Hermite Reduction for D-finite Functions via Integral Bases

Trager's Hermite reduction solves the integration problem for algebraic functions via integral bases. A generalization of this algorithm to D-finite functions has so far been limited to the Fuchsian case. In the present paper, we remove this restriction and propose a reduction algorithm based on integral bases that is applicable to arbitrary D-finite functions.


Introduction
Let R be a certain class of functions in one variable x with the derivation D x .For example, R can be the field of rational functions or algebraic funtions.In the context of symbolic integration, the integrability problem consists in deciding whether a given element f ∈ R is of the form f = D x (g) for some g ∈ R. If such a g exists, we say that f is integrable in R. A relaxed form of the integrability problem is the decomposition problem, which consists in constructing for a given f ∈ R elements g, r ∈ R such that f = D x (g) + r and r is minimal in a certain sense.Ideally the "certain sense" should be such that r = 0 whenever f is integrable.If f ∈ R depends on a second variable t, one can also consider the creative telescoping problem: given an element f ∈ R, the task is to construct c 0 , . . ., c r ∈ R, not all zero, such that c i is free of x for all i ∈ {0, . . ., r} and c r D r t (f ) + • • • + c 0 f = D x (g) for some g ∈ R.
The operator L = c r D r t + • • • + c 0 , if it exists, is called a telescoper for f , and g is called a certificate for L.
Zeilberger first showed the existence of telescopers for D-finite functions [22].Almkvist and Zeilberger [4] solved the integrability problem and the creative telscoping problem for hyperexponential functions.Using the adjoint Ore algebra, Abramov and van Hoeij [2] solved the integrability problem for D-finite functions.Chyzak [12] extended the method of creative telescoping from hyperexponential functions to general D-finite functions.During the past ten years, a reduction-based telescoping approach has become popular, which can find a telescoper without computing the corresponding certificate.This improves the efficiency of telescoping algorithms because certificates tend to have much larger size than telescopers and sometimes are not needed.This approach was first formulated for rational funtions [5] and then extended to hyperexponential functions [6], algebraic functions [10], Fuchsian D-finite functions [11], and D-finite functions [21,7].The reduction-based telescoping algorithms for algebraic functions and for Fuchsian D-finite functions employ the notion of integral bases, while the known reduction-based telescoping algorithms applicable to arbitrary D-finite functions work differently.
The notion of integrality proposed by Kauers and Koutschan [17] for Fuchsian D-finite functions has recently been generalized by Aldossari [3] to arbitrary D-finite functions, so that the question arises whether there is also a reductionbased telescoping algorithm for arbitrary D-finite functions based on integral bases.The purpose of the present paper is to answer this question affirmatively.This paper is based on the results of Chapter 6 of the second author's Ph.D. thesis [13].
First we recall integral bases for D-finite functions [17,3] in Section 2. Then we extend Hermite reduction for the Fuchsian case to the non-Fuchsian case in Section 3, which reduces the pole orders of D-finite functions at finite places.Instead of using polynomial reduction [6,10,11,8] to reduce the pole order at infinity, we present a Hermite reduction at infinity.Combining Hermite reduction at finite places and at infinity, we are able to determine the integrability of D-finite functions in Section 4 and present a reduction-based telescoping algorithm for D-finite functions in Section 5.

Integral Bases
Below we recall the value functions and integral bases for arbitrary linear differential operators [17,14,15,3].Let C be a field of characteristic zero and C be the algebraic closure of C. Let C(x)[D] be an Ore algebra, where D is the differentiation with respect to x and satisfies the commutation rule Dx = xD +1.For an operator The elements of A are called D-finite functions.When there is no ambiguity, an equivalence class P + L in A is also denoted by P .Every element of A can be uniquely represented by For each α ∈ C, an operator L of order n admits n linearly independent solutions of the form (1) . Such objects are called generalized series solutions at α, see [16,18].For α = ∞, the operator L admits n linearly independent solutions of the form for some x if α = ∞).The valuation ν α (t) of a term t := z r exp(p(z −1/s )) log(z) ℓ is the real part of the local exponent r.The valuation ν α (f ) of a nonzero generalized series f at α is the minimum of the valuations of all the terms appearing in f (with nonzero coefficients).The valuation of 0 is defined as ∞.A generalized series f at α is called integral if ν α (f ) ≥ 0. A non-integral series f is said to have a pole at the reference point and its pole order at α is defined as −ν α (f ).Note that in this terminology, it may happen that ν The valuation of a series only depends on its local exponent and not on its exponential part.This valuation is the same as in [3,Definition 5.4].
