) N n,4 (d 1 ; m) := (z 4 +2d 1 z 2 +1) m+1 m n π = 2 3m+3/2 (1 + a) 2 j (1 + d m+1/2 1 ) j j= ( )( )( )( ) 1 2m 2j 1 m n + j 2j m m j 2j j j For m +1 n 2m +1 we have z 2n dz (33) (z 4 +2d 1 z 2 +1) m+1 π = 2 3m+3/2 (1 + d 1 ) m+1/2 n m 1 ( )( )( )( ) 1 2m 2j 1 m n + j 2j m 2 j (1 + d 1 ) j m j 2j j 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,sothat z 2n m n dz ( ) m n + j (z 4 +2d 1 z 2 =2 m 4 j z 2(m j) dz +1) m+1 2j ((1 + d 1 )z 2 +2) m+1 j= 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 u m j 1/2 du (1 + u) m+1 ( )( )( ) 1 2m 2j 2j m = π 2 (2m+j+3/2) (1 + d 1 ) (m j+1/2), m j j j

7 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 655 wherewehaveused u r 1/2 (1 + u) s du = π 2 2(s 1) ( 2r r )( 2(s r 1) s r 1 )( ) 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) := z 2n dz (bz 4 +2az 2 + c) m+1 (34) N n,4 (a, b, c; m) ( { = π c(c/b) m n 8(a + } ) 2m+1 1/2 bc) m n = 2 ( 2m 2 m and for m +1 n 2m +1, )( m n + 2 )( 2 N n,4 (a, b, c; m) ( { = π c(c/b) m n 8(a + } ) 2m+1 1/2 bc) )( m ) 1 ( a +1), bc (35) n m 1 = 2 ( 2m 2 m )( m n + 2 )( 2 )( m ) 1 ( a +1) bc 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 ( a bc ; m 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) form +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 z 2n dz N n,8 (a 1,a 2 ; m) = (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 +4)z 2 +8

8 656 GEORGE BOROS AND VICTOR H MOLL 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 2j m + j 2j m n m j 2j 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, z 2n dz (z 8 + a 2 z 6 +2a 1 z 4 + a 2 z 2 +1) m+1 m m = t,j (m, n; a 1,a 2 )+ =n j= 2m+1 =m+1 2m+1 =n m 1 j= Proof The reduction formula yields z 2n dz (41) (z 8 + a 2 z 6 +2a 1 z 4 + a 2 z 2 +1) m+1 2m+1 ) =2 3m+2 =n 2 2 ( 4m n +2 n We then use Corollary 32 to evaluate (41) m 1 j= t,j (m, n; a 1,a 2 ), t,j (m, n; a 1,a 2 ) z 2 dz (c 2 z 4 +2c 1 z 2 +8) m+1 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 G(a, b) := π/2 dθ a2 cos 2 θ + b 2 sin 2 θ (52) ie, that a 1 =(a + b)/2, b 1 = ab, (53) G(a 1,b 1 ) = G(a, b) 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

9 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 657 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 dθ 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 p 1 p (56) P (z) = b j z 2(p 1 j) and Q(z) = a j z 2(p j) 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) (59) for j 2p 1, andalso (51) Let (511) α p (i) = d 1 = 1 2 c j = { 2 2i 1 p+1 i =1 1+ p 2p 1 = +i 1 i = p = =1 d if i = a + i = a 2 p, a j b p 1 j+ ( +2i 2 ) 1 d+i if 1 i p, α p (i) 2 2i Q(1) 2(1 i/p) for 1 i p 1, and [ p 1 i ( ) ] p 1 + i (512) b + i = Q(1) 2i/p+1/p 2 (c + c 2p 1 ) 2i = for i p 1 Finally, define the polynomials p 1 p (513) P + (z) = b + i z2(p 1 i) and Q + (z) = a + i z2(p i) = Then T 2p (R(z)) := P + (z)/q + (z) satisfies (55), ie, (514) P (z) Q(z) dz = = P + (z) Q + (z) dz

