ASYMPTOTICS OF SIGNED ENTIERE SERIES

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ASYMPTOTICS OF SIGNED ENTIERE SERIES

Luc Abergel

To cite this version:

Luc Abergel. ASYMPTOTICS OF SIGNED ENTIERE SERIES. 2019. �hal-02293730�

ASYMPTOTICS OF SIGNED ENTIERE SERIES.

Luc Abergel

Abstract:

This article focuses on the effects of several pertubations that would be applied to the exponential for large negative values. It’s about identifying which ones are important and what are their effects.

Contents

1 Case of a rational fraction. 3

2 Integral form and applications. 5

3 First properties. 5

4 Some technical results. 8

5 Practical calculations. 8

6 Logarithmic cases. 12

7 Principle of superposition of perturbations. 15

8 Generalizations. 22

Acknowledgement:

A big thank you to Bruno Vallette for his advice on both the feedback and the possible use of this article.

Classification : 58J37

Introduction

The purpose of this work is to study some perturbations in the seriese^−x=

∞

X

n=0

(−1)ⁿxⁿ n! . So we want to give an asymptotic at +∞of

∞

X

n=0

(−1)ⁿxⁿ

n! u(n) for explicit sequences made using expressions such as (n+a)^k ou ln^k(n+a).

The basic difficulty lies in the fact that an error on a term un gives a perturbation of the sum whose order of magnitude isxⁿ. The case of thee^−xseries shows that such an error then hides the asymptotic of the sum .

We therefore do not expect a priori to obtain any meaningful results for this question. We can even think that in fact such series have erratic behavior in +∞.

It turns out that this is not the case for the sequenses envisaged, and that on the contrary, there are very regular results, with even a simple and regular dependence on the parameters (a and k in the examples cited). We will see that if the serieun is not defined for alln, choosing to look at the sum from a certain rank or for the nvalues for which the terms un are defined does not change anything about asymptotics.

Another surprising result, we will also see that there is an order structure on successive perturbations that could be applied to thee^−x series. For example, we can give an infinite number of terms for a perturbation such as (n+a)^k and that in the case of 2 perturbations (n+a1)^k1 and (n+a2)^k2, only one of them plays a role on an asymptotic of the sum, the other contributing only through a constant by which the effect of the major peeturbation is multiplied.

The working method is to give an integral representation of un then the sum

∞

X

n=0

(−1)ⁿxⁿ

n! u(n). It is not so obvious to do, for example by the fact that we could not treat sequences that would not tend towards

1Professor, Lyc´ee Janson de Sailly, PARIS, email : lucabergel@cegetel.net

0 by this methodstricto sensu.

We’ll start with a case that’s very simple to obtain and motivates the work that follows, then we’ll present definitions that allow for the full development of certain sequences, then we’ll deal with concrete examples, and we will end with the study of the case of superposition of perturbations which is the essence of this article, as well as examples illustrating the effectiveness of this approach.

Notations : We will note indifferently Γ(a) = (a−1)! fora /∈ −NandB(x, y) = Z 1

0

t^x−1(1−t)^y−1 dt.

1 Case of a rational fraction.

We’re going to study here

∞

X

n=0

(−1)ⁿxⁿ

n! R(n) whereR is a rational fraction. We will note here that ifu is polynmial, thenF_u(x) =O(ρ^x)

(this being done by treating the case of the polynomialX(X−1)· · ·(X−k+ 1)).

1.1 A first case.

We’re studying the caseu(x) =_x+a¹ with <(a)>0.

On the one hand, we write Z ∞

x

t^a−1e^−tdt=O(x^a−1e^−x), on the other hand

Z x 0

t^a−1e^−tdt=

∞

X

n=0

(−1)ⁿx^n+a n!(n+a) . This shows the formula :

∞

X

n=0

(−1)ⁿxⁿ

n!(n+a) =x^−aΓ(a) +O(x^a−1e^−x).

The error term is well inO(ρ^x).