For each α ∈ C ∪ {∞}, an operator where ′ is the derivation with respect to x.Let f 1 , . . ., f n be a basis of Sol α (L) as in the form of (1) (or (2) Then val α (P ) is the minimum valuation of all series P • f at α, where f runs through all series solutions in Sol α (L).So this definition of value functions is independent of the choice of the basis of Sol α (L).An element P ∈ A is called (locally) integral at α ∈ C ∪ {∞} if val α (P ) ≥ 0. If P is not locally integral at α, then P is said to have a pole at α and its pole order at α is defined as − val α (P ).An element P ∈ A is called (globally) integral if val α (P ) ≥ 0 for all α ∈ C, i.e., P is locally integral at all finite places.When L is Fuchsian, this notion of integrality falls back to the Fuchsian case discussed in [11,17].
The set of all globally integral elements f ∈ A = C(x)[D]/ L forms a C[x]module.A basis of this module is called a (global) integral basis for A. Such bases exist and the algorithm for computing integral bases in the Fuchsian case [17] applies to the setting of the non-Fuchsian case literally.More properties can be found in [3].
For a fixed α ∈ C, the valuation ν α of a nonzero rational function For each α ∈ C ∪ {∞}, the valuation ν α of a rational function is the same as the valuation of its Laurent series expansion at α.The set C(x) α = {f ∈ C(x) | ν α (f ) ≥ 0} forms a subring of C(x).The set of all elements f ∈ A that are locally integral at some fixed α ∈ C ∪ {∞} forms a C(x) α -module.A basis of this module is called a local integral basis at α of A. Such a basis can also be computed [17,3].
An integral basis {ω 1 , . . ., ω n } is always a C(x)-vector space basis of A. A key feature of integral bases is that they make poles explicit.Writing an element f ∈ A as a combination f = n i=1 f i ω i for some f i ∈ C(x), we have that f has a pole at α ∈ C if and only if at least one of the f i has a pole there.Furthermore, ⌊val α (f )⌋ is a lower bound for the valuations of all the f i 's at α. Proof.
(1): The direction "⇐" is obvious.To show "⇒", suppose that f is integral at α. Then there exist f1 , . . ., fn ∈ C(x) α such that f = n i=1 fi ω i .Thus n i=1 ( fi − f i )ω i = 0, and then fi = f i for all i, because {ω 1 , . . ., ω n } is a C(x)-vector space of A. As elements of C(x) α , the f i 's are integral at α. Hence the f i ω i 's are integral at α.
Let W = (ω 1 , . . ., ω n ) be a vector space basis of A over C(x).For f ∈ A, denote its derivative D •f by f ′ .Let e ∈ C[x] be and M = (m i,j ) n i,j=1 ∈ C[x] n×n be such that eW ′ = M W and gcd(e, m 1,1 , m 1,2 , . . ., m n,n ) = 1.If W is an integral basis and L is Fuchsian at all finite places, then e must be squarefree, see [11,Lemma 3].If W is a local integral basis at infinity and L is Fuchsian at infinity, then deg x (m i,j ) < deg x (e) for all i, j, see [11,Lemma 4].However, these two facts are no longer true in the non-Fuchsian case, as the following examples show: has only one singular point 0 in C, which is an irregular singular point.Infinity is a regular singular point.So L is Fuchsian at all points in C ∪ {∞} except 0 and this implies that L is a non-Fuchsian operator.At the point 0, there are two linearly independent solutions y 1 (x) = 1 and y 2 (x) = exp(x −2 ) in Sol 0 (L).An integral basis for A = C(x)[D]/ L is given by ω 1 = 1 and ω 2 = x 3 D, which is also a local integral basis at infinity.Then with e = x 3 .In this example, e is not squarefree.
