Additive Decompositions in Primitive Extensions
aa r X i v : . [ c s . S C ] F e b Additive Decompositions in Primitive Extensions ∗ Shaoshi Chen, Hao Du, and Ziming Li KLMM, Academy of Mathematics and Systems ScienceChinese Academy of Sciences, Beijing, 100190, China School of Mathematical Sciences, University of Chinese Academy of SciencesBeijing 100049, (China) [email protected], [email protected], [email protected]
August 14, 2018
Abstract
This paper extends the classical Ostrogradsky–Hermite reduction forrational functions to more general functions in primitive extensions ofcertain types. For an element f in such an extension K , the extendedreduction decomposes f as the sum of a derivative in K and anotherelement r such that f has an antiderivative in K if and only if r = 0;and f has an elementary antiderivative over K if and only if r is a linearcombination of logarithmic derivatives over the constants when K is alogarithmic extension. Moreover, r is minimal in some sense. Additivedecompositions may lead to reduction-based creative-telescoping methodsfor nested logarithmic functions, which are not necessarily D -finite. Symbolic integration, together with its discrete counterpart symbolic summa-tion, nowadays has played a crucial role in building the infrastructure for apply-ing computer algebra tools to solve problems in combinatorics and mathematicalphysics [15, 16, 27]. The early history of symbolic integration starts from thefirst tries of developing programs in LISP to evaluate integrals in freshman cal-culus symbolically in the 1960s. Two representative packages at the time wereSlagle’s SAINT [28] and Moses’s SIN [19] which were both based on integraltransformation rules and pattern recognition. The algebraic approach for sym-bolic integration is initialized by Ritt [25] in terms of differential algebra [14],which eventually leads to the Risch algorithm for the integration of elementary ∗ S. Chen was supported by the NSFC Grants 11501552, 11688101 and by the Fund of theYouth Innovation Promotion Association, CAS. F isa field together with a derivation ′ that is an additive map on F satisfying theproduct rule ( f g ) ′ = f ′ g + f g ′ for all f, g ∈ F . A given element f in F is said tobe integrable in F if f = g ′ for some g ∈ F . The problem of deciding whether agiven element is integrable or not in F is called the integrability problem in F .For example, if F is the field of rational functions, then for f = 1 /x we canfind g = − /x , while for f = 1 /x no suitable g exists in F . When f is notintegrable in F , there are several other questions we may ask. One possibilityis to ask whether there is a pair ( g, r ) in F × F such that f = g ′ + r , where r is minimal in some sense and r = 0 if f is integrable. This problem is calledthe decomposition problem in F . Extensive work has been done to solve theintegrability and decomposition problems in differential fields of various kinds.Abel and Liouville pioneered the early work on the integrability problem inthe 19th century [25]. In 1833, Liouville provided a first decision procedure forsolving the integrability problem on algebraic functions [18]. For other classesof functions, complete algorithms for solving the integrability problem are muchmore recent: 1) the Risch algorithm [23, 24] in the case of elementary functionswas presented in 1969; 2) the Almkvist–Zeilberger algorithm [2] (also knownas the differential Gosper algorithm) in the case of hyperexponential functionswas given in 1990; 3) Abramov and van Hoeij’s algorithm [1] generalized theprevious algorithm to the general D -finite functions of arbitrary order in 1997.The decomposition problem was first considered by Ostrogradsky [20] in1845 and later by Hermite [13] for rational functions. The idea of Ostrogradskyand Hermite is crucial for algorithmic treatments of the problem, since it avoidsthe root-finding of polynomials and only uses the extended Euclidean algorithmand squarefree factorization to obtain the additive decomposition of a rationalfunction. This reduction is a basic tool for the integration of rational functionsand also plays an important role in the base case of our work. We will referthis reduction as to the rational reduction in this paper. The rational reduc-tion has been extended to more general classes of functions including algebraicfunctions [29, 9], hyperexponential functions [12, 4], multivariate rational func-tions [5, 17], and more recently including D -finite functions [10, 30]. Blendingreductions with creative telescoping [2, 31] leads to the fourth and most recentgeneration of creative telescoping algorithms, which are called reduction-basedalgorithms [3, 4, 5, 9, 10].The telescoping problem can also be formulated for elementary functions.Two related problems are how to decide the existence of telescopers for el-ementary functions and how to compute one if telescopers exist. Reductionalgorithms have been shown to be crucial for solving these two problems. Thisnaturally motivates us to design reduction algorithms for elementary functions.2n this paper, we extend the rational reduction to elements in straight andflat towers of primitive extensions (see Definition 3.5). Our extended reductionssolve the decomposition problems in such towers without solving any Rischequations (Theorems 4.8 and 5.15), and determine elementary integrability insuch towers when primitive extensions are logarithmic (Theorem 6.1).The remainder of this paper is organized as follows. We present basic no-tions and terminologies on differential fields, and collect some useful facts aboutintegrability in primitive extensions in Section 2. We define the notions ofstraight and flat towers, and describe some straightforward reduction processesin Section 3. Additive decompositions in straight and flat towers are given inSections 4 and 5, respectively. The two decompositions are used to determineelementary integrability in Section 6. Examples are given in Section 7 to illus-trate that the decompositions may be useful to study the telescoping problemfor elementary functions that are not D -finite. Let ( F, ′ ) be a differential field of characteristic zero, and let C F denote the sub-field of constants in F . Let E be a differential field extension of F . An element z of E is said to be primitive over F if z ′ belongs to F . If z is primitive andtranscendental over F with C F ( z ) = C F , then it is called a primitive monomial over F , which is a special instance of Liouvillian monomials [8, Definition 5.1.2].Let z be a primitive monomial over F in the rest of this section. An element f ∈ F ( z ) is said to be proper with respect to z or z -proper for brevity if thedegree of its numerator in z is lower than that of its denominator. In particular,zero is z -proper. It is well-known that f can be uniquely written as the sum ofa z -proper element and a polynomial in z . They are called the fractional andpolynomial parts of f , and denoted by fp z ( f ) and pp z ( f ), respectively.Let p be a polynomial in F [ z ]. The degree and leading coefficient of p are denoted by deg z ( p ) and lc z ( p ), respectively. By [8, Thereom 5.1.1], p issquarefree if and only if gcd( p, p ′ ) = 1. A z -proper element is z -simple if itsdenominator is squarefree. Note that z -simple elements are not necessarily z -proper in [8], but they are assumed to be z -proper in this paper without loss ofgenerality.For S ⊂ E , we use S ′ to denote the set { f ′ | f ∈ S } . If S is a C E -linearsubspace, so is S ′ . For m ∈ N , let F [ z ] ( m ) = { p ∈ F [ z ] | deg z ( p ) < m } . In particular, F [ z ] (0) = { } .For f ∈ F ( z ), Algorithm HermiteReduce in [8, page 139] computes a z -simple element g ∈ F ( z ) and a polynomial p ∈ F [ z ] such that f ≡ g + p mod F ( z ) ′ . This algorithm is an extension of the rational reduction by Ostro-gradsky and Hermite. For rational functions, we have p = 0 since all polynomialshave polynomial antiderivatives. Algorithm HermiteReduce is fundamental forour approach to additive decompositions in primitive extensions.3 emma 2.1.