1.2 Generalization to the case a / ∈ − N .

Proposition : The caseu(x) = _(x+a)¹ witha /∈ −N. We have the following asymptotic :

∞

X

n=0

(−1)ⁿxⁿ n!(n+a) ∼

∞

Γ(a)

x^a sia /∈ −N.

Proof :

As example, we will only deal with the case of−1<<(a)≤0.

Leta=−1 +bwith<(b)>0.

We write

∞

X

n=0

(−1)ⁿxⁿ n!(n+a) =

∞

X

n=0

(−1)ⁿxⁿ

n!(n−1 +b) = 1

−1 +b +

∞

X

n=0

−(−1)ⁿxⁿ⁺¹ (n+ 1)!(n+b). Then

∞

X

n=0

(−1)ⁿxⁿ n!(n+a) = 1

b−1 −x

∞

X

n=0

(−1)ⁿxⁿ n!

1

(n+ 1)(n+b) = 1 b−1 −x

∞

X

n=0

(−1)ⁿxⁿ n!

1

n+ 1− 1 n+b

1 b−1. Let

∞

X

n=0

(−1)ⁿxⁿ n!(n+a)= 1

b−1− x b−1

Γ(1)

x −x^−bΓ(b) +O(ρ^x)

, and we deduce :

∞

X

n=0

(−1)ⁿxⁿ

n!(n+a)=x^−b+1Γ(b)

b−1+O(ρ^x).

This is the formula for 1.1 but in the more general case<(a)>−1.

The casea /∈ −Nis also written in the same way.

1.3 Case of a multiple pole.

Caseu(x) =_(x+a)¹ _k witha /∈ −Nandp∈N. We have the following asymptotic :

∞

X

n=0

(−1)ⁿxⁿ n!(n+a)^k ∼

∞

Γ(a) x^a

ln^k−1(x)

(k−1)! sia /∈ −N. Proof :

Again, we will only deal with one case, the k = 2 case, the working method becoming generalized with- out any difficulty for the other values ofk.

We star with

∞

X

n=0

(−1)ⁿx^n+a−1

n!(n+a) = Γ(a)

x +O(ρ^x) that we integrate on [0, x].

1.4 Case of a rational fraction.

The case of a rational fraction is now available.

Note the case whereuis polynomial :

We’re dealing with the caseu_k(x) =x(x−1)...(x−k+ 1) that gives F_uk(x) =

∞

X

n=0

(−1)ⁿxⁿ (n−k)!. And soFu_k(x) = (−x)^ke^−x=O(ρ^x).

The case of a rational fraction.

If we giveu(x) = P(x) Π

1≤i≤p(x+a_i)^pi withP polynomial, with integer exponentsp_i strictly positive, if we have<(a₁)<<(a_i) pour touti6= 1, and if we finally notice p=p₁, a=a₁ so

∞

X

n=0

(−1)ⁿxⁿ n! u(n)∼

∞AΓ(a) x^a

ln^p−1(x) (p−1)!. withA= _Π^P(−a)

1≤i≤p(a_i−a)^pi. Proof :

It’s enough to decompose into simple elements and apply the previous results.

Generalization in the case of a rational fraction that may have negative integer poles.

Again by decomposition into simple elements, it is enough to treat the case of a monome.

Le cas

∞

X

n=p+1

(−1)ⁿxⁿ n!(n−p)^k. We have the asymptotic

∞

X

n=p+1

(−1)ⁿxⁿ n!(n−p)^k ∼

∞(−1)^k+1x^p p!

ln^k(x) k! .

Proof :

We therefore want to study, for example, the case of

∞

X

n=p+1

(−1)ⁿxⁿ

n!(n−p)^k = (−x)^p+1

∞

X

n=0

(−1)ⁿxⁿ

n!(n+ 1)^k+1(n+ 2)· · ·(n+p+ 1) According to the case of a rational factorized fraction, we obtain an equivalent by looking at the part relating

to the pole _(n+1)¹_k+1 :

1

(n+1)^k+1(n+2)···(n+p+1) =_p!¹_(n+1)¹k+1 +· · ·.