Infinity is an irregular singular point.So L is not Fuchsian at infinity.There are two linearly independent solutions y 1 (x) = 1 and y In this example, e = 1 and the condition deg x (m i,j ) < deg x (e) fails.
is a local integral basis at α.Given an integral basis and a local integral basis at infinity, Trager [20] presented an algorithm for computing an integral basis that is normal at infinity in the algebraic function field.The same procedure also applies in the present situation, see [3,Algorithm 5.20].

Hermite Reduction
Hermite reduction, first introduced by Ostrogradsky in 1845 [19], is a classical symbolic integration technique that reduces rational functions to integrands with only simple poles.Hermite reduction was extended by Trager [20] from the field of rational functions to that of algebraic functions via integral bases.Trager's Hermite reduction solved the integration problem for algebraic functions.This work was extended to the case of Fuchsian D-finite functions [11].We shall further extend Hermite reduction to general D-finite functions, including the non-Fuchsian case.To reduce the pole order at infinity, we develop a Hermite reduction at infinity for D-finite functions, which plays the same role as polynomial reduction [6,10,11,8].In this section, Hermite reduction at finite places and at infinity are formulated in the same framework.More precisely, we shall use a local integral basis at α ∈ C ∪ {∞} to reduce the pole orders of D-finite functions at α.
For convenience, we introduce some notations for the valuations of a matrix with rational coefficients.For each α ∈ C ∪ {∞}, the valuation of a matrix T ∈ C(x) n×n at α, denoted by ν α (T ), is defined as the minimal valuation at α of all entries in this matrix.The degree of T ∈ C(x) n×n , denoted by deg x (T ), is defined as −ν ∞ (T ).In particular, the degree of a rational function

The Local Case
Let L ∈ C(x)[D] be of order n and let A = C(x)[D]/ L .For an arbitrary but fixed point α ∈ C ∪ {∞}, let W = (ω 1 , . . ., ω n ) be a local integral basis at α of A and then there exists a matrix T ∈ C(x) n×n such that W ′ = T W .We write z = x − α (or z = 1 x if α = ∞).Let λ = −ν α (T ) be the pole order of T at α. Then λ ∈ Z and there exists a matrix M = (m i,j ) n i,j=1 ∈ C(x) n×n α such that where After differentiating both sides of (4) and multiplying by z d , we get 6) can be reduced modulo z: It follows that b i ≡ µ(d + µ) −1 a i mod z is the unique solution of (7) in C(x) α / z .If λ ≥ −µ, then multiplying (5) by z λ+µ and reducing this equation modulo Let ψ i := z d+λ+µ ωi z d+µ ′ for i = 1, . . ., n.To perform Hermite reduction, we have to show that (8) always has a solution (b 1 , . . ., b n ) in C(x) α / z λ+µ+1 n .In the Fuchsian case, Chen et al. [11] proved that λ = 1.When α ∈ C, they showed that {ψ 1 , . . ., ψ n } forms a local integral basis at α and hence (8) has a solution and that solution is unique.When α = ∞, instead of solving the modular system (8), they introduced the polynomial reduction to reduce the degree in x.We shall show that {ψ 1 , . . ., ψ n } still forms a local integral basis at infinity.Then the polynomial reduction can be formulated as Hermite reduction at infinity, as suspected by one of the anonymous referees of [11].