10 658 GEORGE BOROS AND VICTOR H MOLL 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):= j= 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 z 2 dz L := 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 2j z 2(p 1 j) dz 2j p i= α p(i)z 2(p i) = p 1 α p (p) j=1 ( p + j 2 2 )λ 2(j 1) 2p 2j+1 2j 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) 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 Q m+1 (z) ( z 2p Q(1/z) ) m+1 and An algorithm for integration Let x =(a, b) witha =(a 1,,a p 1 ), b = (b,,b p 1 ), and let O + 2p = R + p 1 R p + Wethenhaveamap 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 )ofpointsino + 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

11 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 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 2 b + b 2 (a 1 + a 2 +2) 1/3 using Theorem 51 The map Φ 6 preserves the integral b z 4 + b 1 z 2 (62) + b 2 U 6 (a 1,a 2,b ; b 1,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 ) R5 + 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 {(b n,bn 1,bn 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 b z 4 + b 1 z 2 + b 2 R(z) := 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 3andy = a 2 3 the leading order term is T 1 (x, y) = 1728(x y) 2 This

12 66 GEORGE BOROS AND VICTOR H MOLL 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 3andy = a 2 3 with leading term T 2 (x, y) := (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) withn 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 = π is computed by Mathematica 3 using (32) in 1 seconds The direct calculation too 1227 seconds and 64 extra seconds to simplify the answer 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( ) (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 m ( )( ) 2m 2 m + (72) P m (a) = 2 2m 2 (a +1) 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) = 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 N m,4 (a; m) = 1 z m 1/2 dz 2 (z 2 +2az +1) m+1, which is [8] ( ) 2m m

13 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 661 Table 1 n a n 1 a n 2 b n b n 1 b n Example 5 A symmetric function of degree 6 The integral I = x 8 (x 6 +4x 4 +4x 2 +1) 5 dx = 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) (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) (x 4 +3x 2 +1) + 31x 2 (x 4 +3x 2 +1) Each of these terms is now computable yielding I = π 16 Example 6 Nonsymmetric functions of degree 6 In this case we can use the scheme (61) to produce numerical approximations to the integral For example, the evaluation of 45z z z 6 + z 4 + 3z 2 +1 dz is shown in Table 1 Thus L and 45x x x 6 + x 4 + 3x 2 +1 dx π 2 = Example 7 Symmetric functions of degree 8 These integrals can be evaluated using Theorem 41 For example: dz (z 8 +5z 6 +14z 4 +5z 2 +1) 4 = ( ) π ( ) 7/2

14 662 GEORGE BOROS AND VICTOR H MOLL 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 +2)3/2 a n+1 2 = an 1 an 3 +6(an 1 + an 3 )+2(an 2 + 1) a n 1 + an 2 + a3 3 +2, a n+1 a n 1 + a n = (a n 1 + an 2 +, an 3 +2)1/2 b n+1 b n = + bn 1 + bn 2 + bn 3 (a n 1 + an 2 +, an 3 +2)3/4 b n+1 1 = bn 3 (3an 1 + an 2 +6)+bn 2 (an 1 +4)+bn 1 (an 3 +4)+bn (3an 3 + an 2 +6) (a n 1 + an 2 +, an 3 +2)5/4 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 +2)3/4 b n+1 3 = b n + bn 3 (a n 1 + an 2 + an 3 +2)1/4, with initial conditions a 1,a 2,a 3,b,b 1,b 2,b 3Then b x 6 + b 1 x 4 + b 2 x 2 (73) + b 3 U 8 (a 1,a 2,a 3,b,b 1,b 2,b 3 ) := x 8 + a 1 x 6 + a 2 x 4 + a 3 x 2 +1 dx is invariant under these transformations Note Numerical calculations show that (a n 1,an 2,an 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 z 18 dz I := (z z 1 +15z 8 +4z 6 +15z 4 +14z 2 +1) 3 as I = 25π( ) (74) Here p =3,n=9,andm =2,son>(m +1)p 1 and we need to apply the transformation z 1/z to reduce the value of n Indeed, we have z 16 dz I = (z z 1 +15z 8 +4z 6 +15z 4 +14z 2 +1) 3, and Theorem 21 now yields 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 z 1 dz I := Q 2 (z),