The equivalent sought is therefore (−x)^{p+1 1}_p!^lnk_k!^(x)Γ(1)_x . So

∞

X

n=p+1

(−1)ⁿxⁿ n!(n−p)^k ∼

∞(−1)^p+1x^p p!

ln^k(x) k! .

It should be noted, given the equivalent obtained, that the result is the same forX

n6=p

(−1)ⁿxⁿ n!(n−p)^k.

An example where several poles have even real parts. Let’s deal with the caseF(x) =

∞

X

n=0

(−1)ⁿxⁿ n!(n²+ 1).

We write _n2¹+1 = ⁱ₂h

1

x+i−_x−i¹ i

. This gives usF(x) =₂ⁱ

Γ(i)x⁻ⁱ−Γ(−i)xⁱ

+O(ρ^x), or, by writing Γ(i) =ρe^iθ :

F(x) =

+∞ρsin(ln(x)−θ) +O(r^x) avecr <1.

2 Integral form and applications.

Notations and working context :

- We consider functions defined from a certain rankn₀ and we will talk about sequences by looking at the values of such functions over fairly large integers.

- For a given sequenceuwe noteF_u(x) =

∞

X

n≥n0

(−1)ⁿxⁿ n! u(n).

For a given sequenceu: - If we can write u(x) =

Z 1 0

t^xπ(t)dt for a function π and for anyx≥0, then we’ll say that the u is density sequence. It will be said to be associated withπthat we will noteπu.

- If we can write u(x) = Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!

π(t)dt for an integer p, then we will say that u is deduced from a density sequence.

In practice we will assumeπmonotonous around 0.

The working context will be that of density sequences or suites that are deduced from them. We will naturally see these last ones appear by operations on density sequences.

It should be noted that the case of a defined sequence, for example for x ≥ 1, is not that of a density sequence, but only of a sequence that is deduced from it, as will be seen later.

In the case of a density sequence,Fu(x) = Z 1

0

e^−txπu(t)dt, it is then the study of this integral that gives an asymptotic ofFuin + +∞. So we’ll look for an asymptotic atO(ρ^x) withρ <1.

Therefore, sometimes we will write’=’ for two terms that differ by aO(ρ^x) (always withρ <1).

Finally, let’s note that the values ofu(x) forxlarge essentially involve the behavior ofπin 1, while those of Fuin +∞that of πin 0.

3 First properties.

- Derivation :

Ifuis a density sequence associated toπ_u, thenu⁰ also is, associated toπ_u⁰ = ln(t)π_u(t).

- Primitivation : If U(x) =

Z x 0

u(t)dt, then U(t) = Z 1

0

(1−t^x) πu(t)

ln(1/t) dt. This sequence is therefore deduced from a density sequence.

- Truncation :

The truncated sequenceT u(0) = 0 etT u(n) =u(n) forn≥1 verifies FT u(x) =

Z 1 0

(e^−xt−1)πu(t)dt.

Ifπis integrable in 0 (so thatu(0) is defined), noting Π_u the primitive ofπ_u which is vanishes in 1, F_{T u}(x) =x

Z 1 0

e^−xtΠ_u(t)dt.

We will see just after that that it means that this sequence is deduced from a density sequence.

This is immediate after an integration by parts.

- Shift :

We note S^α the operator who has a sequence u associates the sequence x → u(x+α) where α is real. Ifuis a density sequence, thenS^α(u) is associated tot^απu(t).

And so FS^α(u)(x) = Z ∞

0

e^−xtt^απu(t)dt.

Proof :

We writeS^α(u)(x) = Z 1

0

t^xt^απu(t)dt.

This then givesF_Sα(u)(x) = Z ∞

0

e^−xtt^απu(t)dt.

Remark :

For example, if we consider the sequence u(x) = _ln(x+2)¹ defined for x ≥ 0, the sequence S⁻¹(u) is only defined from the rank 1. It is therefore not a priori a density sequence, but it is deduced by truncation from theusequence from the rank 1.

It is therefore necessary to considerF_{T S}⁻¹(u)(x) =x Z 1

0

e^−xtΠu(t)dt with Πu(t) =− Z 1

t

πu(s) s ds.