In the non-Fuchsian case, it may happen that λ > 1, see Examples 2 (for α = 0) and 3 (for α = ∞).Another difference is that {ψ 1 , . . ., ψ n } may not be a local integral basis at α anymore, see the following Example 4. Fortunately, the linear system (8) still has a unique solution in C(x) α / z λ+µ+1 n as we shall prove in this section.There are two steps.First we show that {ψ 1 , . . ., ψ n } is linearly independent over C(x) and then we find a rational solution (b 1 , . . ., b n ) whose entries admit nonnegative valuation at α.So the b i 's belong to C(x) α .Taking b i modulo z λ+µ+1 gives a unique solution of (8).Once we know that (8) has a solution, equating the coefficients of the ω i 's on both sides, we can find its solution b = (b 1 , . . ., b n ) by solving the following linear system of congruence equations: where I n is the identity matrix in C[x] n×n .
Example 4. Continue Example 2. For α = 0 and λ = 3, let In this example, ψ 1 , ψ 2 are integral elements but do not form a local integral basis at 0, because 1 When d > 1, even though the coefficient matrix is not invertible over C(x) 0 / x 3 , this equation still has a unique solution Let O α denote the set of all elements in C(x)[D]/ L that are locally integral at α ∈ C ∪ {∞}.Even though {ψ 1 , . . ., ψ n } may not be a local integral basis at α, it is not so far away.In Example 4, we have C(x) In general, if we represent a locally integral element at α as a linear combination of {ψ 1 , . . ., ψ n } with coefficients in C(x), the pole orders at α of these coefficients are at most λ + µ.
Proof.We prove this proposition using the same technique as in [10, Lemma 10].To show n i=1 C(x) α ψ i ⊆ O α , we only need to show that for every i = 1, . . ., n, the element ψ i is integral at α.After differentiating, we get Suppose to the contrary that there exists an element f ∈ O α \ 1 z λ+µ n i=1 C(x) α ψ i .Furthermore, we can find such an element f of the form Then also their sum ′ must be integral, where h = n i=1 c i ω i .Since {ω 1 , . . ., ω n } is an integral basis at α, by Lemma 1 we have 0 ≤ val α (h) < 1.There exists a generalized series solution . For this fixed series y i , let T be the dominant term of h • y i , i.e., among all terms with minimal r the one with the largest exponent ℓ.Let k = deg x (p) and c = lc x (p) be the degree and the leading coefficient of p in x respectively.Then where "• • • " denotes some terms of valuation higher than r The above calculation reveals that z d−1 z −d−µ h ′ = f + g is not integral at α, which contradicts our assumption on the integrality of f .Hence O α ⊆ Proof.By Proposition 5, the C(x) α -module generated by 1 z λ+µ ψ 1 , . . ., 1 z λ+µ ψ n contains a submodule O α of rank n.So {ψ 1 , . . ., ψ n } is linearly independent over C(x).Then there exist unique t 1 , . . ., t n ∈ C(x) such that n i=1 z λ+µ a i ω i = n i=1 t i ψ i .To find a solution b i , we have to show that t i ∈ C(x) α for all i = 1, . . ., n.If so, b i ≡ t i mod z λ+µ+1 is the unique solution of (12).Since a i ∈ C(x) α and the ω i 's are integral at α, the element Since {ψ 1 , . . ., ψ n } are linearly independent over C(x), we have t i ∈ C(x) α for all i.Thus t i ∈ C(x) α as claimed.