15 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 663 where Q(z) =z 2 +6z z 16 24z z z z 8 24z 6 +93z 4 +6z 2 +1 The factorization Q(z) = (1+z 2 ) 2 T (z)t ( z) with 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 z z z 4 43z 5 +34z 6 55z (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 z 1 (4 + z 2 )(z 6 +36z 4 +96z ) dz I = (z 2 +1) 2 (z 8 +3z 6 +8z 4 +3z 2 +1) 2, the integrand of which can be expanded to yield 9 dz I = (z 2 +1) 2 75 dz z z z z (z 8 +3z 6 +8z 4 +3z 2 +1) 2 dz 91z z z (z 8 +3z 6 +8z 4 +3z 2 +1) dz Every piece is now computable, with the final result I = ( ) π 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 +8(2 + d 3 +6d 4 )z (8 + d 4 )z This new integral is symmetric provided [ ] [ ] [ [ ] d (75) = + d d ] 3 + d 7 4 Introduce the new parameters ( ) ( ) 8 8 e j = d j for 2 j 4 and e 1 = d j 2 4

16 664 GEORGE BOROS AND VICTOR H MOLL Then (75) yields (76) [ ] [ [ ] e1 3 8 = e e 2 4] 3 + e 7 4 Thus, if the original denominator has the form 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 ) +2(15 + 3d 3 8d 4 )z 8, the integral P (z) m+1 dz ( D 4 (z) ) 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 z 4 (77) R 1 (z) = (z 16 + z 14 + z z 1 +2z z 6 + z 4 + z 2 +1) 2 The main theorem yields R 2 (z) = 124z z z z 1 +6z 12 + z 14 (78) 2 7 (16z 8 +36z 6 +27z 4 +36z ) 2 so that R 1 (z) dz = R 2 (z) dz Letting f[n] :=N n,8 [1,n,27/32, 9/4] we obtain R 1 (z) dz = 2 15 (f[] + 6f[1] f[2] + 496f[3]) and conclude that z 4 dz (z 16 + z 14 + z z 1 +2z z 6 + z 4 + z 2 +1) 2 = ( )π Example 12 We classify the symmetric denominators of degree 32 that yield computable integrals These functions depend on 8 parameters 8 (79) D 8 (d 1,,d 8 ; z) = 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 d 1 = d 5 +64d 6 312d d 8, d 2 = d 5 11d d d 8, d 3 = d 5 +64d 6 329d 7 224d 8, d 4 = 496 8d 5 19d 6 +8d d 8,

17 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 665 and the symmetric E 8 is E 8 (d 5,d 6,d 7,d 8 ; z) = 32768(1 + z 16 ) + ( d 8 )(z 2 + z 14 ) + ( d d 8 )(z 4 + z 12 ) + ( d d d 8 )(z 6 + z 1 ) + ( d d d d 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 d d 3 d 4 = d (71) d d 5 d 6 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 ( ) 16 8 Then (711) e 1 e 2 e 3 e 4 e 5 e 6 = e e 8 and as in the case of degree 16 we can compute a 2-parameter family of symmetric integrals of degree 32 Appendix A Two binomial sums The closed-form evaluation of sums involving binomial 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 (A1) N j= ( )( ) 2N +1 N j 2j = ( ) 2N 4 N

18 666 GEORGE BOROS AND VICTOR H MOLL 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 2j 2j = j= j= = N ( ) 2N +1 = (x +1) N j 2j 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 y = ( ) 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) +1 4 (N+1+) x 2N+1+ ; x +1 = similarly x = 4y and i = 2N 1 yields (1 + x +1) 2N+1 ( ) 2N = ( 1) +1 4 (N+1+) x 2N+1+ x +1 =2N+1 Thus (1 + x +1) 2N+1 (1 x +1) 2N+1 2N ( ) (2N +1 2) 2 = ( 1) 4 N x x +1 = N ( ) 2N = 4 N x = Lemma A2 Let N N Then N ( ) 2N +1 (1 + z 2 ) N j = 2j j= N ( ) N + j 4 j z 2(N j) 2j Proof The coefficient of z 2 on the left-hand side is N ( )( ) 2N +1 N j, 2j j= and the corresponding coefficient on the right-hand side is ( ) 2N 4 N Theresult then follows from Lemma A1 j= j= Lemma A3 Let, N N with N Then ( )( ) 2N N j = 22 1 N 2j N ( + N 1 N )