Cases of sequences that are deduced from density sequences.

We also want to study sequences that do not tend towards 0 in +∞. We will be able to show them as sequences deduced from density sequences.

given a density ofπ, we’re interested in the sequence u(x) = Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!

π(t)dtand we want to studyF_u in +∞.

This will require some definitions.

We recall the notation ∆(P) =P(X+ 1)−P(X) for a polynomial.

- For a functionπthat can be integrated on ]0.1[, we define the operator Φ by Φ(π)(t) = 1

t Z t

0

π(s)ds.

- For a positive integerp, we noteQ_α,p the polynomial defined byS_α,p(X) =

p

X

i=0

b_α,i(−X)ⁱ i!

withb_α,i= ∆ⁱ((X+α)^p)(0). In the case ofα= 0, we will simply noteb_i=b_α,iandS_p=S_0,p.

Let us note a remark that will be used later :

Note on coefficientsb_α,i:

p

X

i=0

(−1)ⁱbα,i

α+i i

= (−1)^p.

Proof :

This comes from ^−α−1_i

= (−1)^{i α+i}_i

and from (X+α)^p=

p

X

i=0

bα,i

X i

which is applied inX=−α−1.

Let beπan integrable function on ]0,1[.

Ifu(x) = Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!

π(t)dt, then :

- u(x) = (−x)^p Z 1

0

t^xΦ^p(π)(t)dt.

- Fu(x) = (−1)^p Z 1

0

e^−xtSp(xt)Φ^p(π)(t)dt.

Proof : First point :

We integrate by parts Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!

π(t)dt.

We then obtain

"

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!

tΦ(π)(t)

#¹

0

− Z 1

0

x t^x−1−

p−1

X

i=1

(xln(t))ⁱ⁻¹ t(i−1)!

!

tΦ(π)(t)dt,

then (−x) Z 1

0

t^x−

p−2

X

i=0

(xln(t))ⁱ i!

!

Φ(π)(t)dt, and we conclude by induction.

Second point :

NoteNi(X) = X(X−1)···(X−i+1)

i! = ^X_i

. We writeX^p=

p

X

i=0

biNi(X) avecbi= ∆ⁱ(X^p)(0).

So we haveu(x) = (−1)^p

p

X

i=0

biNi(x) Z 1

0

t^xΦ^p(π)(t)dt.

Forvi(x) =Ni(x) Z 1

0

t^xΦ^p(π)(t)dt, we have F_vi(x) =X

n≥0

(−1)ⁿxⁿ n! N_i(n)

Z 1 0

tⁿΦ^p(π)(t)dt=X

n≥i

(−1)ⁿxⁿ i!(n−i)!

Z 1 0

tⁿΦ^p(π)(t)dt, and then Fvi(x) =

Z 1 0

X

n≥i

(−1)ⁿ(xt)ⁿ

i!(n−i)! Φ^p(π)(t)dt= Z 1

0

e^−xt(−xt)ⁱ

i! Φ^p(π)(t)dt.

SoFu(x) = (−1)^p Z 1

0

e^−xt

p

X

i=0

bi

(−xt)ⁱ

i! Φ^p(π)(t)dt= (−1)^p Z 1

0

e^−xtSp(xt)Φ^p(π)(t)dt.

Application to an integral representation ofS^αuand ofF_Sαu. For such a sequenceu, per shift we have :

- S^α(u)(x) = (−1)^p(x+α)^p Z 1

0

t^x+αΦ^p(π)(t)dt.

- FS^αu(x) = (−1)^p Z 1

0

e^−xtSα,p(xt)t^αΦ^p(π)(t)dt.

Proof :

First point is obvious.

For the second, we must write (X+α)^p=

p

X

i=0

aα,i

X i

withaα,i= ∆^p((X+α)^p)(0).

Thus we obtain the result by the relationSα,p(X) =

p

X

i=0

aα,i

(−X)ⁱ i! .

4 Some technical results.

Let us mention here some asymptotics that will be used, without dwelling on the demonstrations.