According to Theorem 6, we can perform one step of Hermite reduction for D-finite functions as described in the beginning of this section.The element b i in C(x) α / z λ+µ+1 is of the form So in Equation ( 4), for α ∈ C, if d > max{1, λ}, then we can guarantee that the coefficients of 1 x 4 and use Hermite reduction at 0 to reduce the power of x in its denominator.So we start with z = x, µ = −1, d = 4, a 1 = −2x 2 − x 4 and a 2 = −2 + 3x 2 − 3x 4 .From (9), we get By Theorem 6, we know that this equation has a unique solution.Indeed, we find a solution b 1 = 2 3 x 2 , b 2 = 4 3 x 2 .Then one step of the Hermite reduction at 0 simplifies f to with λ = 2. Consider the D-finite function and use Hermite reduction at infinity to reduce its degree in x.So we start with z = 1 x , µ = 1, d = 3, a 1 = 4 and a 2 = 1 x 2 .From (9), we get This coefficient matrix is not invertible over C(x) ∞ / x −4 .However, by Theorem 6, we know this equation has a unique solution.Indeed, we find a solution Then one step of the Hermite reduction at infinity simplifies f to For a rational function g ∈ C(x) and any point α ∈ C, if ν α (g) = 0, we have ν α (g ′ ) = ν α (g) − 1.So the pole order of a rational function increases by exactly one under each derivation.In the D-finite case, the pole order increases by at least one.A lower bound of its pole order under each derivation is given in [3,Lemma 5.7].
Proof.Let y i be a generalized series solution in Sol α (L) such that val α be the dominant term of f • y i , i.e., among all terms with minimal r the one with the largest exponent ℓ.Let k = deg x (p) and c = lc x (p).Then Note that r − k s + µ ≤ r + µ and r = 0 (by the assumption val α (g) = 0).So the valuation of the term D • T in g ′ • y i is less than or equal to r + µ, which implies that val α (g ′ ) ≤ val α (g) + µ.
Let W = {ω 1 , . . ., ω n } be a local integral basis at infinity.Let λ ∈ Z and M ∈ C(x) n×n be such that . By repeating the reduction at infinity, we can reduce the degree in x as far as possible and decompose f ∈ A as where g ∈ A, h i ∈ C(x) with deg x (h i ) < max{0, λ} for all i and the coefficients of g are polynomials.The following lemma derives an upper bound for the degree of any hypothetical integral of h in A.

The Global Case
To avoid algebraic extensions of the base field, Hermite reduction can be performed simultaneously at all roots of some squarefree polynomial.Let W = {ω 1 , . . ., ω n } be an integral basis of n×n be such that eW ′ = M W and gcd(e, m 1,1 , m 1,2 , . . ., m n,n ) = 1.Let v be a nontrivial squarefree polynomial and λ ∈ N be an integer such that v λ | e and gcd( If λ = 0, then gcd(e, v) = 1.Multiplying ( 15) by uv d and reducing this equation modulo v yield 15) by uv d+λ−1 and reducing this equation modulo One can adapt the argument in the local case to show that ( 17) always has a unqiue solution (b As an analog of Proposition 5, for each root α ∈ C of v, we have Thus the linear system Equating the coefficients of the ω i 's on both sides of (17), the vector b = (b 1 , . . ., b n ) can be found by solving the following linear system of congruence equations: where I n is the identity matrix in C[x] n×n .By repeated application of the above Hermite reduction step, we can reduce the pole orders at finite places as far as possible, i.e., we can decompose any f ∈ A as where g ∈ A, h 1 , . . ., h n , d ∈ C[x], d is squarefree, gcd(d, e) = 1 and the coefficients of g are proper rational functions.
Lemma 11.Let h ∈ A be as in (20).
, where q 1 , . . ., q n , u ∈ C[x] and u | gcd(e, e ′ ).Furthermore, we have If α ∈ C is not a root of e, then b i has no pole at α for all i.Otherwise, suppose b i has a pole at α for some i ∈ {1, . . ., n}.Then H has a pole at α, because {ω 1 , . . ., ω n } is an integral basis.But then by Lemma 9, h would have a pole of order greater than 1, which is impossible because gcd(d, e) = 1 and d is squarefree.Therefore b i is integral at α for all i and hence d is a constant.
