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SIAM REVIEW c© 2008 Society for Industrial and Applied MathematicsVol. 50, No. 2, pp. 275–293

Asymptotic Expansions ofMellin Convolution Integrals∗

Jose L. Lopez†

Abstract. We present a new method for deriving asymptotic expansions of∫∞0 f(t)h(xt)dt for small

x. We only require that f(t) and h(t) have asymptotic expansions at t = ∞ and t = 0,respectively. Remarkably, it is a very general technique that unifies a certain set of asymp-totic methods. Watson’s lemma and other classical methods, Mellin transform techniques,McClure and Wong’s distributional approach, and the method of analytic continuationturn out to be simple corollaries of this method. In addition, the most amazing thingabout it is that its mathematics are absolutely elemental and do not involve complicatedanalytical tools as the aforementioned methods do: it consists of simple “sums and sub-tractions.” Many known and new asymptotic expansions of important integral transformsare trivially derived from the approach presented here.

Key words. asymptotic expansions of integrals, Mellin convolution integrals, Mellin transforms

AMS subject classifications. 41A60, 30B40, 46F10

DOI. 10.1137/060653524

1. Introduction. The subject of asymptotics can be divided into two main areas.The first area is concerned with solutions of differential equations. The most famousbook where this topic is presented is that of Olver [20]. Olver introduces in his book a“universal” method to obtain asymptotic expansions of solutions of linear differentialequations of the second order, including error bounds. The second area deals withfunctions that are expressible in the form of definite integrals or contour integrals. Themost complete and modern book in this area is perhaps Wong’s [24]. Other excellentbooks are, for example, the classical Blestein and Handelsman [2] or the more recentbook of Paris and Kaminski [21], where a new perspective on asymptotics from Mellin–Barnes integrals is introduced. See also other reference books cited in [2], [21], and[24]. As distinct from the first area, in this second area we cannot speak about a“universal” method. On the contrary, many asymptotic methods have been designedto obtain asymptotic expansions of different kinds of integrals: Watson’s lemma forLaplace transforms, Laplace’s method, saddle point methods and steepest descentsfor contour integrals, summability methods for Fourier transforms, stationary phasemethods, Mellin transform techniques, distributional methods, analytic continuationmethods, etc. This variety of methods gives a certain ad hoc aspect to the asymptotictheory of integrals.

In the last few decades, some investigations have tried to unify the classical meth-ods for integrals by looking for a common root. The first idea was suggested in 1963

∗Received by the editors March 2, 2006; accepted for publication (in revised form) December 1,2006; published electronically May 5, 2008. This work was supported by the Direccion General deCiencia y Tecnologıa (MTM2004-05221).

http://www.siam.org/journals/sirev/50-2/65352.html†Departamento de Matematica e Informatica, Universidad Publica de Navarra, 31006-Pamplona,

Spain ([email protected]).275


276 JOSE L. LOPEZ

by Erdelyi and Wyman [4], [25]. In their work, they show that Darboux’s method,Watson’s lemma, steepest descents, and stationary phase can be viewed as particularcases of the method of Laplace. More recently, following the work of Wong [24], thisauthor suggested that Watson’s lemma and integration by parts should be consideredas “fundamental classical methods”: steepest descents, Laplace’s method, and Per-ron’s method, for example, are based on Watson’s lemma, whereas stationary phaseor summability methods, for example, are based on the integration by parts technique[13].

Despite the effort of those authors, it seems impossible to design a unique asymp-totic method valid for any kind of integral containing an asymptotic parameter x,∫

Γ f(x, t)dt. Nevertheless, a first step toward this goal can be taken: we show herethat a simple method is possible to bring a certain order and shed some light on the“unification” of asymptotic methods of integrals of the form

(1) I(x) ≡∫ ∞

0f(t)h(xt)dt.

The ideas developed in this paper may be generalized to complex x, but for the sakeof clarity, we restrict ourselves to positive values of x. Without loss of generality wecan think of x as a small parameter. (If x is large, perform the change of variablet → t/x and replace the roles of f and h in (1).) Many integral transforms can beput in the form (1): Laplace, Fourier, Stieltjes, Hankel, Poisson, Glasser, Lambert,and so on [26].

If we want to approximate (1) for small x, we may think that only the behaviorof h(t) near the origin is relevant. Then we require for h(t) an expansion at t = 0,

(2) h(t) =n−1∑k=0

bktk+β + hn(t),

substitute this expansion for h(xt) in (1), and interchange summation and integration.We obtain, formally, an asymptotic expansion for small x:

(3) I(x) =n−1∑k=0

[bk

∫ ∞0

f(t)tk+βdt

]xk+β +

∫ ∞0

f(t)hn(xt)dt.

In fact, classical methods such as Watson’s lemma, Laplace’s method, and saddlepoint techniques are based on this idea. This is the message established in [4], [13],and [25]. After some manipulations of the integrals to put them in the form (1) andsome hypotheses on f and h, those methods show that the above expansion is notonly formal, but a valid asymptotic expansion.

On the other hand, we may write (1) in a different form:

(4) I(x) ≡ x−1∫ ∞

0f

(t

x

)h(t)dt.

Written in this form, it seems plausible that only the behavior of f(t) at infinity isrelevant to approximate I(x) when x→ 0. Then we require for f(t) an expansion att =∞:

(5) f(t) =n−1∑k=0

aktk+α + fn(t), t→∞.


ASYMPTOTIC EXPANSIONS OF MELLIN CONVOLUTION INTEGRALS 277

Substituting this expansion in (4) and interchanging summation and integration weobtain the formal expansion

(6) I(x) =n−1∑k=0

[ak

∫ ∞0

t−k−αh(t)dt]xk+α−1 +

∫ ∞0

fn(t)h(xt)dt.

If the negative moments of h(t) exist, we can think of this formula as generating a newfamily of classical methods. Then, somehow, “the two natural and easy possibilities”to obtain asymptotic expansions of I(x) for small x are as follows:

(i) Try an expansion of h(t) at t = 0 if the positive moments of f(t) exist (classicalmethods I).

(ii) Try an expansion of f(t) at t = ∞ if the negative moments of h(t) exist(classical methods II).

From (3) and (6) we see that classical methods require the existence of either allthe positive moments of f(t) or all the negative moments of h(t). But what happensif f(t) does not converge fast enough to 0 when t→∞ and h(t) does not converge fastenough to 0 when t→ 0? Under these circumstances, the coefficients of the expansion(3) or (6) are not defined and the classical expansion makes no sense. Importantexamples of failure are Laplace transforms for small parameter [24, Chap. 6, sec. 5],Stieltjes transforms for large parameter [24, Chap. 6, sec. 2], the elliptic integrals[14], [15], the Epstein–Hubbel integral [8], the Appell function [10], and the Poissontransform for small parameter [12], among others.