Sit^απ(t) is integrable on ]0,1] et siρ <1, soρ^x=o(Fu(x)).

Furthermore, ifπ1∼

0 π2 in 0, thenFu₁ ∼

+∞Fu₂. For the first point :

Z 1 0

e^−xtπ(t)dt= Z 1

0

t^−αe^−xtt^απ(t)dt≥a^−αe^−xa Z 1

a

t^απ(t)dtfora >0, this leading termρ^x if we chooseasmall enough.

For the second :

Ifa(t) =o(π(t)) in 0, let’s check that Z 1

0

e^−xta(t)dt=o Z 1

0

e^−xtπ(t)dt

. We write|a(t)| ≤επ(t) over ]0, α] and so

Z 1 0

e^−xta(t)dt≤ε Z 1

0

e^−xtπ(t)dt+e^−αx Z 1

α

a(t)dt.

The increase in the first point then makes it possible to conclude.

And finally fora >0 (voir [2]):

∗ Z 1

0

e^−xtt^a−1ln^b(1/t)dt∼

∞Γ(a)ln^b(x) x^a

= 1 x^a

p

X

i=0

i b

ln^b−i(x)(−1)ⁱΓ⁽ⁱ⁾(a) +o ln^b−p x^a

!

∗ ∗ Z 1/e

0

e^−xtt^a−1ln^b(1/t) ln^c(ln(1/t))dt∼

∞Γ(a)ln^b(x) ln^c(ln(x)) x^a

= 1 x^a

X

0≤i+k≤pandj≤q

i b

ln^b−i−k(x) ln^c−j(ln(x))(−1)ⁱΓ^(i+k)(a) +o ln^b−pln^c−q(ln(x)) x^a

!

5 Practical calculations.

5.1 Case u(x) =

with a and k positive.

Foru(x) =_x+1¹ we haveπu(t) = 1[0,1]. We noticeπor evenπuthis function.

By calculating the derivative, we have (−1)^kk!

(x+ 1)^k+1 = Z 1

0

t^xln^k(t)dt.

That is, foru_k(x) =_(x+1)¹ _k,π_uk(t) = ^(−1)k−1_(k−1)!^lnk−1(t) on the interval ]0,1].

The case of a positive integer exponentkwas treated as such.

This is widespread in : ifuk(x) =_(x+1)¹ _k withk∈R⁺, thenπu_k(t) = ^lnk−1_(k−1)!^(1/t)1_]0,1].

This relation is easily verified by putingt=e^−u in 1 (k−1)!

Z 1 0

t^xln^k−1(1/t)dt, remembering the rating (k−1)! = Γ(k).

This results in the following formulaF_uk(x) = 1 (k−1)!

Z 1 0

e^−xtln^k−1(t)dt.

We will now present here the casevk(x) =_(x+a)¹ _k witha >0.

We just saw that ifuk(x) = _(x+1)¹ _k, then πu_k(t) =^lnk−1_(k−1)!^(1/t)1_]0,1]. By the shift operator we obtainvk=S^a−1(uk).

Soπv_k(t) =t^{a−1 lnk−1}_(k−1)!^(1/t)1_]0,1], and thenF_vk(x) = 1

(k−1)!

Z 1 0

e^−xtt^a−1ln^k−1(1/t)dt.

For real positiveaandk, we have the asymptotic

∞

X

n=0

(−1)ⁿxⁿ

n!(n+a)^k =x^−a

p

X

i=0

k−1 i

ln^k−1−i(x)(−1)ⁱΓ⁽ⁱ⁾(a) +o ln^k−p(x) x^a

! .

And in particular

∞

X

n=0

(−1)ⁿxⁿ n!(n+a)^k ∼

∞

Γ(a) x^a

ln^k−1(x) (k−1)!. This is directly deduced from the integral expression and the∗relation.

Important remark :

It should be noted that everything that has been mentioned is true for an exponent k > −1 since the functions that appear to be integrable on ]0.1[.