Note that ν α (gcd(e, e ′ )) = ν α (e) − 1 if α is a root of e and ν α (gcd(e, e ′ )) = 0 if α is not a root of e.So u is a common multiple of the denominators of the b i 's.
be the same differential operator as in Examples 2 and 4. Since e = x 3 , we have u = gcd(e, e ′ ) = x 2 .We could find a Hermite remainder such that the denominator of its integral is u, for example,

Additive Decompositions
Now we combine the Hermite reduction at finite places and at infinity to decompose a D-finite function f as f = g ′ + h such that f is integrable if and only if the remainder h is zero.To achieve this goal, first we confine all remainders into a finite-dimensional vector space and then find all possible integrable functions in this vector space.This procedure is similar to the hyperexponential case [6], the algebraic case [10], the Fuchsian case [11] and the D-finite case [21,7].It provides an alternative method for solving the accurate integration problem for D-finite functions [2].Since there may not exist a basis of A = C(x)[D]/ L that is a local integral basis at all α ∈ C ∪ {∞}, we need two bases to perform Hermite reduction at finite places and at infinity, respectively.Let W = (ω 1 , . . ., ω n ) ∈ A n be an integral basis of A that is normal at infinity.There exists T = diag x τ1 , . . ., x τn ∈ C(x) n×n with τ i ∈ Z such that V := T W is a local integral basis at infinity.Let e, a ∈ C[x] and M, B ∈ C[x] n×n be such that eW ′ = M W and aV ′ = BV .

Since the derivative of
e T M T −1 V, we may assume that a = x λ e for some λ ∈ N.For any integers µ, δ ∈ Z with µ ≤ δ, define a subspace of Laurent polynomials in C[x, x −1 ] as follows: Theorem 13.Let W, V ∈ A n be as described above.Then any element f ∈ A can be decomposed into Proof.Let h ∈ A be a Hermite remainder as in (20).By the extended Euclidean algorithm, we compute r i , s i ∈ C[x] such that h i = r i e + s i d and deg x (r i ) < deg x (d).Then h decomposes as Writing h in vector form, by (20) we decompose f ∈ A as where g In the next step, we shall reduce the degree of S and confine S to a finite-dimensional vector space over C that is independent of f .We rewrite the last summand in (23) with respect to the new basis V : where S = x λ ST −1 ∈ x µ C[x] n with µ = min{−τ 1 , . . ., −τ n , 0}.Since V is a local integral basis at infinity, using Hermite reduction at infinity in Section 3.1, we obtain from (14) that where This implies that deg For the last assertion, assume that f is integrable (the other direction of the equivalence holds trivially).Then Lemma 11 implies that d ∈ C, and therefore R must be zero because deg x (R) < deg x (d).Hence the last summand in (21) and the left hand side of (25) are also integrable.We want to find its integral by estimating the valuation of this integral at all points in C ∪ {∞}.Since W is a global integral basis, using Lemma 11 again, we know where b ∈ C[x] n and u = gcd(e, e ′ ).Then Finally we have c ∈ C[x] n µ ′ ,δ ′ .The remaining step is to reduce all integrable D-finite functions to zero.Note that in Theorem 13, U is a C-vector space of dimension n(δ ′ − µ ′ + 1) with a basis x j v i u i = 1, . . ., n; j = µ ′ , . . ., δ ′ , where V = (v 1 , . . ., v n ).Let K = 1 x λ e bV b ∈ C[x] n µ,δ .Differentiating all elements in the basis of U and using Gauss elimination, we can find a basis of U ′ and decompose K = (U ′ ∩ K) ⊕ N V as a direct sum of two subspaces, where N V is a complement of U ′ ∩ K in K.This means f in (21) can be further decomposed as where g = g + g 1 with g 21) is called an additive decomposition of f with respect to x.When L is a Fuchsian operator, the additive decomposition of f was obtained in [11,Theorem 23].
In practice, we may choose a fixed complement of K ∩ U ′ in K. To do this, we define a term over position order on the set From now on, let N V denote the standard complement of K ∩ U ′ in K.This definition of N V is essentially the same as in [11], because there is a bijection from a D-finite function to its coefficients of the v i 's.