McClure and Wong (M&W) solved this problem for certain families of functionsf(t) and h(t) by using the theory of distributions and analytic continuation techniques[18], [19], [24, Chaps. 5, 6], or using also convolutions of distributions [23], [24, Chap. 6,sec. 7]. Different and more general proofs using only analytic continuation (AC) wereproposed in [22], [17], and [16]. A different solution to this problem was proposedby Handlesman and Lew using the method of Mellin transforms (MTs) [5], [6], [7],[24, Chap. 3]. Their technique writes I(x) in the Mellin transformed space and usesanalytic continuation and the Cauchy residue theorem. M&W, AC, and MT are allmore difficult techniques than classical methods I or II.

Having reached this point, a little bit of simplification would be very welcome.This is exactly the main idea to be presented in this work: the appearance of diver-gences in cases (i) or (ii) above is an artificial problem. An unnecessary problem iscreated when expanding h or f in (2) or (5) up to n terms, substituting this expansionin (1), and interchanging sum and integral. And, if we do not create the problem,we will not have to solve it by using M&W, AC, MT, or any other repairing tool.So, do not expand only h or only f up to n terms. The idea is as simple as this:expand h and f simultaneously and substitute these expansions in (1) in such a waythat you do not create any divergence. The aim of this paper is to show that in factthis idea not only works, but it also generates an extraordinarily simple method whichcontains, as straightforward corollaries, classical methods I and II, M&W theory, theAC method, and MT techniques. It will be shown that this procedure works for afamily of functions f and h at least as large as the families considered in the classi-cal methods, M&W theory, MT techniques, or in the AC method. Moreover, all ofthe aforementioned methods can be seen as one unique method viewed from differentangles, the easiest angle being the one presented here, the easiest perspective comingby the hand of simplicity.

In the next section we give some definitions and technical results and MT tech-niques, the AC method, and M&W theory are briefly resumed. Section 3 presents the


278 JOSE L. LOPEZ

main result of the paper: a unified and simple method to obtain asymptotic expan-sions of I(x) for small x. In section 4, error bounds for the remainders are derived.Section 5 rederives some classical results, MT techniques, the AC method, and M&Wtheory as corollaries of the fundamental theorem of section 3. An example whichshows the applicability of the method is given in section 6. Section 7 contains someconclusions and final remarks.

2. Preliminaries. This section recalls very briefly the MT techniques, M&Wtheory, and the AC method. Some definitions and two important formulas are neededto formulate accurately the concepts mentioned in the introduction.

2.1. Definitions and Technical Results.Definition 1. We denote by F the set of functions f ∈ L1

Loc(0,∞) such that(i) f has an asymptotic expansion at infinity,

(7) f(t) =n−1∑k=0

aktαk

+ fn(t), n = 1, 2, 3, . . . ,

where, for k = 0, 1, 2, . . . , {ak} and {αk} are sequences of complex and real numbers,respectively, with αk strictly increasing and fn(t) = O(t−αn) as t→∞;

(ii) f(t) = O(t−a) as t→ 0+ with a ∈ R.Definition 2. We denote by H the set of functions h ∈ L1

Loc(0,∞) such that(i) h has an asymptotic expansion at t = 0+,

(8) h(t) =n−1∑k=0

bktβk + hn(t), n = 1, 2, 3, . . . ,

where, for k = 0, 1, 2, . . . , {bk} and {βk} are sequences of complex and real numbers,respectively, with βk strictly increasing and hn(t) = O(tβn) as t→ 0+;

(ii) h(t) = O(t−b) when t→∞ with b ∈ R.Definition 3. Let g ∈ L1

Loc(0,∞). We denote by M [g; z] the Mellin transform ofg,∫∞

0 tz−1g(t)dt (when this integral exists), or its analytic continuation as a functionof z.

The following remark and the two following formulas are proved in [16] for theparticular case αk = k + α, βk = k + β, α, β ∈ R. The proof in the general case is astraightforward generalization and is omitted here.

Remark 1. In the foregoing discussion we require the parameters a, b, α0, andβ0 to satisfy, without loss of generality, the following relations.

Condition I. a− β0 < 1 < b+ α0.Condition II. −β0 < b and a < α0.The Mellin transform M [f ; z] of every function f ∈ F exists and defines a mero-

morphic function of z in the half plane �z > a. More precisely, for any n ∈ N,

(9) M [f ; z] =

∫ ∞0

tz−1f(t)dt for a < �z < α0,

∫ 1

0tz−1f(t)dt−

n−1∑k=0

akz − αk

+∫ ∞

1tz−1fn(t)dt for a < �z < αn,

∫ ∞0

tz−1fn(t)dt for αn−1 < �z < αn.



(a) (b)

Fig. 1 The different strips of analyticity of the different representations of the Mellin transformsM [f ; z] and M [h; z] given in (9) and (10), respectively. In (a), the strips (i), (ii), and (iii)are the regions of analyticity of the respective lines in the right-hand side of (9). In (b), thestrips (i), (ii), and (iii) are the regions of analyticity of the respective lines in the right-handside of (10).

Observe that M [f ; z] has simple poles at the points z = αk, k = 0, 1, 2, . . . , withresidues −ak (see Figure 1(a)).

The Mellin transform M [h; z] of every function h ∈ H exists and defines a mero-morphic function of z in the half plane �z < b. More precisely, for any m ∈ N,(10)

M [h; z] =

∫ ∞0

tz−1h(t)dt for − β0 < �z < b,

∫ 1

0tz−1hm(t)dt+

m−1∑k=0

bkz + βk

+∫ ∞

1tz−1h(t)dt for − βm < �z < b,

∫ ∞0

tz−1hm(t)dt for − βm < �z < −βm−1.

Observe that M [h; z] has simple poles at the points z = −βk, k = 0, 1, 2, . . . , withresidues bk (see Figure 1(b)).

2.2. MT Techniques. Roughly speaking, the MT technique proceeds as follows.Let h ∈ H and f ∈ F and let c be any real number satisfying −β0 < c < b and1 − α0 < c < 1 − a. If M [f ; 1 − c − i·] ∈ L1(−∞,∞) or M [h; c + i·] ∈ L1(−∞,∞),then I(x) may be written in the form [24, Chap. 3]

(11) I(x) =1

2πi

∫ c+i∞

c−i∞x−zM [f ; 1− z]M [h; z]dz.