In conclusion :

Pour u(x) = _(x+a)¹ k with k >−1 and a >0, we have π_u(t) =t^a−1ln^k−1(1/t)

(k−1)!

Fu(x)∼

∞

Γ(a) x^a

ln^k−1(x) (k−1)!

5.2 Case x

with k positive.

We will present here the caseu(x) =x^−k etu(0) = 0.

So we’re looking atFu(x) =

∞

X

n=1

(−1)ⁿxⁿ n!n^k .

Here the shift and truncation operators will intervene.

We know that forv(x) = (x+ 1)^−k we have - v(x) =

Z 1 0

t^xln^k−1(1/t) (k−1)! dt.

- Fv(x) = Z 1

0

e^−xtln^k−1(1/t) (k−1)! dt.

By shifting and then truncation (see page 5), we haveFu(x) =x Z 1

0

e^−xtΠ(t)dtwith Π(t) =− Z 1

t

ln^k−1(1/s) s(k−1)! ds.

By truncation, we have thus shownF_u(x) =−x Z 1

0

e^−xtln^k(1/t) k! dt.

Forkreal positive, we have the asymptotic

+∞

X

n=1

(−1)ⁿxⁿ n!n^k =−1

k!

p

X

i=0

k i

ln^k−i(x)(−1)ⁱΓ⁽ⁱ⁾(1) +O

ln^k−p(x) .

And in particular

∞

X

n=1

(−1)ⁿxⁿ n!n^k ∼

∞−ln^k(x) k! . This is directly deduced from the integral expression and the relation∗ In conclusion :

Foru(x) = _n¹_k with k real positive and x >0, we have Fu(x) =−x

Z 1 0

e^−xtln^k(1/t) k! dt F_u(x)∼

∞−ln^k(x) k!

5.3 Case of a negative exponent.

We will study the caseu(x) = (x+a)^k withk >0.

We will start with the caseu(x) = (x+ 1)^k then conclude with the shift operator.

Integral representation ofu(x) = (x+ 1)^p−k : If 0< k <1 and ifpis a positive integer, then

(x+ 1)^p−k= Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!ln^k−p−1(1/t)

(k−p−1)! dt+P_k(x).

wherePk is polynomial.

Proof :

It should be noted first thatt→ t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!ln^k−p−1(1/t)

(k−p−1)! is integrable over ]0,1[.

Let us note up(x) = (x+ 1)^p−k and vp(x) = Z 1

0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!ln^k−p−1(1/t)

(k−p−1)! dt+Qk(x) with Qk a polynomial whose derivative isPk.

A derivation calculation left to the reader shows thatu⁰_p(x) = (p−k)u_p−1(x) etv_p⁰(x) = (p−k)v_p−1(x)+Pk(x).

The casep=0 corresponds tou(x) = _(x+a)¹−k with −k >−1 which has been treated in section 5.1 (under the title important remark).

It is then sufficient to adjustP_k(0) to state thatu_k =v_k.

We can therefore conclude by induction that the Proposition is verified for anyp∈N. Remark :

We have already reported that for a polynomialufunction,Fu(x) =O(ρ^x).

The appearance of the polynomialsPkis therefore of no importance to the asymptotics sought, even if clearly we could calculate these polynomials.

The general caseuk(x) = (x+a)^k. Fork /∈N^∗ we have

Fu_k(x)∼

∞

Γ(a) x^a

ln^−k−1(x) (−k−1)!.

Proof :

We’ll actually take 0< k <1 and process the caseu(x) = (x+a)^k−p. Let’s poseπ_k(t) =^lnk−p−1_(k−p−1)!^(1/t).

In the casea= 1 asuis deduced from a density sequence, we know that Fu(x) = (−1)^p

Z 1 0

e^−xtSp(−xt)Φ^p(πk)(t)dt.

So by the shift operator (see page 4) Fu(x) = (−1)^p

Z 1 0

e^−xtSa−1,p(−xt)t^a−1Φ^p(πk)(t)dt.