Let g be a z -simple element in F ( z ) . Then g = 0 if g ∈ F ( z ) ′ + F [ z ] . Proof.
Suppose that g = 0. Since g is z -proper, there exists a nontrivial ir-reducible polynomial p ∈ F [ z ] dividing the denominator of g . Since g ∈ F ( z ) ′ + F [ z ] , there exist a ∈ F ( z ) and b ∈ F [ z ] such that g = a ′ + b. Theorder of g at p is equal to −
1. But the order of a ′ at p is either nonnegative orless than − b at p is nonnegative,a contradiction.Every element f ∈ F ( z ) is congruent to a unique z -simple element g mod-ulo F ( z ) ′ + F [ z ] by Algorithm HermiteReduce and Lemma 2.1. We call g the Hermitian part of f with respect to z , denoted by hp z ( f ). The map hp z is C F -linear on F ( z ). Its kernel is equal to F ( z ) ′ + F [ z ]. Thus, two elementshave the same Hermitian parts if they are congruent modulo F ( z ) ′ + F [ z ]. Thisobservation is frequently used in the sequel.Now, we collect some basic facts about primitive monomials. They areeither straightforward or scattered in [8]. We list them below for the reader’sconvenience. Lemma 2.2. If p ∈ F [ z ] and p ∈ F ( z ) ′ , then there exists c ∈ C F such that lc z ( p ) ≡ cz ′ mod F ′ . Proof.
Assume p = r ′ for some r ∈ F ( z ). Then r ∈ F [ z ] by Theorem 4.4.2 (i)in [8]. Set d = deg z ( p ) and ℓ = lc z ( p ). Then deg z ( r ) ≤ d + 1 by Lemma 5.1.2in [8]. Assume that r ≡ az d +1 + bz d mod F [ z ] ( d ) for some a, b ∈ F . Then r ′ ≡ a ′ z d +1 + (( d + 1) az ′ + b ′ ) z d mod F [ z ] ( d ) . Since p = r ′ , we have that a ′ = 0 and ℓ = ( d + 1) az ′ + b ′ . Hence, ℓ ≡ cz ′ mod F ′ with c = ( d + 1) a ∈ C F .The next lemma will be used to decrease the degree of a polynomial modulo F ( z ) ′ . Its proof is a straightforward application of integration by parts. Lemma 2.3.
We have f ′ z d ≡ F ( z ) ′ + F [ z ] ( d ) for all f ∈ F and d ∈ N . Recall that an element f in F is said to be a logarithmic derivative in F if f = a ′ /a for some nonzero element a ∈ F . Lemma 2.4. If f is a C F -linear combination of logarithmic derivatives in F ( z ) ,then f = hp z ( f ) + r, where r is a C F -linear combination of logarithmic deriva-tives in F .Proof. It suffices to assume that f is a logarithmic derivative in F ( z ), becausethe map hp z is C F -linear.If f = 0, then we choose r = 0, which equals 1 ′ /
1. Otherwise, there existtwo monic polynomials u, v ∈ F [ z ] and w ∈ F such that f = u ′ /u − v ′ /v + w ′ /w by the logarithmic derivative identity in [8, page 104]. Note that u ′ /u − v ′ /v is z -simple by Lemma 5.1.2 in [8] and w ′ /w is in F . Thus, hp z ( f ) = u ′ /u − v ′ /v and r = w ′ /w . 4 Primitive extensions
Let ( K , ′ ) be a differential field of characteristic zero. Set C = C K . Considera tower of differential fields K ⊂ K ⊂ · · · ⊂ K n , (3.1)where K i = K i − ( t i ) for all i with 1 ≤ i ≤ n . The tower given in (3.1) is saidto be primitive over K if t i is a primitive monomial over K i − for all i with1 ≤ i ≤ n . The notation introduced in (3.1) will be used in the rest of thepaper. Remark 3.1.
The derivatives t ′ , . . . , t ′ n are linearly independent over C , since C K n = C in (3.1) . The following lemma tells us a way to modify the leading coefficient of apolynomial in K n − [ t n ] via integration by parts and Algorithm HermiteReduce . Lemma 3.2.
Let the tower (3.1) be primitive with n ≥ . Then, for all ℓ ∈ K n − and d ∈ N , there exist a t n − -simple element g ∈ K n − and a polynomial h ∈ K n − [ t n − ] such that ℓt dn ≡ ( g + h ) t dn mod K ′ n + K n − [ t n ] ( d ) . Proof.