A displacement of the integration contour to the straight line �z = d < c and the useof the Cauchy residue theorem gives

I(x) =∑

d<�z<cRes{x−zM [f ; 1− z]M [h; z]; z = 1− αk,−βk}

+1

2πi

∫ d+i∞

d−i∞x−zM [f ; 1− z]M [h; z]dz.


280 JOSE L. LOPEZ

Then, from (9) and (10), when αk − βj = 1 ∀ k, j ∈ N⋃{0}, and for appropriate n

and m ∈ N [24, Chap. 3],

(12)

∫ ∞0

h(xt)f(t)dt =n−1∑k=0

akM [h; 1− αk]xαk−1 +m−1∑j=0

bjM [f ;βj + 1]xβj

+1

2πi

∫ d+i∞

d−i∞x−zM [f ; 1− z]M [h; z]dz.

If αk − βj = 1 for some k, j ∈ N⋃{0}, the pole z = 1−αk of M [f ; 1− z] and the

pole z = −βj of M [h; z] coalesce and, then, the integrand x−zM [f ; 1 − z]M [h; z] in(11) has a double pole. In this case the first line in the right-hand side of (12) mustbe replaced by

limz→0

{xβj

[akx−zM [h; 1 + z − αk] + bjM [f ; z + βj + 1]

]}.

Formally, the sum (12) yields an asymptotic expansion of I(x) for small x. Thedifficulty of this method lies in the technical results required to write I(x) in theform (11) and on the proof of the asymptotic character of (12). Moreover, from theexpansion given above it is not always clear how to obtain appropriate error boundsfor the remainder

∫ d+i∞d−i∞ x−zM [f ; 1− z]M [h; z]dz.

2.3. McClure and Wong’s Distributional Theory. Roughly speaking, M&Wtheory proceeds as follows. Consider the tempered distributions f , t−k−α

+ , and fnassociated to the corresponding functions f(t), t−k−α, and fn(t) in (7) for the par-ticular case αk = k + α, k = 0, 1, 2, . . . . Consider first the case 0 < α < 1. Thosedistributions act over functions h ∈ S[0,∞) (the Schwarz class of C(∞)[0,∞) rapidlydecreasing functions) in the following way [24, Chap. 6]:

(13)〈f , h〉 =

∫ ∞0

f(t)h(t)dt, 〈fn, h〉 = (−1)n∫ ∞

0fn,n(t)h(n)(t)dt,

⟨t−k−α+ , h

⟩=

1(α)k

∫ ∞0

t−αh(k)(t)dt

for k = 0, 1, 2, . . . , where

(14) fn,n(t) ≡ (−1)n

(n− 1)!

∫ ∞t

(u− t)n−1fn(u)du.

From [24, Chap. 6, Lem. 1], we have that these distributions are related by the equality

(15) f =n−1∑k=0

akt−k−α+ +

n−1∑k=0

(−1)k

k!M [f ; k + 1]δ(k) + fn,

where δ(k) is the kth derivative of the delta distribution at the origin:⟨δ(k), h

⟩=

(−1)kh(k)(0). Applying (15) to specific kernels h(xt) ∈ S[0,∞) and using (13), wecan derive asymptotic expansions of certain integral transforms I(x). For example,if h(t) = e−t, we derive the asymptotic expansion of the Laplace transform near theorigin for functions f(t) ∈ F [24, Chap. 6, Thm. 13]:∫ ∞

0e−xtf(t)dt =

n−1∑k=0

akΓ(1− k − α)xk+α−1 +n−1∑k=0

(−1)kM [f ; k + 1]

k!xk−1

+ xn∫ ∞

0e−xtfn,n(t)dt.



The case α = 1 is more complicated. In this case, the second line of (13) is replacedby

(16)⟨t−k−1+ , h

⟩= − 1

k!

∫ ∞0

h(k+1)(t) log t dt, k = 0, 1, 2, . . . .

Formula (15) must be also replaced by [24, Chap. 6, Lem. 2]

(17) f =n−1∑k=0

akt−k−1+ +

n−1∑k=0

(−1)k

k!ckδ

(k) + fn,

with

ck ≡ limz→k

[M [f ; z + 1] +

akz − k

]+ ak(γ + ψ(k + 1)).

The asymptotic expansion of the Laplace transform for α = 1 is also given in [24,Chap. 6, Thm. 13]. Asymptotic expansions of Stieltjes, Fourier, and Hilbert trans-forms are also derived in [24, Chap. 6] using this technique. Asymptotic expansions ofthe three standard elliptic integrals and the Appell function F1 were derived in [10],[14], [15] using this technique.

The complexity of M&W’s method lies in the derivation of (15) and (17) andtheir a posteriori implementation in specific kernels h(t). Moreover, the calculationof a general error bound for the remainder is still a challenge.

Similar results may be found in [23] and [24, Chap. 6, sec. 7] using convolutionsof distributions. Essentially, that method (called the regularization method) requiresboth f, g ∈ F and f, g ∈ H. For brevity, we do not reproduce them here and insteadrefer the reader to [24, Chap. 6, sec. 7].

2.4. The AC Technique. As well as M&W’s method, the AC method consideredin [17] requires f ∈ F for the particular case αk = k+α, k = 0, 1, 2. But for h it onlyrequires h ∈ C(∞)[0,∞) and not the more stringent condition h ∈ S[0,∞) requiredin M&W’s method. This method uses AC techniques instead of distributions. Never-theless, it gives rise to a particular case of (12) with αk = k + α, k = 0, 1, 2, . . . ,M [h; 1 − αk] replaced by 1

(α)k

∫∞0 t−αh(k)(t)dt, bj replaced by h(j)(0)/j!, and an

“M&W form” for the remainder [17, Theorems 1 and 2]:

∫ ∞0


ak(α)k

xk+α−1∫ ∞

0t−αh(k)(t)dt+

n−1∑k=0

M [f ; k + 1]k!

h(k)(0)xk

+ (−1)nxn−1∫ ∞

0fn,n

(t

x

)h(n)(t)dt if 0 < α < 1

and∫ ∞0

h(xt)f(t)dt = −n−1∑k=0

akk!xk∫ ∞

0h(k+1)(t) log tdt

+n−1∑k=0

[ak(ψ(k + 1) + γ − log x) + lim

z→k+1

(M [f ; z] +

akz − k − 1

)]h(k)(0)k!

xk

+ (−1)nxn−1∫ ∞

0fn,n

(t

x

)h(n)(t)dt if α = 1.