19 TRANSFORMATIONS AND INTEGRATION OF RATIONAL FUNCTIONS 667 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 ) 2j N F (; j) = ), N2 2 1( +N 1 N and define, with the pacage EKHAD, the function j(2j 1) G(; j) = F (; j) 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) overj 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 (A3) p p ( ) 2p 2 d p+1 z 2 (1 + z 2 ) p j 2j = j= p+1 p = z 2p + 2 2i 1 z 2(p i) j=1 d j i=1 p+1 i j=1 j + i 1 i ( ) j +2i 2 j 1 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), p+1 p i ( )( ) [LHS] (2i) = d r 2r 2 r j 1 2j r j 1+i r=p+1 i j= It is easy to chec that the coefficients of z 2p match It suffices to show that for each i such that i p 1andforeachr such that p +1 i r p +1we have p i ( 2r 2 2j 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 d j+i ( ) r + p i 2 r p + i 1 The suggestions of the referees and the editor are gratefully acnowledged References [1] BOROS, G and MOLL, V: A criterion for unimodality Elec Jour of Combinatorics, 6 (1999), #R1 MR 99:517 [2] BOROS, G and MOLL, V: An integral hidden in Gradshteyn and Ryzhi JourComp Appl Math 16, , 1999 MR 2c:3324 [3] BOROS, G and MOLL, V: A rational Landen transformation Contemporary Mathematics 251, 83-91, 2 MR 21h:3329

20 668 GEORGE BOROS AND VICTOR H MOLL [4] BORWEIN,JandBORWEIN,P: Pi and the AGM Canadian Mathematical Society Wiley- Interscience Publication MR 89a:11134; MR 99h:11147 [5] BRONSTEIN, M: Symbolic Integration I Transcendental functions Algorithms and Computation in Mathematics, 1 Springer-Verlag, 1997 MR 98a:6814 [6] GEDDES, K, CZAPOR, SR and LABAHN, G: Algorithms for Computer Algebra Kluwer, Dordrecht The Netherlands, 1992 MR 96a:6849 [7] GAUSS, KF: Arithmetische Geometrisches Mittel, 1799 In Were, 3, Konigliche Gesellschaft der Wissenschaft, Gottingen Reprinted by Olms, Hildescheim, 1981 [8] GRADSHTEYN, IS and RYZHIK, IM: Table of Integrals, Series and Products Fifth Edition, ed Alan Jeffrey Academic Press, 1994 MR 94g:8 [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 MR 5:2417 [1] HERMITE, C: Sur l integration des fractions rationelles Nouvelles Annales de Mathematiques (2 eme serie) 11, , 1872 [11] HOROWITZ, E: Algorithms for partial fraction decomposition and rational function integration Proc of SYMSAM 71, ACM Press, , 1971 [12] LAZARD, D and RIOBOO, R: Integration of Rational Functions: Rational Computation of the Logarithmic Part Journal of Symbolic Computation 9, , 199 MR 91h:6891 [13] MCKEAN, H and MOLL, V: Elliptic Curves: Function Theory, Geometry, Arithmetic Cambridge University Press, 1997 MR 98g:1432 [14] NEWMAN, D: A simplified version of the fast algorithm of Brent and Salamin Math Comp 44, 27-21, 1985 MR 86e:653 [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, , [16] PETKOVSEK, M, WILF, HS and ZEILBERGER, D: A=B A K Peters, Wellesley, Massachusetts 1996 MR 97j:51 [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, [18] TRAGER, BM: Algebraic factoring and rational function integration Proc SYMSAC 76, [19] WILF, HS: generatingfunctionology Academic Press, 199 MR 91g:58 Department of Mathematics, Xavier University, New Orleans, Louisiana address: gboros@xulamathedu Department of Mathematics, Tulane University, New Orleans, Louisiana address: vhm@mathtulaneedu