By Algorithm
HermiteReduce , there are f, g ∈ K n − with g being t n − -simple, and h ∈ K n − [ t n − ] such that ℓ = f ′ + g + h . Then ℓt dn = f ′ t dn +( g + h ) t dn .Applying Lemma 2.3 to the term f ′ t dn , we see that the lemma holds.Let ≺ be the purely lexicographic ordering on the set of monomials in t , t , . . . , t n with t ≺ t ≺ . . . ≺ t n . For all i with 0 ≤ i ≤ n − p ∈ K i [ t i +1 , . . . , t n ] with p = 0, the head monomial of p , denoted by hm( p ), isdefined to be the highest monomial in t i +1 , . . . , t n appearing in p with respectto ≺ . The head coefficient of p , denoted by hc( p ), is defined to be the coefficientof hm( p ), which belongs to K i . The head coefficient of zero is set to be zero.The monomial ordering ≺ induces a partial ordering on K i [ t i +1 , . . . , t n ], whichis also denoted by ≺ . Example 3.3.
Let ξ = t t t . Viewing ξ as an element of K [ t , t , t ] , wehave hm( ξ ) = ξ and hc( ξ ) = 1 , while, viewing ξ as an element of K [ t , t ] , wehave hm( ξ ) = t t and hc( ξ ) = t , The next lemma will be used in Section 5. We present it below because itholds for primitive towers.
Lemma 3.4.
Let n ≥ . For a polynomial p ∈ K n − [ t n ] , there are polynomials p i ∈ K i [ t i +1 , . . . , t n ] such that p ≡ P n − i =0 p i mod K ′ n , and that hc( p i ) is t i -simple for all i with ≤ i ≤ n − . Moreover, deg t n ( p i ) ≤ deg t n ( p ) for all i with ≤ i ≤ n − . roof. We proceed by induction on n . If n = 1, then set p = p , because thereis no requirement on hc( p ). Assume that n > n − p ∈ K n − [ t n ] and d = deg t n ( p ). By Lemma 3.2, p ≡ ( g + h ) t dn mod K ′ n + K n − [ t n ] ( d ) , where g ∈ K n − is t n − -simple and h ∈ K n − [ t n − ]. Then there exist h j ∈ K j [ t j +1 , . . . , t n − ] such that h = P n − j =0 h j + u ′ for some u in K n − and hc( h j ) is t j -simple for all j when 1 ≤ j ≤ n − h n − = g . By Lemma 2.3, p ≡ n − X j =0 h j t dn mod K ′ n + K n − [ t n ] ( d ) . (3.2)We need to argue inductively on d . If d = 0, then it is sufficient to set p j = h j for all j with 0 ≤ j ≤ n −
1, as K n − [ t n ] (0) = { } . Assume that d > K n − [ t n ] ( d ) . By (3.2) and the inductionhypothesis on d , we have p ≡ n − X j =0 h j t dn + n − X j =0 ˜ p j mod K ′ n , where ˜ p j is in K j [ t j +1 , . . . , t n ], hc(˜ p j ) is t j -simple when j ≥
1, and deg t n (˜ p j ) < d .Set p j = h j t dn + ˜ p j . Then p ≡ P n − j =0 p j mod K ′ n . Since hc( p j ) is hc( h j ) if h j = 0and hc( p j ) is hc(˜ p j ) if h j = 0, the requirements on each hc( p j ) with j ≥ d is completed, and so is the induction on n . Definition 3.5.
The tower given in (3.1) is said to be straight if hp t i − ( t ′ i ) = 0 for all i with ≤ i ≤ n . The tower is said to be flat if t ′ i ∈ K for all i with ≤ i ≤ n . Example 3.6.
Let K = C ( x ) with the usual derivation in x . Let log( x ) = Z x − dx and Li( x ) = Z log( x ) − dx. Then the tower K ⊂ K (log( x )) ⊂ K (log( x ) , Li( x )) is straight, while the tower K ⊂ K (log( x )) ⊂ K (log( x ) , log( x + 1)) is flat. They contain no new constants by Theorem 5.1.1 in [8]. In this paper, we consider additive decompositions for elements in eitherstraight or flat towers with K = C ( t ).6 emma 3.7. Let the tower (3.1) be primitive. Assume that K is equal to C ( t ) with the usual derivation in t . Then hp t ( t ′ ) is nonzero. Moreover, hp t ( t ′ i ) isnonzero for all i with ≤ i ≤ n if (3.1) is flat.Proof. By the rational reduction, t ′ = u ′ + v for some u, v ∈ K with v being t -simple. Then v is nonzero, since C = C K n . The second assertion can be provedsimilarly. Example 3.8.
Let K − = C , K = K − ( t ) with the usual derivation in t ,and each t i be logarithmic in (3.1) for all i with ≤ i ≤ n . By Lemma 2.4, t ′ i = hp t i − ( t ′ i ) + r i , where r i ∈ K n − for all i with ≤ i ≤ n . The tower isstraight if and only if t ′ i ∈ K i − \ K i − for all i with ≤ i ≤ n . In addition, t ′ ∈ K \ K − as K = K − ( t ) . In this section, we assume that the tower (3.1) is straight and that K = C ( t )with the usual derivation with respect to t . The subfield C of constants isdenoted by K − in recursive definitions and induction proofs to be carried out.Our idea is reducing a polynomial in K n − [ t n ] to another of lower degree viaintegration by parts, whenever it is possible. The notion of t n -rigid elementsdescribes r ∈ K n − such that rt dn cannot be congruent to a polynomial of degreelower than d modulo K ′ n . Definition 4.1.
An element r ∈ K − is said to be t -rigid if r = 0 . Let r ∈ K n − with f = fp t n − ( r ) and p = pp t n − ( r ) . We say that r is t n -rigid if f is t n − -simple, f = c hp t n − ( t ′ n ) for any nonzero c ∈ C , and lc t n − ( p ) is t n − -rigid. Note that zero is t n -rigid, because hp t n − ( t ′ n ) is nonzero. Example 4.2.
Let t = x , t = log( x ) and t = Li( x ) . Let ℓ = 1 x + k and ℓ = 1 t + k + ℓ t + xt + x . Then ℓ is t -rigid if k = 0 and ℓ is t -rigid if k k = 0 . The next lemma, together with Lemma 2.2, reveals that a nonzero polyno-mial p in K n − [ t n ] with a t n -rigid leading coefficient has no antiderivative in K n . Lemma 4.3.