282 JOSE L. LOPEZ

This method is generalized in [22] and [16] to the case h ∈ H. A formula similarto (12) is obtained in [22] and [16] for the particular case αk = k+α and βk = k+ β,k = 0, 1, 2, . . . . Also, a different form for the remainder is obtained there.

3. The “Sum Up and Subtract” Method. We have seen in the previous sectionthat the M&W method is a particular case of the AC method considered in [17],which is a particular case of the method introduced in [22] and revisited in [16]. Theasymptotic formula given in [22] and [16] is just (12) for the particular case αk = k+αand βk = k + β, k = 0, 1, 2, . . . , and a different form for the remainder. Then, inprinciple, we may think that the AC method is a particular case of the MT method.But the MT method requires the additional hypothesis M [f ; 1− c− i·] ∈ L1(−∞,∞)or M [h; c + i·] ∈ L1(−∞,∞), which is not required in the AC method. We presenthere a trivial proof of (12) without the restrictions αk = k + α or βj = j + β whichdoes not require M [f ; 1−c−i·] ∈ L1(−∞,∞) or M [h; c+i·] ∈ L1(−∞,∞). Moreover,it gives a simpler expression for the remainder from which a universal error boundis derived. Then we derive a method which results in a generalization of the M&W,AC, and MT methods.

We define α−1 ≡ a and β−1 ≡ −b and observe that α−1 < α0 and β−1 < α0 (seeRemark 1). We make the following observation stated in the form of a lemma.

Lemma 1. For any n ∈ N⋃{0}, ∃ m ∈ N

⋃{0} such that αn−1 − βm < 1 <

αn − βm−1.Proof. The lemma is true for n = 0 with m = 0: α−1 − β0 < 1 < α0 − β−1 (see

Remark 1). From this we have that αn − β−1 > 1 ∀ n ≥ 0. Now, for a given n ∈ N,take the unique m ∈ {0, 1, 2, . . .} such that αn − βm−1 > 1 and αn − βm ≤ 1. Thenαn−1 − βm < 1.

The main result of this paper is stated in the following two theorems.Theorem 1. Let f ∈ F and h ∈ H. Then, for any n, m ∈ N such that

αn−1 − βm < 1 < αn − βm−1,

(18)

∫ ∞0


akM [h; 1− αk]xαk−1 +m−1∑j=0

bjM [f ;βj + 1]xβj

+∫ ∞

0fn(t)hm(xt)dt.

If αk − βj = 1 for some pair (k, j), then, in this formula, the sum of terms

akM [h; 1− αk]xαk−1 + bjM [f ;βj + 1]xβj

must be replaced by

(19)limz→0

{xβj


]}= xβj

{limz→0

[akM [h; 1 + z − αk] + bjM [f ; z + βj + 1]

]− akbj log x

}.

Proof. Define f0(t) = f(t) and h0(t) = h(t). From Lemma 1, for any k ∈ N⋃{0}

it is always possible to find a j ∈ N⋃{0} such that αk−1 − βj < 1 < αk − βj−1. For

the given (k, j) we have that the following integral exists:

(20)∫ ∞

0fk(t)hj(xt)dt.



We start at k = j = 0 (the inequalities α−1−β0 < 1 < α0−β−1 hold) and switchon the following algorithm which increases (k, j) step by step from (0, 0) up to (n,m):

(a) For a given (k, j) satisfying αk−1 − βj < 1 < αk − βj−1 do the following. Ifαk − βj < 1, go to (b). If αk − βj > 1, go to (c). If αk − βj = 1, go to (d).

(b) Use fk(t) = akt−αk + fk+1(t) in (20) and the third line of (10):

∫ ∞0

fk(t)hj(xt)dt = akxαk−1M [h; 1− αk] +

∫ ∞0

fk+1(t)hj(xt)dt.

Go to (a) with k replaced by k + 1.(c) Use hj(xt) = bj(xt)βj + hj+1(xt) in (20) and the third line of (9):

∫ ∞0

fk(t)hj(xt)dt = bjxβjM [f ;βj + 1] +

∫ ∞0

hj+1(xt)fk(t)dt.

Go to (a) with j replaced by j + 1.(d) Use first fk(t) = akt

−αk + fk+1(t) and then hj(xt) = bj(xt)βj + hj+1(xt) in(20):

(21)

∫ ∞0

fk(t)hj(xt)dt

=∫ ∞

0

[akt−αkhj(xt) + bjx

βj tβjfk+1(t)]dt+

∫ ∞0

hj+1(xt)fk+1(t)dt.

Define the function

Fk,j(z, t) ≡ tz[akt−αkhj(xt) + bjx

βj tβjfk+1(t)], z ∈ C.

Then∫ ∞

0fk(t)hj(xt)dt =

∫ ∞0

Fk,j(0, t)dt+∫ ∞

0fk+1(t)hj+1(xt)dt.

On the one hand, hj(t) = bjtβj + O(tβj+1) when t → 0+. On the other hand,

fk+1(t) = −akt−αk + O(t−αk−1) when t → 0+. Hence, Fk,j(z, ·) ∈ L1[0,∞) forMax{αk − βj+1, αk−1 − βj} − 1 < �z < Min{αk − βj−1, αk+1 − βj} − 1. Choosetwo numbers z0 and z1 satisfying Max{αk − βj+1, αk−1 − βj} − 1 < z0 < 0 and0 < z1 < Min{αk − βj−1, αk+1 − βj} − 1. Then we have that, for z0 ≤ �z ≤ z1,

|Fk,j(t, z)| ≤ Gk,j(t) ≡{|Fk,j(t, z0)| for t ∈ [0, 1],|Fk,j(t, z1)| for t ∈ [1,∞),

and that Gk,j(t) ∈ L1[0,∞). Using the dominated convergence theorem we have that

∫ ∞0

Fk,j(0, t)dt = limz→0

∫ ∞0

Fk,j(z, t)dt

= xβj limz→0

∫ ∞0

[akt

z−αkx−zhj(t) + bjtz+βjfk+1(t)

]dt.


284 JOSE L. LOPEZ

From αk = βj + 1 and from (9) and (10) we have that M [h; z + 1−αk] and M [f ; z +βj+1] have a common strip of analyticity: a−βj−1 < �z < βj+b. From ConditionsI and II we have a − β0 − 1 < 0 < β0 + b, and then the point z = 0 belongs to thatstrip of analyticity. Then∫ ∞

0fk(t)hj(xt)dt = xβj lim

z→0

{akx−zM [h; z + 1− αk] + bjM [f ; z + βj + 1]

}+∫ ∞

0hj+1(xt)fk+1(t)dt.