We clearly have the implication : Ifπ(t)∼

0

ln^α(1/t)

α! then Φ(π)(t)∼

0

ln^α(1/t) α! . So here, Φ^p(πk)(t)∼

0

ln^k−p−1(1/t) (k−p−1)! . Let’s writeSa−1,p(X) =

p

X

i=0

ba−1,i

(−X)ⁱ i! . As

Z 1 0

e^−xt(xt)ⁱt^a−1Φ^p(π_k)(t)dt∼

∞xⁱΓ(a+i) x^a+i

ln^k−p−1(x) (k−p−1)!, we thus obtainFu(x) = (−1)^p

p

X

i=0

(−1)ⁱb_a−1,iΓ(a+i) i!x^a

ln^k−p−1(x)

(k−p−1)!+o ln^k−p−1(x) x^a

!

at +∞.

We write ^Γ(a+i)_i! = (a+i−1)···a

i! Γ(a) = ^i+a−1_i Γ(a).

The relation

p

X

i=0

(−1)ⁱba−1,i

i+a−1 i

= (−1)^p quoted and established in page 5 then makes it possible to conclude.

In conclusion :

u(x) = (x+a)^k with k >0, u(x) =

Z 1 0

t^x−

p−1

X

i=0

(xln(t))ⁱ i!

!ln^k−p−1(1/t)

(k−p−1)! dt+Qk(x)with Qk polynomial For k /∈N^∗ we have Fu(x)∼

∞

Γ(a) x^a

ln^−k−1(x) (−k−1)!

5.4 Case of a negative pole a.

Asymptotic ofF_u(x) = X

n≥n0

(−1)ⁿxⁿ

n! (n+a)^k aveck >0.

Letn₀−1< a≤n₀ and letu(x) = _(x−a)¹ _k If n0−1< a < n0 thenFu(x)∼

∞(−1)ⁿ⁰x^aΓ(−a)^ln_(−k−1)!^−k−1(x) Ifa=n0 then

Fu(x)∼

∞(−1)ⁿ⁰⁺¹xⁿ⁰ n0!

ln^−k(x) (−k)!

Proof :

We puta=n0−εwith 0≤ε <1.

If we noten=m+n0, we rewriteFu(x) asFu(x) = (−x)ⁿ⁰ X

m≥0

(−1)^mx^m

m!(m+ 1)· · ·(m+n₀)(m+ε)^k.Ifk is an integer, by the case of a rational fraction treated at the beginning, we have

Fu(x)∼

∞(−x)ⁿ⁰KFv(x) wherev(x) =_(x+ε)¹ _k andK= [(1−ε)(2−ε))· · ·(n0−ε)]⁻¹.

For the other values ofk, we will find this result by the principle of no superposition of perturbations in the example paragraph.

- Casn0−1< a < n0: AsFv(x)∼

∞ Γ(ε)

x^ε

ln^−k−1(x)

(−k−1)! and by the relation Γ(ε) = (ε−1)· · ·(ε−n0)Γ(ε−n0), we thus obtain the relationFu(x)∼

∞(−1)ⁿ⁰x^aΓ(−a)^ln_(−k−1)!^−k−1(x). - Casea=n0 :

In this case v(x) =_x¹k andFv(x)∼

∞−ln^−k(x) (−k)! . SoFu(x)∼

∞(−1)^n0+1x_nⁿ⁰

0! ln^−k(x)

(−k)! .

It should be noted that, as in the case of rational fractions, we have the same equivalent if we study Fu(x) = X

n6=n₀

(−1)ⁿxⁿ n! u(n).

5.5 Conclusion.

We consideru(x) = _(x+a)¹ k with xand k real We define n₀= 0 ifa >0 and n₀−1< a≤n₀ if a≤0

Fu(x)∼

∞(−1)ⁿ⁰Γ(a) x^a

ln^k−1(x)

(k−1)! sia /∈ −N F_u(x) =O(ρ^x) sia∈ −N

6 Logarithmic cases.

6.1 Cas ln(x + a).

We will present here the caseu(x) = ln(x+a).

Caseu(x) = ln(x+a).