Let r ∈ K n − be t n -rigid. If r ≡ ct ′ n mod K ′ n − (4.1) for some c ∈ C , then both r and c are zero. roof. We proceed by induction on n . If n = 0, then r = 0 by Definition 4.1.Thus, ct ′ ≡ K ′− . Consequently, c = 0 because K ′− = { } and t ′ = 1.Assume that n > n −
1. Set f = fp t n − ( r ).Then f = hp t n − ( r ), since f is t n − -simple by Definition 4.1. Applying the maphp t n − to (4.1), we have f = c hp t n − ( t ′ n ) by Lemma 2.1. Hence, c = 0 and f = 0 by Definition 4.1.Set p = pp t n − ( r ). Then (4.1) becomes p ≡ K ′ n − , which, togetherwith Lemma 2.2, implies that lc t n − ( p ) ≡ ˜ ct ′ n − mod K ′ n − for some ˜ c ∈ C . Itfollows from the induction hypothesis that lc t n − ( p ) is zero, and so is p . Thus, r is zero.In K n − [ t n ], we define a class of polynomials that have no antiderivatives in K n . Definition 4.4.
For n ≥ , a polynomial in K n − [ t n ] is said to be t n -straight if its leading coefficient is t n -rigid. Proposition 4.5.
Let p ∈ K n − [ t n ] be a t n -straight polynomial. Then p = 0 if p ∈ K ′ n .Proof. If n = 0, then p = 0 by Definition 4.1. Otherwise, lc t n ( p ) ≡ ct ′ n mod K ′ n − for some c ∈ C by Lemma 2.2. Then lc t n ( p ) = 0 by Lemma 4.3.Consequently, p = 0.Next, we reduce a polynomial to a t n -straight one. Lemma 4.6.
For p ∈ K n − [ t n ] , there exists a t n -straight polynomial q ∈ K n − [ t n ] with deg t n ( q ) ≤ deg t n ( p ) such that p ≡ q mod K ′ n . Proof. If p = 0, then we choose q = 0. Assume that p is nonzero. We proceedby induction on n .If n = 0, then p ≡ K ′ , as every element of K − [ t ] has an antideriva-tive in the same ring.Assume that n > n −
1. Let p ∈ K n − [ t n ]with degree d and leading coefficient ℓ . By Algorithm HermiteReduce , thereare f, g, u, v ∈ K n − with g, v being t n − -simple, and h, w ∈ K n − [ t n − ] suchthat ℓ = f ′ + g + h and t ′ n = u ′ + v + w. We are going to concoct a new expression for ℓ such that ℓ = ( ct n + a ) ′ + r, (4.2)where c ∈ C , a ∈ K n − and r ∈ K n − is t n -rigid. The expression helps usdecrease degrees. To this end, we consider two cases. Case 1.
Assume g = cv for any c ∈ C \ { } . By the induction hypothesis, thereexists a t n − -straight polynomial ˜ h ∈ K n − [ t n − ] such that h = b ′ + ˜ h for some b ∈ K n − . Then ℓ = a ′ + r , where a = f + b and r = g + ˜ h .8 ase 2. Assume g = cv for some c ∈ C \ { } . By the induction hypothesis, thereexists a t n − -straight polynomial ˜ h ∈ K n − [ t n − ] such that h − cw = b ′ + ˜ h forsome b ∈ K n − . Then ℓ = ( ct n + a ) ′ + r , where a = f − cu + b and r = ˜ h .In both cases, r is t n -rigid by Definition 4.1.If d = 0, then p = ℓ . By (4.2), we have p ≡ r mod K ′ n . Let q = r , which is t n -straight by Definition 4.4.Assume that d > K n − [ t n ] ( d ) is congruent toa t n -straight polynomial modulo K ′ n . By (4.2), Lemma 2.3 and the equality ct ′ n t dn = (cid:16) cd +1 t d +1 n (cid:17) ′ , we have p ≡ rt dn + ˜ q mod K ′ n for some ˜ q ∈ K n − [ t n ] ( d ) . If r = 0, then set q = rt dn + ˜ q. Otherwise, applyingthe induction hypothesis on d to ˜ q yields a t n -straight polynomial q with p ≡ q mod K ′ n . Example 4.7.
Consider the integral Z log( x ) Li( x ) dx. With the notation introduced in Example 4.2, we reduce the integrand t t . Wehave that lc t ( t t ) = t . Since t is not t -rigid, t t can be reduced. In fact, t t = x ′ t t . By Lemma 2.3 and a straightforward calculation, we get t t = (cid:0) xt t − xt − x t (cid:1) ′ + 2 xt t + x t . Since x/t is t -rigid, we have that (2 x/t ) t + ( x /t ) is t -straight. Hence, t t has no antiderivative in C ( x, t , t ) by Proposition 4.5. Below is an additive decomposition in a straight tower.
Theorem 4.8.
For f ∈ K n , the following assertions hold.(i) There exist a t n -simple element g ∈ K n and a t n -straight polynomial p ∈ K n − [ t n ] such that f ≡ g + p mod K ′ n . (4.3) (ii) f ∈ K ′ n if and only if both g and p in (4.3) are zero.(iii) If f ≡ ˜ g + ˜ p mod K ′ n , where ˜ g ∈ K n is a t n -simple element and ˜ p ∈ K n − [ t n ] , then g = ˜ g and deg t n ( p ) ≤ deg t n (˜ p ) . Proof. (i) By Algorithm
HermiteReduce , there exist a t n -simple element g ∈ K n and a polynomial h ∈ K n − [ t n ] such that f ≡ g + h mod K ′ n . By Lemma 4.6, h can be replaced by a t n -straight polynomial p .9ii) Since f ∈ K ′ n , the congruence (4.3) becomes g + p ≡ K ′ n . Apply-ing the map hp t n to the new congruence, we have g = 0, because g = hp t n ( g + p ).Thus, p = 0 by Proposition 4.5.(iii) Since g − ˜ g ≡ ˜ p − p mod K ′ n , we have g = ˜ g by Lemma 2.1. If deg t n (˜ p ) < deg t n ( p ), then p − ˜ p is t n -straight, because lc t n ( p − ˜ p ) equals lc t n ( p ). So p − ˜ p = 0by Proposition 4.5, a contradiction. Example 4.9.