Using x−z = 1− z log x+O(z2) when z → 0 and

M [h; z + 1− αk] =∫ ∞

0tz−αkhj(t)dt =

bjz

+O(1) when z → 0,

we find that the above expression can also be written in the form∫ ∞0

fk(t)hj(xt)dt = xβj{

limz→0

[akM [h; 1 + z − αk] + bjM [f ; z + βj + 1]

]− akbj log x

}

+∫ ∞

0hj+1(xt)fk+1(t)dt.

Go to (a) with k replaced by k + 1 and j replaced by j + 1.This algorithm generates (18)–(19).Theorem 2. Within the hypothesis of Theorem 1, the expansion (18) is an

asymptotic expansion for small x:

(22)∫ ∞

0fn(t)hm(xt)dt = O(xβm + xαn−1) when x→ 0 and αn = βm + 1

and

(23)∫ ∞

0fn(t)hm(xt)dt = O(xβm log x) when x→ 0 and αn = βm + 1.

Proof. On the one hand, from (i) of Definition 1, there is a c1n > 0 and a t10 suchthat |fn(t)| ≤ c1nt

−αn for t ≥ t10. From (i) of Definition 2, there is a c2m > 0 and a t20such that |hm(xt)| ≤ c2m(xt)βm for xt ≤ t20. For small enough x we have that t10 < t20/xand we can choose a t0 ∈ [t10, t

20/x]. Then

∫ ∞0

fn(t)hm(xt)dt =∫ t0

0fn(t)hm(xt)dt+

∫ ∞t0

fn(t)hm(xt)dt,

(24)∣∣∣∣∫ ∞

0fn(t)hm(xt)dt

∣∣∣∣ ≤ c2mxβm

∫ t0

0|fn(t)|tβmdt+ c1nx

αn−1∫ ∞xt0

|hm(t)|t−αndt

≤ (xβm + xαn−1)[c2m

∫ t0

0|fn(t)|tβmdt+ c1n

∫ ∞0|hm(t)|t−αndt

].

On the other hand, from (i) of Definition 1, there is a c1n > 0 and a t10 such that|fn(t/x)| ≤ c1nx

αnt−αn for t/x ≥ t10. From (i) of Definition 2, there is a c2m > 0 and



a t20 such that |hm(t)| ≤ c2mtβm for t ≤ t20. For small enough x we have that t10x < t20

and we can choose a t0 ∈ [t10x, t20]. Then,∫ ∞

0fn(t)hm(xt)dt =

1x

∫ ∞0

fn(t/x)hm(t)dt

=1x

{∫ t0

0fn(t/x)hm(t)dt+

∫ ∞t0

fn(t/x)hm(t)dt},

(25)∣∣∣∣∫ ∞

0fn(t)hm(xt)dt

∣∣∣∣ ≤ c2mxβm

∫ t0/x

0|fn(t)|tβmdt+ c1nx

αn−1∫ ∞t0

|hm(t)|t−αndt

≤ (xβm + xαn−1)[c2m

∫ ∞0|fn(t)|tβmdt+ c1n

∫ ∞t0

|hm(t)|t−αndt].

The integrals between brackets in the last line of (24) are finite for αn < βm + 1(and αn−1− βm < 1 < αn− βm−1). The integrals between brackets in the last line of(25) are finite for αn > βm + 1 (and αn−1− βm < 1 < αn− βm−1). Then (22) followsfrom (24) and (25).

If αn = βm+ 1, the second integral in the last line of (24) and the first integral inthe last line of (25) are divergent and those inequalities are true but useless. In thiscase, from the first line of (24) and hm(t) = hm+1(t) + bmt

βm ,

∫ ∞xt0

|hm(t)|t−αndt ≤∫ ∞

1|hm(t)|t−αndt+

∫ 1

xt0

(|hm+1(t)|+ |bmtβm |)t−αndt

≤∫ ∞

1|hm(t)|t−αndt+

∫ 1

0|hm+1(t)|t−αndt+ |bm log(xt0)|.

The integrals in the last line above are finite and (23) follows.Remark 2. The above theorem applies also to integrals of the form

∫∞ch(xt)f(t)dt

with 0 < c <∞ and with the same hypotheses for f and h except for one concerningthe asymptotic behavior of f at t = 0:∫ ∞

c

h(xt)f(t)dt =∫ ∞

0h(xt)fc(t)dt

with fc(t) = f(t)χ(c,∞)(t), χ(c,∞)(t) being the characteristic function of the interval(0,∞): χ(c,∞)(t) = 1 if t ∈ (c,∞) and χ(c,∞)(t) = 0 if t /∈ (c,∞). In this case we havefc(t) = O(t−a) as t → 0+ with a < 0 and |a| as large as we wish. The theorem alsoapplies to integrals of the form

∫ dch(xt)f(t)dt by writing

∫ d

c

h(xt)f(t)dt =∫ ∞c

h(xt)f(t)dt−∫ ∞d

h(xt)f(t)

and using the above considerations.The coefficients of the expansion are given in terms of Mellin transforms of f and

h. Several representations of those Mellin transforms are given in (9) and (10), buta simpler representation which does not use the concept of analytic continuation ispossible when f and h are differentiable.

Lemma 2. Suppose that αk = k + α with α + b > 1 and that βj = j + β witha− β < 1. Define h(t) ≡ t−βh(t) and f(t) ≡ t−αf(t−1).


286 JOSE L. LOPEZ

(i) If h ∈ C(∞)[0,∞) with h(j)(t) = O(t−β−b) when t → ∞ ∀ j = 0, 1, 2, . . . andα− β /∈ Z, then

M [h; 1− k − α] =(−1)j

(β − k − α+ 1)j

∫ ∞0

tβ+j−α−kh(j)(t)dt,

with j = �k + α− β� (this means αk − βj < 1 < αk − βj−1).(ii) If f ∈ C(∞)[0,∞) with f (k)(t) = O(ta−α) when t → ∞ ∀ k = 0, 1, 2, . . . and

α− β /∈ Z, then

M [f ; 1 + j + β] =(−1)k

(α− β − j − 1)k

∫ ∞0

tk+α−β−j−2f (k)(t)dt,

with k = �β + j + 2− α� (this means αk−1 − βj < 1 < αk − βj).(iii) If the hypotheses of both (i) and (ii) hold but with α− β ∈ Z and b+ β > 1,

α− a > 1, then

limz→0

{xβj


]}= −xβ+j

{akj!