Note that Q 2 belongs to a C-vector space C where τ = max{τ 1 , .
and hence U ′ is generated by The leading terms of all elements in K ∩U ′ are ω 1 or xω 1 .Since lt((x− 4 9 )ω 2 ) = xω 2 is different from all these terms, by Theorem 13 we know f is not integrable in A = C(x)[D]/ L and (29) is an additive decomposition of f with respect to x.

Example 15. Let
Combining Hermite reduction at all finite places in Example 7 and Hermite reduction at infinity, we get

Applications
Let K(x)[∂ t , D x ] with K = C(t) be an Ore algebra, in which D x is the differentiation with respect to x and ∂ t is either the differentiation with respect to t or the shift t → t + 1.Let I be a left ideal of K(x)[∂ t , D x ] generated by L and ∂ t − u t with L, u t ∈ K(x)[D x ].The quotient A = K(x)[∂ t , D x ]/I is a finite-dimensional vector space over K(x) and a basis is given by {1, D x , . . ., D n−1 x }, where n is the order of L. Every element f in A can be uniquely written as P f + I for some P f ∈ K(x)[D x ].The map sending f to P f + L gives an isomorphism from A to K(x)[D x ]/ L as a K(x)[D x ]-module.Using this isomorphism, for any f ∈ A, we can apply our additive decomposition to test whether f is integrable (in x).If f ∈ A is not integrable, one can ask to find a nonzero operator T ∈ C(t)[∂ t ] (free of x) such that T (f ) is integrable.Such an operator T if it exists is called a telescoper for f .Applying the additive decomposition with respect to x in Section 4 to ∂ i t • f ∈ A yields that where g i , h i ∈ A, and ∂ i t • f is integrable in A if and only if h i = 0.If there exist c 0 , c 1 , . . ., c r ∈ K such that r i=0 c i h i = 0, then T = r i=0 c i ∂ i t is a telescoper for f (because ∂ t and D x commute).Such a telescoper if it exists is of minimal order.This approach is the method of reduction-based telescoping and was developed for various classes of functions [5,6,9,10,11,7,21].If ∂ t = D t is the differentiation with respect to t, then telescopers always exist [22].We implemented our algorithm in Maple.More examples and our code are available in [1].Similar to the Fuchsian case [11,Lemma 24], for any i ∈ N, the derivative D i t • f has an additive decomposition (27) of the form where  Next we consider the derivative D t • f which has an additive decomposition  Now we can find a telescoper S n − 2n − 1.This was obtained by the algorithm in [7].
be the set of all finite C-linear combination of generalized series solutions of L at α. Then Sol α (L) is a C-vector space of dimension n.Throughout the paper, we assume that for each α ∈ C ∪ {∞}, all series of Sol α (L) have p ∈ C[x], µ ∈ C and b ∈ C[[x]][y] (this can always be achieved by a suitable choice of C).If all series of Sol α (L) have p = 0 and s = 1, then L is called Fuchsian at α.The operator L is simply called Fuchsian if it is Fuchsian at all α ∈ C ∪ {∞}.In this case, the elements of A are called Fuchsian D-finite functions.For simplicity, we assume throughout that C is a subfield of C. Given two complex numbers a, b ∈ C, we say a ≥ b if and only if Re(a) ≥ Re(b).For each α ∈ C ∪ {∞}, let z = x − α (or z = 1

3
An integral basis ofA = K(x)[D x ]/ L with K = C(t) is ω = 1 and a local integral basis at infinity is v = x −1 ω.As the integrand H corresponds to 1 ∈ A, its representation in the bases is f = ω = xv.The additive decomposition of f is + 1)x − t 2t 4 x(2x − t) v.
with deg x (R i ) < deg x (d) and Q i ∈ N V .Then by (28) we obtain an upper bound for the order of telescopers, which is a generalization of [11, Corollary 25].Corollary 16.Every f ∈ A has a telescoper of order at most n deg x (d) + dim x (N V ), which is bounded by n(deg x (d) + max{τ, 0} + max{λ + deg x (e), deg x (B)}).