Consider the integral Z x ) + log( x ) Li( x ) dx. The integrand is f := 1 /t + t t , in which the notation is introduced in Exam-ple 4.2. By Algorithm HermiteReduce , we have f = ( − t /t ) ′ + x/t + t t . By Theorem 4.8 and Example 4.7, f has no antiderivative in C ( x, t , t ) . In this section, we let the tower (3.1) be flat. The ground field K will bespecialized to C ( t ) later in this section. We are not able to fully carry out thesame idea in Section 4, because hp t i − ( t ′ i ) = 0 for all i = 2 , . . . , n . This spoilsLemma 4.3 and Proposition 4.5. So we need to study integrability in a flat towerdifferently.This section is divided into two parts. First, we extend Lemma 2.3 to thedifferential ring K [ t , . . . , t n ]. Second, we present a flat counterpart of theresults in Section 4. Let us denote K [ t , . . . , t n ] by R n . For a monomial ξ in t , . . . , t n , the C -linearsubspace { p ∈ R n | p ≺ ξ } is denoted by R ( ξ ) n . The notion of scales is motivatedby the following example. Example 5.1.
Let n = 2 , and ξ = 1 , ξ = t and ξ = t . And let ℓ = t ′ + t ′ .Using integration by parts, we find three congruences ℓξ ≡ K ′ , ℓξ ≡ − t ′ t mod K ′ , ℓξ ≡ − t ′ t mod K ′ . The first and third congruences lead to monomials lower than ξ and ξ , respec-tively. But the second one leads to t , which is higher than ξ . The notion ofscales aims to prevent the second congruence from the reduction to be carriedout. efinition 5.2. Let p ∈ R n \ { } and hm( p ) = t e · · · t e n n . The scale of p withrespect to n is defined to be s if e = 0 , . . . , e s − = 0 and e s > . Let p ∈ K .The scale of p with respect to n is defined to be n . The scale of p with respectto n is denoted by scale n ( p ) . Example 5.3.
Let ξ = 1 , ξ = t t and ξ = t . Regarding ξ , ξ and ξ as elements in K [ t , t , t ] , we have that scale ( ξ ) = 3 , scale ( ξ ) = 1 and scale ( ξ ) = 3 ; while, regarding them as elements in K [ t , t , t , t ] , we havethat scale ( ξ ) = 4 , scale ( ξ ) = 1 and scale ( ξ ) = 3 . Notably, if p ∈ K , then the scale of p with respect to n is equal to n , whichvaries as n does. Otherwise, the scale is fixed by hm( p ) no matter in which ring p lives.The next lemma extends Lemma 2.3 and indicates what kind of integrationby parts will be used for reduction. Lemma 5.4.
Let ξ be a monomial in t , . . . , t n and f ∈ K . Then the follow-ings hold.(i) f ′ ξ ≡ K ′ n + R ( ξ ) n . (ii) Let s = scale n ( ξ ) . Then, for all c , . . . , c s ∈ C , ( c t ′ + · · · + c s t ′ s ) ξ ≡ K ′ n + R ( ξ ) n . Proof. (i) It follows from integration by parts and the fact that ξ ′ belongsto R ( ξ ) n .(ii) Set L = 0 and L i = P ij =1 c j t j for i = 1, . . . , n .If ξ = 1, then s = n and L ′ n ξ ∈ K ′ n . The assertion clearly holds. Assumethat ξ = t e s s · · · t e n n with e s >
0. Then L ′ s ξ = L ′ s − ξ + c s t ′ s ξ. Note that L ′ s − ξ belongs to K ′ n + R ( ξ ) n by a direct use of integration by parts. Set η = ξ/t e s s .Then the term c s t ′ s ξ is equal to c s e s +1 (cid:0) t e s +1 s (cid:1) ′ η . Integration by parts leads to c s t ′ s ξ ≡ − c s e s + 1 t e s +1 s η ′ mod K ′ n . (5.1)Then η = 1 if e j = 0 for all j with j > s . So c s t ′ s ξ belongs to K ′ n by (5.1).Otherwise, e j > j with s < j ≤ n . Then each monomial in t e s +1 s η ′ is of total degree P nj = s e j and is of degree e s + 1 in t s . So t e s +1 s η ′ ≺ ξ .Consequently, c s t ′ s ξ ∈ K ′ n + R ( ξ ) n by (5.1).In the rest of this section, we let K = C ( t ) with the usual derivation in t .By Lemma 3.7, we may further assume that t ′ i is nonzero and t -simple for all i with 1 ≤ i ≤ n . Definition 5.5.
For every k with ≤ k ≤ n , an element of K is said to be k -rigid if either it is equal to zero or it is t -simple and is not a C -linearlycombination of t ′ , . . . , t ′ k . roposition 5.6. For p ∈ R n , there exists q ∈ R n such that p ≡ q mod K ′ n and that hc( q ) is s -rigid, where s = scale n ( q ) . Moreover, q (cid:22) p .Proof. Set q = 0 if p = 0. Assume p = 0 and ξ = hm( p ). By the rationalreduction, hc( p ) = f ′ + g for some f, g ∈ K with g being t -simple. Then p = f ′ ξ + gξ mod R ( ξ ) n . By Lemma 5.4 (i), p ≡ gξ + r mod K ′ n for some r ∈ R ( ξ ) n . Set s = scale n ( ξ ). If g is nonzero and s -rigid, then set q = gξ + r .Otherwise, p ≡ ˜ r mod K ′ n for some ˜ r ∈ R ( ξ ) n by Lemma 5.4 (ii). The propositionfollows from a direct Noetherian induction on hm(˜ r ) with respect to ≺ . Example 5.7.