∫ ∞0

h(j+1)(t) log(t

x

)dt+

bjk!

∫ ∞0

f (k+1)(t) log t dt},

with k = β + j + 1− α (this means αk − βj = 1).Proof. To prove (i), write

M [h; 1− αk] =∫ ∞

0t−k−αhj(t)dt =

∫ ∞0

tβ−k−α

[h(t)−

j−1∑l=0

bltl

]dt

and integrate by parts j times.To prove (ii), write

M [f ; 1 + βj ] =∫ ∞

0tj+βfk(t)dt =

∫ ∞0

tα−β−j−2

[f(t)−

k−1∑l=0

altl

]dt

and integrate by parts k times.To prove (iii), write

limz→0

{xβj


]}=∫ ∞

0

[akt−αkhj(xt) + bjx

βj tβjfk+1(t)]dt

= xβ

{ak

∫ ∞1tβ−α−k

[h(xt)−

j−1∑l=0

blxltl

]dt+ bjx

j

∫ ∞1tα−β−j−2

[f(t)−

k−1∑l=0

altl

]dt

}

+ xβ

{ak

∫ 1

0tβ−α−k

[h(xt)−

j∑l=0

blxltl

]dt+ bjx

j

∫ 1

0tα−β−j−2

[f(t)−

k∑l=0

altl

]dt

}

and integrate by parts j + 1 times in the first and third integrals and k + 1 times inthe second and fourth integrals.



4. Error Bounds. Theorem 2 does not offer a precise bound for the remainder.We show in this section that a precise bound for the remainder may be obtained ifthe bound |fn(t)| ≤ c1nt

−αn holds ∀ t ∈ (0,∞) and not only for t ∈ [t10,∞) and thebound |hm(t)| ≤ c2mt

βm holds ∀ t ∈ (0,∞) and not only for t ∈ (0, t20] (see the proofof Theorem 2).

In order to obtain a precise bound for the remainder∫∞

0 fn(t)hm(xt)dt, let theremainder fn(t) in the expansion (7) satisfy the bound |fn(t)| ≤ Fnt

−αn ∀ t ∈ (0,∞)and let the remainder hm(t) in (8) satisfy the bound |hm(t)| ≤ Hmt

βm ∀ t ∈ (0,∞)for some positive constants Fn and Hm.

Consider first the case βm + 1 = αn. Write

(26)∫ ∞

0fn(t)hm(xt)dt =

∫ 1

0fn(t)hm(xt)dt+

∫ ∞1

fn(t)hm(xt)dt.

If βm > αn − 1, perform the change of variable t → t/x and use hm(t) = hm−1(t) −bm−1t

βm−1 in the second integral in the right-hand side above. If βm < αn − 1, usefn(t) = fn−1(t)− an−1t

−αn−1 in the first integral in the right-hand side above. Usingthe inequalities αn−1−βm < 1 < αn−βm−1, |fn(t)| ≤ Fnt

−αn , and |hm(t)| ≤ Hmtβm

in (26) and straightforward operations we obtain

(27)∣∣∣∣∫ ∞

0fn(t)hm(xt)dt

∣∣∣∣ ≤{C1n,mx

αn−1 if βm > αn − 1,C2n,mx

βm if βm < αn − 1,

with

(28) C1n,m ≡ Fn

[Hm

1 + βm − αn+|bm−1|+Hm−1

αn − βm−1 − 1

]

and

(29) C2n,m ≡ Hm

[Fn

αn − βm − 1+|an−1|+ Fn−1

βm + 1− αn−1

].

Consider now the case βm + 1 = αn. Write(30)∫ ∞

0fn(t)hm(xt)dt =

∫ 1

0fn(t)hm(xt)dt+

∫ 1/x

1fn(t)hm(xt)dt+

∫ ∞1/x

fn(t)hm(xt)dt.

Perform the change of variable t → t/x in the last integral, use fn(t) = fn−1(t) −an−1t

−αn−1 in the first integral in the right-hand side, and use hm(t) = hm−1(t) −bm−1t

βm−1 in the last integral. Using the inequalities αn−1 − βm < 1 < αn − βm−1,|fn(t)| ≤ Fnt

−αn , and |hm(t)| ≤ Hmtβm and straightforward operations we obtain

(31)∣∣∣∣∫ ∞

0fn(t)hm(xt)dt

∣∣∣∣ ≤ [C3n,m + FnHm| log x|]xβm if βm = αn − 1,

with

(32) C3n,m ≡ Fn

|bm−1|+Hm−1

αn − βm−1 − 1+Hm

|an−1|+ Fn−1

βm + 1− αn−1.

In [9] we introduced certain families of functions f and h quite common in practicewhich satisfy the bounds |fn(t)| ≤ Fnt

−αn and |hm(t)| ≤ Hmtβm ∀ t ∈ (0,∞).


288 JOSE L. LOPEZ

Moreover, for those families of functions f and h, the constants Fn and Hm can beeasily obtained from f and h.

Family 1. Let f ∈ F be a real function with αn = n + α. We consider thefunction f(u) ≡ u−αf(u−1). If f(w) is a bounded analytic function in the region Uof the complex w-plane consisting of all points w located at a distance < r0 from thepositive real axis, Ur0 = {w ∈ C, |�w| < r0 if �w ≥ 0 and |w| < r0 if �w < 0}, then

(33) |fn(t)| ≤ 2Mr−nt−n−α,

where M is a bound of |f(w)| in Ur0 and 0 < r < r0.Family 2. Let h ∈ H be a real function with βm = m + β. We consider the

function h(u) ≡ uβh(u). If h(w) is a bounded analytic function in the region of thecomplex w-plane consisting of all points w located at a distance < r0 from the positivereal axis, Ur0 , then

|hm(t)| ≤ 2M r−mtm+β ,

where M is a bound of |h(w)| in Ur0 and 0 < r < r0.Family 3. If f(t) is real and the expansion (7) satisfies the error test, that is, if

sign(fn(t)) = −sign(fn+1(t)) ∀ n = 0, 1, 2, . . . and ∀ t ∈ (0,∞), then

|fn(t)| ≤ |an|t−αn and |fn(t)| ≤ |an−1|t−αn−1 .

Family 4. If h(t) is real and the expansion (8) satisfies the error test, then

|hm(t)| ≤ |bm|tβm and |hm(t)| ≤ |bm−1|tβm−1 .

From these bounds, the following observation is obvious.Remark 3. If f belongs to Family 1, then the bounds (27) and (31) hold with

Fn = 2Mr−n. If h belongs to Family 2, then the bounds (27) and (31) hold withHm = 2M r−m. If f is real and the expansion (7) satisfies the error test, then thebounds (27) and (31) hold replacing Fn by |an|. If g is real and the expansion (8)satisfies the error test, then the bounds (27) and (31) hold replacing Hm by |bm|.

5. Universality. Many classical techniques, the MT techniques, M&W theory,and the AC method are easy corollaries of Theorems 1 and 2.

Corollary 1 (classical methods I). If tnf(t) ∈ L1(0,∞) ∀ n ≥ 0, then, inDefinition 1, ak = 0 ∀ k,

M [f ; 1 + βk] =∫ ∞

0f(t)tβkdt, and fn(t) = f(t).

Then, from Theorem 1,

∫ ∞0

h(xt)f(t)dt =m−1∑k=0

[bk

∫ ∞0

f(t)tβkdt]xβk +

∫ ∞0

f(t)hm(xt)dt.

An important example is Watson’s lemma: set f(t) = e−t and x = x−1 in theabove formula with x→∞, to get

x

∫ ∞0

h(t)e−xtdt =∫ ∞

0h(xt)e−tdt =

m−1∑k=0

bkΓ(βk + 1)xβk

+ x

∫ ∞0

e−xthm(t)dt.



Corollary 2 (classical methods II). If t−nh(t) ∈ L1(0,∞) ∀ n ≥ 0, then, inDefinition 2, bk = 0 ∀ k,

M [h; 1− αk] =∫ ∞

0h(t)t−αkdt, and hm(t) = h(t).

Then, from Theorem 1,∫ ∞0


[ak

∫ ∞0

h(t)t−αkdt]xαk−1 +

∫ ∞0

fn(t)h(xt)dt.

Corollary 3 (MT techniques). Formula (12) is just (18) with a different ex-pression for the remainder. Apart from the conditions required for f and h in The-orem 1 above, the MT technique requires also the integrability of M [f ; 1 − c − y·] orof M [h; c + y·] in order to write I(x) in the form (11). Then expansion (12) followsfrom calculating the poles and residues of M [f ; 1 − z] and M [h; z]. But what we seein Theorem 1 is that it is not necessary to write I(x) in the form (11) and, therefore,the integrability of M [f ; 1 − c − y·] or of M [h; c + y·] is not necessary. In fact, thelocation of the poles and the value of the residue of M [f ; 1−z] and M [h; z] are of fun-damental importance to derive the asymptotic expansion of I(x) in the MT technique.But from the second lines of formulas (9) and (10) we see that that information isalready contained in the expansions (7) and (8) of f(t) and h(t) and the expansion ofI(x) follows directly from (1).

Corollary 4 (M&W method). If h ∈ S[0,∞) ⊂ C(∞)[0,∞) and αk = k + α,then βk = k (k = 0, 1, 2, . . .), m = n in Lemma 1, and

bk =h(k)(0)k!

.

Integrating by parts we have that∫ ∞0

fn(t)hn(xt) = (−x)n∫ ∞

0fn,n(t)h(n)

n (xt)dt = (−x)n∫ ∞

0fn,n(t)h(n)(xt)dt,

where fn,n(t) is defined in (14). On the other hand, if 0 < α < 1,

M [h; 1− k − α] =∫ ∞

0hk(t)t−k−αdt =

1(α)k

∫ ∞0

t−αh(k)k (t)dt =

⟨t−k−α+ , h

⟩.

(Observe that the definition of⟨t−k−α+ , h

⟩by means of integration by parts in M&W’s

theory coincides with the definition of analytic continuation given by the Mellin trans-form of h.) Therefore, from (18),(34)∫ ∞

0h(xt)f(t)dt =

n−1∑k=0

ak(α)k

xk+α−1∫ ∞

0t−αh(k)(t)dt+

n−1∑k=0

M [f ; k + 1]k!

h(k)(0)xk

+ (−1)nxn−1∫ ∞

0fn,n

(t

x

)h(n)(t)dt,

which is a generalization of the expansions given in [24, Chap. 6] for 0 < α < 1. Ifα = 1 (αk − βk = 1 ∀ k), then from (19),

limz→0

{xk[akx−zM [h; z − k] + bkM [f ; z + k + 1]

]}= limz→0

{xk[

ak(−z)k+1

∫ ∞0

tzh(k+1)(t)dt+M [f ; z + k + 1]

k!h(k)(0)

]}.


290 JOSE L. LOPEZ

It is obvious that the first term in the last line above has a pole at z = 0:

tz

(−z)k+1=

1k!

[1z

+ log t+ ψ(k + 1) + γ

]+O(z) as z → 0.

Moreover, the residue of that term at z = 0 is akh(k)(0)/k!:

ak(−z)k+1

∫ ∞0

tzh(k+1)(t)dt =akk!

{[1z

+ ψ(k + 1) + γ

]h(k)(0)

−∫ ∞

0h(k+1)(t) log tdt

}+O(z) as z → 0.

But also, from the second line of (9), the Mellin transform M [f ; z+ k+ 1] has a poleat z = 0 with residue ak. Using this last identity and taking the limit z → 0 we obtain

(35)

∫ ∞0

h(xt)f(t)dt = −n−1∑k=0

akk!xk∫ ∞

0h(k+1)(t) log tdt

+n−1∑k=0

[ak(ψ(k + 1) + γ − log x) + lim

z→k+1

(M [f ; z] +

akz − k − 1

)]h(k)(0)k!

xk

+ (−1)nxn−1∫ ∞

0fn,n

(t

x

)h(n)(t)dt,

which is a generalization of the expansions given in [24, Chap. 6] for α = 1.The method of regularization given in [24, Chap. 6, sec. 7], is also a corollary

of Theorem 1. The expansions derived in [23], [24, Chap. 6, sec. 7] may be cast inthe form (18)–(19). The conditions required for f and h there are more stringentthan those of Theorem 1 above: in that theory both f and h must belong to F and Hsimultaneously.

Corollary 5 (AC techniques). The expansions derived in [17, Theorems 1and 2] by means of analytic continuation are nothing more than expansions (34) and(35). The conditions required for f(t) and h(t) in [17] are more stringent than thoseof Theorem 1 above: f ∈ F , h ∈ C(∞)[0,∞) ⊂ H. Moreover, the expansions derivedin [22] and [16] are just the expansion given in Theorem 1 for the particular caseαk = k + α and βk = k + β.