Let K = C ( x ) , t = log( x ) , t = log( x + 1) . and p = t t + (2 /x ) t t + ((2 / ( x + 1)) t . Then hc( p ) = 1 , which is not -rigid. Since t t = x ′ t t , integration by partsleads to p = (cid:0) xt t (cid:1) ′ + q , where q = (cid:0) x − (cid:1) t t − xx +1 t + x +1 t . We canthen reduce q further, because hc( q ) = (2 t − x ) ′ , which is not -rigid either.Repeating this reduction a finite number of times, we see that p is equal to thederivative of ( x + 1) t t − xt t − xt + (2 x + 2) t + 4 xt − x. A flat analogue of straight polynomials is given below.
Definition 5.8.
A polynomial in C [ t ] is said to be t -flat if it is zero. For n ≥ , p ∈ K n − [ t n ] is called a t n -flat polynomial if there exist p i ∈ K i [ t i +1 , . . . , t n ] for all i with ≤ i ≤ n − such that p = P n − i =0 p i , hc( p i ) is t i -simple for all i ≥ , and hc( p ) is s -rigid, where s = scale n ( p ) . The sequence { p i } i =0 , ,...,n − is called a sequence associated to p . Example 5.9.
Let n = 3 and t = x , t = log( x ) , t = log( x + 1) and t = log( x + 2) . Consider p ∈ K [ t ] p = 1 t t |{z} p + 1 t t t | {z } p + 1 x + k t + xt t | {z } p , where k ∈ Z . Obviously, hc( p ) is t -simple and hc( p ) is t -simple. Moreover, scale ( p ) = 3 and hc( p ) is -rigid if k / ∈ { , , } . So p is t -flat if k / ∈ { , , } . We are going to extend the results in Section 4 to the flat case, based on thefollowing technical lemma.
Lemma 5.10.
Let n ≥ and p ∈ K n − [ t n ] be t n -flat. Set ℓ to be lc t n ( p ) . Then fp t n − ( ℓ ) is t n − -simple and pp t n − ( ℓ ) is t n − -flat. Moreover, pp t n − ( ℓ ) − ct ′ n is t n − -flat for all c ∈ C if n > . roof. The lemma is trivial if p = 0.Assume that p is nonzero and d = deg t n ( p ). Let { p i } i =0 , ,...,n − be a se-quence associated to p , and let ℓ i be the coefficient of t dn in p i . Evidently, ℓ = P n − i =0 ℓ i , fp t n − ( ℓ ) = ℓ n − and pp t n − ( ℓ ) = P n − i =0 ℓ i . Moreover, ℓ i = 0if deg t n ( p i ) < d , and hc( ℓ i ) = hc( p i ) otherwise. This is because ≺ is purelylexicographic with t i +1 ≺ · · · ≺ t n for all i with 0 ≤ i ≤ n −
1. Thus, fp t n − ( ℓ )is t n − -simple and pp t n − ( ℓ ) is t n − -flat by Definition 5.8.It remains to show the second assertion. Assume n >
1. Thenpp t n − ( ℓ ) − ct ′ n = ℓ n − + · · · + ℓ + ˜ ℓ with ˜ ℓ = ℓ − ct ′ n , (5.2)and hc( ℓ j ) is t j -simple for all j with 1 ≤ j ≤ n − s = scale n ( p ) and ˜ s = scale n − (˜ ℓ ). It suffices to prove that hc(˜ ℓ ) is˜ s -rigid by (5.2) and Definition 5.8. Case 1. ℓ / ∈ K . Then s < n ,hm( p ) = t e s s · · · t e n − n − t dn and hm( ℓ ) = t e s s · · · t e n − n − , where e s >
0. Moreover, s = scale n − ( ℓ ) , hm( ℓ ) = hm(˜ ℓ ) , and hc( p ) = hc( ℓ ) = hc(˜ ℓ ) . In particular, ˜ s = s . Hence, hc(˜ ℓ ) is ˜ s -rigid, because hc( p ) is s -rigid. Case 2. ℓ ∈ K with ℓ = 0. Then hm( p ) = t dn and s = n . Moreover, ˜ s = n − ℓ ∈ K . Note that p is t n -flat. So hc( p ) is not a C -linear combinationof { t ′ , . . . , t ′ n − , t ′ n } , and neither is ℓ because ℓ = hc( p ). Consequently, ˜ ℓ is not a C -linear combination of { t ′ , . . . , t ′ n − } , and neither is hc(˜ ℓ ), becausehc(˜ ℓ ) = ˜ ℓ . Thus, hc(˜ ℓ ) is ( n − Case 3. ℓ = 0. Then ˜ s = n − ℓ ) = ˜ ℓ = − ct ′ n , which is ( n − Lemma 5.11.
Let n ≥ and p ∈ K n − [ t n ] be t n -flat. If lc t n ( p ) ≡ ct ′ n mod K ′ n − (5.3) for some c ∈ C , then both p and c are zero.Proof. If n = 1, then the tower K ⊂ K is also straight, and p is t -straight byDefinition 4.4 and Lemma 3.7. Both p and c are zero by Lemma 4.3.Assume that n > n −
1. Set ℓ = lc t n ( p ). Applyingthe map hp t n − to (5.3), we have hp t n − ( ℓ ) = 0. Then fp t n − ( ℓ ) = 0 by the firstassertion of Lemma 5.10. Consequently, we have ℓ = pp t n − ( ℓ ). Let q = ℓ − ct ′ n .Then q is t n − -flat by the second assertion of Lemma 5.10. On the other hand,13 ∈ K ′ n − by (5.3). Then lc t n − ( q ) ≡ ˜ ct ′ n − mod K ′ n − for some ˜ c ∈ C byLemma 2.2. So q = 0 by the induction hypothesis. Accordingly, ℓ = ct ′ n ∈ K . (5.4)Let { p i } i =0 , ,...,n − be a sequence associated to p . By (5.4), we have hc( p ) = ct ′ n , because ct ′ n is not t i -simple for all i with 1 ≤ i ≤ n . Hence, hm( p ) is a non-negative power of t n and scale n ( p ) = n . Then ct ′ n is n -rigid by Definition 5.8.We have c = 0. By (5.4), we conclude that ℓ is zero, and so is p .The following proposition corresponds to Proposition 4.5. Proposition 5.12.