6. Examples. Formula (18) can be applied straightforwardly to many examplesof important integral transforms and special functions. Some examples previouslyconsidered by using the much more complicated distributional techniques, Mellintransforms, or AC methods are the Laplace, Stieltjes, and Lambert transforms forlarge or small argument, fractional integrals for large argument, the Poisson transformfor large time, hypergeometric functions for large variables, elliptic integrals for largeor small parameters, the Appell function F1 or the Lauricella functions near theinfinity, and thermonuclear reaction rates for small Sommerfeld parameter [3], [8], [9],[10], [11], [12], [14], [15], [24, Chap. 6]. The asymptotic expansions obtained therecan be derived trivially as simple corollaries of Theorems 1 and 2. Other interestingexamples not yet considered are the Appell function F2 for large variables, the Appellfunctions F1 and F2 near their branch points, or the Hurwitz–Lerch zeta functionnear its singular point.

We give details here of the example of the exponential integral near the origin.The exponential integral E1(z) is defined in [1, eq. (5.1.1)]. After a straightforward



change of the integration variable it reads

(36) E1(z) = e−z∫ ∞

0f(t)h(xt)dt, �z > 0,

with x = |z|,

f(t) =1

1 + t, h(t) = e−γt,

and γ ≡ eiArg(z). We can apply Theorem 1 to the integral in the right-hand side of(36) with αk = k + 1, βk = k, m = n, x = |z|, ak = (−1)k, bk = (−γ)k/k!,

M [f ;w] =π

sin(πw)and M [h;w] =

Γ(w)γw

.

Then (18) and (19) give the following asymptotic expansion for small z:

(37) ezE1(z) =n−1∑k=0

zk

k![ψ(k + 1)− log z] +Rn(z),

with

Rn(z) =∫ ∞

0fn(t)hn(xt)dt.

Two consecutive derivatives of the function f(t) have opposite sign for any t > 0.Using the Lagrange formula for the remainder of the Taylor expansion of f(u) =f(u−1) at u = 0 we see that the expansion of the function f(t) in inverse powers of tsatisfies the error test (belongs to Family 3) and then |fn(t)| ≤ t−n−1 for t > 0. Thefunction f(t) also belongs to Family 1 and then, from (33), we can obtain anotherbound for fn(t), although less competitive: as a function of the complex variable u,the function f(u) is analytic in the region Ur0 considered in Family 1 with r0 = 1. Wechoose r < r0 in (33) in such a way that the maximum M of |f(u)| in Ur0 satisfies thecondition that Mr−n attains its smallest possible value: M = n − 1, r = (n − 1)/n,and Mr−n = n

(nn−1

)n−1. Then, another bound (less competitive than the previous

one) is |fn(t)| ≤ 2n(

nn−1

)n−1t−n−1 for t > 0.

From the Lagrange formula for the Taylor remainder hn(t) of the expansion ofh(t) at t = 0 we have that hn(t) = h(n)(ξ)tn/n!, with ξ ∈ (0, t). Then, for �z > 0 wehave |hn(t)| ≤ tn/n! for t > 0.

In this example βn = αn − 1 and then in (31) and (32) we set Fn = 1 andHn = 1/n!. Formula (31) reads

|Rn(z)| ≤ 1(n− 1)!

|z|n[2 +

1n

(2 + | log z|)].

From this bound we see that expansion (37) is convergent ∀ z and then

E1(z) = e−z∞∑k=0

zk

k!ψ(k + 1)− log z,

to be compared with the expansion in [1, eq. (5.1.11)].


292 JOSE L. LOPEZ

7. Concluding Remarks. When the positive moments of f(t) or the negativemoments of h(t) exist, asymptotic expansions of integral transforms

∫∞0 f(t)h(xt)dt

for small x may be obtained by means of classical techniques [24, Chaps. 1 and 2].When neither the positive moments of f(t) nor the negative moments of h(t) exist,we may try other asymptotic methods: MT techniques [24, Chap. 3], M&W theory[24, Chaps. 5 and 6], or AC methods [22], [17], [16].

Theorem 1 above offers a new asymptotic method for∫∞

0 f(t)h(xt)dt, which hasthe following features.

(i) It is very general. It is valid whether the positive moments of f(t) or thenegative moments of h(t) exist or not and it contains some classical meth-ods, MT techniques, M&W theory, and the AC method as straightforwardcorollaries.

(ii) It is trivial. It does not require the complexity of the theory of distributionsor the transformation to the Mellin space used in MT techniques or M&Wtheory, nor the analytic continuation techniques used in the AC method. Itjust follows from an expansion of the integrand and an interchange of sumand integral (in an appropriate order).

(iii) It unifies somehow the asymptotic theory. We see that the asymptotic prin-ciple under Theorem 1 is quite pragmatic:(a) When the positive moments of f(t) exist, expand h(t) at t = 0 and

interchange sum and integral.(b) When the negative moments of h(t) exist, expand f(t) at t = ∞ and

interchange sum and integral.(c) When both the positive moments of f(t) and the negative moments of

h(t) do not exist, interlace the expansions of h(t) at t = 0 and of f(t) att = ∞ and interchange sum and integral in such a way that only finitemoments appear.

Theorem 1 above shows that the asymptotic principia under classical methodsI and II, M&W theory, MT techniques, and the AC method is indeed quite simple:expand the integrand and exchange sum and integral (in a proper way). Just oneasymptotic sequence appears when classical methods apply because just one, f or h,needs to be expanded. On the other hand, two asymptotic sequences appear naturallyin M&W theory and MT and AC techniques if both f and h need to be expandedin order to avoid the appearance of divergent coefficients. Theorem 1 combines allthese situations in a single formulation. If the positive moments of f exist, thenak = 0. If the negative moments of h exist, then bk = 0. In any case, just one asymp-totic sequence appears in (18). If none of these moments exist, then two asymptoticsequences appear in (18).

The presence of double poles in the integrand in (11) in MT theory correspondswith the case αn = βm+1 in our theory or in M&W’s method or in the AC technique.This situation translates into the appearance of log x terms in the expansion (18) andthe coincidence of exponents of x in the two asymptotic sequences of that formula.

MT techniques and M&W theory deal also with the possibility of f(t) having anoscillatory asymptotic expansion at infinity of the form f(t) = eict

∑n−1k=0 akt

−αk +fn(t) instead of (7) (and similarly for h(t) at t = 0+). Theorems 1 and 2 may begeneralized to this kind of function. Moreover, they may be generalized to complexvalues of αk, βk, and x. This is the subject of further investigation.

Acknowledgment. The improvements suggested by the anonymous referees areacknowledged.



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