Let n ≥ and p be a t n -flat polynomial in K n − [ t n ] . If p ∈ K ′ n , then p = 0 .Proof. Since p ∈ K ′ n , we have lc t n ( p ) ≡ ct ′ n mod K ′ n − for some c ∈ C byLemma 2.2. Then p = 0 by Lemma 5.11.The next lemma corresponds to Lemma 4.6. Lemma 5.13.
For p ∈ K n − [ t n ] , there exists a t n -flat polynomial q ∈ K n − [ t n ] such that p ≡ q mod K ′ n . Moreover, deg t n ( q ) is no more than deg t n ( p ) .Proof. By Lemma 3.4, there exist p i ∈ K i [ t i +1 , . . . , t n ] for all i with 1 ≤ i ≤ n − p ∈ R n such that p ≡ n − X i =1 p i + p mod K ′ n . Moreover, hc( p i ) ∈ K i is t i -simple for all i ≥
1. By Proposition 5.6, there exists r ∈ R n with s = scale n ( r ) such that p ≡ r mod K ′ n and that hc( r ) is s -rigid.Set q to be P n − i =1 p i + r . Then q is t n -flat and p ≡ q mod K ′ n . Example 5.14.
Let p be given as in Example 5.9, in which k = 2 . By integra-tion by parts, we have p ≡ p + p + − t ′ t t + xt t | {z } q mod K ′ . Then scale ( q ) = 2 and hc( q ) = − t ′ = − / ( x + 2) , which is -rigid. Hence, p + p + q is t -flat. We are ready to present the main result of this section.
Theorem 5.15.
For f ∈ K n , the following assertions hold.(i) There exist a t n -simple element g ∈ K n and a t n -flat polynomial p ∈ K n − [ t n ] such that f ≡ g + p mod K ′ n . (5.5)14 ii) f ≡ K ′ n if and only if both g and p are zero.(iii) If f ≡ ˜ g + ˜ p mod K ′ n , where ˜ g ∈ K n is t n -simple and ˜ p ∈ K n − [ t n ] , then g = ˜ g and deg t n ( p ) ≤ deg t n (˜ p ) . Proof. (i) Applying Algorithm
HermiteReduce to f with respect to t n , we geta t n -simple element g of K n and an element h of K n − [ t n ] such that f ≡ g + h mod K ′ n . We can replace h with a t n -flat polynomial p by Lemma 5.13.(ii) Assume f ∈ K ′ n . Then (5.5) becomes g + p ≡ K ′ n . Applying themap hp t n to the above congruence yields g = 0 by Lemma 2.1. Thus, p ≡ K ′ n . Consequently, p = 0 by Proposition 5.12.(iii) Since ( g − ˜ g )+ ( p − ˜ p ) ≡ K ′ n and g − ˜ g is t n -simple, we have g = ˜ g by Lemma 2.1. So p − ˜ p ≡ K ′ n . By Lemma 2.2, we have lc t n ( p − ˜ p ) ≡ ct ′ n mod K ′ n − for some c ∈ C . If deg t n (˜ p ) is smaller than deg t n ( p ), thenlc t n ( p ) = lc t n ( p − ˜ p ) ≡ ct ′ n mod K ′ n − . By Lemma 5.11, we conclude p = 0, acontradiction. In this section, we study elementary integrability of elements in a straight orflat tower by Theorems 4.8 and 5.15.
Theorem 6.1.
Let the tower be given in (3.1) , in which C is algebraicallyclosed, K = C ( t ) and t i be a C -linear combination of logarithmic monomialsover K i − for all i with ≤ i ≤ n . Assume that, for f ∈ K n , f ≡ g + p mod K ′ n , (6.1) where g and p are described in (4.3) if (3.1) is straight, and in (5.5) if (3.1) isflat, respectively. Then f is elementarily integrable over K n if and only if g + p is a C -linear combination of logarithmic derivatives in K n .Proof. We denote by L i the C -linear subspace spanned by the logarithmicderivatives in K i for all i with 0 ≤ i ≤ n .Clearly, f is elementarily integrable over K n if g + p ∈ L n .Conversely, there exists r ∈ L n such that f ≡ r mod K ′ n by Liouville’stheorem ([8, Theorem 5.5.1]). By (6.1), g + p ≡ r mod K ′ n . (6.2)Note that hp t n ( g + p ) = g , as g is t n -simple. So g = hp t n ( r ) by (6.2) andLemma 2.1. Hence, g ∈ L n by Lemma 2.4. Set ˜ r to be r − hp t n ( r ). Then˜ r ∈ L n − by Lemma 2.4. Moreover, (6.2) becomes p ≡ ˜ r mod K ′ n . (6.3)We show that (6.2) implies g + p ∈ L n by induction. If n = 0, then p is zero.The assertion holds. Assume that the assertion holds for n −
1. Let d = deg t n ( p )and ℓ = lc t n ( p ). 15 ase 1 . d >
0. Then ℓ = lc t n ( p − ˜ r ), which, together with (6.3) and Lemma 2.2,implies that ℓ ≡ ct ′ n mod K ′ n − for some c ∈ C . Then ℓ is equal to 0 byLemma 4.3 in the straight case and by Lemma 5.11 in the flat case, a contra-diction. Case 2 . d = 0. Then ℓ = p . So ℓ ≡ ˜ r + ct ′ n mod K ′ n − for some c ∈ C by (6.3)and Lemma 2.2. Consequently, ℓ is elementarily integrable over K n − , because˜ r, t ′ n ∈ L n − . Moreover, the above congruence can be rewritten asfp t n − ( ℓ ) + pp t n − ( ℓ ) ≡ ˜ r + ct ′ n mod K ′ n − (6.4)Note that fp t n − ( ℓ ) is t n − -simple and pp t n − ( ℓ ) is t n − -straight (resp. flat) byDefinition 4.4 (resp. Lemma 5.10). By (6.4) and the induction hypothesis, wesee that ℓ ∈ L n − . Therefore, p belongs to L n − . Accordingly, g + p ∈ L n .Determining whether an element r in K n is a C -linear combination of loga-rithmic derivatives amounts to computing partial fraction decompositions andRothstein–Trager resultants by Theorem 4.4.3 in [8]. So elementary integrabil-ity in straight and flat towers can be checked by merely algebraic computationwhenever a decomposition in the form (6.1) is available. Example 6.2.
Let K , t and t be given as in Example 5.7. We compute anadditive decomposition for f = 1 xt + 1 xt + t + t t + 2 x t t + 2 x + 1 t + 1 x + 2 . By Theorem 5.15 and Example 5.7, we have f = a ′ + 1 xt + t | {z } g + 1 xt + 1 x + 2 | {z } p , where a = ( x + 1) t t − xt t − xt + (2 x + 2) t + 4 xt − x . As the Rothstein–Trager resultant of each fraction in g + p has only constant roots, g + p is a C -linear combination of logarithmic derivatives in K . So f is elementarilyintegrable over K by Theorem 6.1. Indeed, Z f dx = a + log( t ) + log( t ) + log( x + 2) . The problem of creative telescoping is classically formulated for D -finite func-tions in terms of linear differential operators [2, 31]. Raab in his thesis [21] hasstudied the telescoping problem viewed as a special case of the parametric inte-gration problem in differential fields. However, there are no theoretical resultsconcerning the existence of telescopers for elementary functions. To be more16recise, let F be a differential field with two derivations D x and D y that com-mute with each other and let F ∂ be the set { f ∈ F | ∂ ( f ) = 0 } for ∂ ∈ { D x , D y } .For a given element f ∈ F , the telescoping problem asks whether there existsa nonzero linear differential operator L = P di =0 ℓ i D ix with ℓ i ∈ F D y such that L ( f ) = D y ( g ) for some g in a specific differential extension E of F . We call L a telescoper for f and g the corresponding certificate for L in E . Usually, wetake E to be the field F itself or an elementary extension of F . In contrast to D -finite functions, telescopers may not exist for elementary functions as shownin the following example. Example 7.1.
Let F = C ( x, y ) and E = F ( t , t ) be a differential field exten-sion of F with t = log( x + y ) and t = log(1 + t ) . We first show that f = 1 /t ∈ F ( t ) has no telescoper with certificate in anyelementary extension of F ( t ) . Since t is a primitive monomial over F , wehave F D y = C ( x ) . We claim that for any i ∈ N , D ix ( f ) can be decomposed as D ix ( f ) = D y ( g i ) + a i t , where g i ∈ F ( t ) , and a i ∈ F satisfies the recurrence relation a i +1 = D x ( a i ) − D y (cid:18) xa i y (cid:19) with a = 1 .For n = 0 , the claim holds by taking g = 0 . Assume that the claim holds forall i < k . Applying the induction hypothesis and Algorithm HermiteReduce to D kx ( f ) yields D kx ( f ) = D x ( D k − x ( f )) = D x (cid:18) D y ( g k − ) + a k − t (cid:19) = D y (cid:18) D x ( g k − ) + a k − xyt (cid:19) + D x ( a k − ) − D y ( xa k − y ) t . This completes the induction. A straightforward calculation shows that a i = A i /y i for some A i ∈ C [ x, y ] \ { } with deg y ( A i ) < i . Using the notion ofresidues in [8, page 118], we have residue t (cid:18) a i t (cid:19) = a i D y ( t ) = ( x + y ) A i y i +1 , which is not in C ( x ) . Then D ix ( f ) is not elementarily integrable over F ( t ) for any i ∈ N by the residue criterion in [8, Theorem 5.6.1]. Assume that f has a telescoper L := P di =0 ℓ i D ix with ℓ i ∈ C ( x ) not all zero. Then L ( f ) iselementarily integrable over F ( t ) . However, L ( f ) = D y d X i =0 ℓ i g i ! + P di =0 ℓ i a i t . ince all of the ℓ i ’s are in C ( x ) and gcd( x + y , y m ) = 1 for any m ∈ N , theresidue of P di =0 ℓ i a i /t is not in C ( x ) , which implies that L ( f ) is not elemen-tarily integrable over F ( t ) , a contradiction.We now show that p = f t + 1 ∈ F ( t )[ t ] has no telescoper with certificatein any elementary extension of F ( t , t ) . Since t is also a primitive monomialover F ( t ) , we have E D y = C ( x ) . Assume that L := P di =0 ℓ i D ix with ℓ i ∈ C ( x ) not all zero is a telescoper for p . Then L ( p ) is elementarily integrable over E .By a direct calculation, we get L ( p ) = L ( f ) t + r with r ∈ F ( t ) . The elementaryintegrability of L ( p ) implies that L ( f ) = cD y ( t ) + D y ( b ) for some c ∈ C ( x ) and b ∈ F ( t ) by the formula (5.13) in the proof of Theorem 5.8.1 in [8, page 157].We claim that c = 0 . Since D ix ( f ) = u i /t i +11 with u i ∈ F [ t ] and deg t ( u i )
Let F = C ( x, y ) and E = F ( t ) be a differential field extensionof F with t = log( x + y ) . Consider the function f = t + 1 − y ( x + y ) t . Sincethe derivatives D ix (1 /t ) = a i /t i +2 with a i ∈ F \ { } are linearly independentover F , we see that /t is not D -finite over F , and nether is f . Note that f can be decomposed as f = D y (1 /t ) + t + 1 . Since t + 1 is D-finite, it has a telescoper, and so does f . In this paper, we developed additive decompositions in straight and flat towers,which enable us to determine in-field and elementary integrability in a straight-forward manner. It is natural to ask whether one can develop an additivedecomposition in a general primitive tower. Moreover, we plan to investigateabout the existence and the construction of telescopers for elementary functionsusing additive decompositions.
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