Ifu(x) = ln(x+a), then

Fu(x)∼

∞− Γ(a) x^aln(x)

Proof :

Letu(x) = ln(x+ 1).

Clearly, by primitization,u(x) = Z 1

0

t^x−1 ln(t) dt.

LetLi(t) = Z t

0

ds ln(s).

Becauseuis deduced from a density sequence, we then haveF_u(x) =− Z 1

0

e^−xtS₁(xt)Φ(Li)(t)dt.

We have the asymptoticLi(t)∼

0 t ln(t),

and then Φ(Li)(t)∼ _ln(t)¹ , so,

Z 1 0

e^−xtLi(t)dt∼

∞x Z 1

0

e^−xt t

ln(1/t) dt(RememberS₁(X) =−X).

Thus, by the relation∗,F_u(x)∼

∞−_xln(x)¹ . Let us treat the casea6= 1 in the same way :

Forv(x) = ln(x+a), by shiftingubyS^a−1,Fv(x) =− Z 1

0

e^−txt^a−1S_1,a−1(xt)Φ(Li)(t)dt, and thenF_v(x) =−

Z 1 0

e^−xt(a−1−xt)Li(t)t^a−1dt, and finally, after calculations left to the reader,Fv(x)∼

∞−_xa^Γ(a)ln(x). In conclusion :

u(x) = ln(x+a)

Fu(x) =− Z 1

0

e^−xt(a−1−xt)Li(t)t^a−1 dt

F_u(x)∼

∞− Γ(a) x^aln(x)

6.2 Case ln

(x + a) with k integer.

Caseu(x) = ln^k(x+a) withkpositive integer.

Ifv_k(x) = ln^k(x+a) for a positive integerk, then F_vk(x)∼

∞(−1)^kka!

x^a

ln^k−1(ln(x)) ln(x) .

Proof :

We’re going to proceed in 3 steps.

1) Study of I_k(x) = Z 1

0

(1−t^x)ln^k(ln(1/t)) ln(1/t) dt.

In this paragraph we will noteck = Γ^(k)(1) = Z ∞

0

ln^k(t)e^−tdt.

By changing the variablee^−t=uwe get Γ^(k)(1) = Z 1

0

ln^k(ln(1/u))du.

We clearly haveI_k⁰(x) = Z 1

0

t^xln^k(ln(1/t))dt.

We writeI_k⁰(x) = Z x

0

t^x−1tln^k(ln(1/t))dtthat we integrate by parts : I_k⁰(x) =

(t^x−1)

x tln^k(ln(1/t)) ¹

0

| {z }

0

− Z 1

0

(t^x−1) x

ln^k(ln(1/t))−ktln^k−1(ln(1/t)) −1/t ln(1/t)

| {z }

ln^k(ln(1/t))+k^lnk−1 (ln(1/t)) ln(1/t)

dt.

ThusxI_k⁰(x) =−I_k⁰(x) +ck−kI_k−1(x).

We then obtain the relation (x+ 1)I_k⁰(x) =ck−kIk−1(x).

We then definePk byIk(x) =Pk(ln(x+ 1)).

The recurrence relation is written asP_k⁰ =ck−kPk−1withPk(0) = 0 and alsoP0=XsinceI0(x) = ln(x+1).

Let’s check by induction the relation

(k+ 1)P_k(X) = Z ∞

0

e^−th

ln^k+1(t)−(ln(t)−X)^k+1i dt:

NoteJk(X) the integral 1 k+ 1

Z ∞ 0

e^−th

ln^k+1(t)−(ln(t)−X)^k+1i dt.

Fork= 0 we clearly have J0(X) =X.

As Jk(0) = 0, to check the relation to the k rank by induction, just check that Jk satisfies the relation J_k⁰(X) =ck−kJ_k−1(X) :

J_k⁰(X) = Z ∞

0

e^−t(ln(t)−X)^k dt=−kJk−1(X) +k Z ∞

0

e^−tln^k(t)dt=−kJk−1+c_k−1. Thus

Z 1 0

(1−t^x)ln^k(ln(1/t))

ln(1/t) dt= 1 k+ 1

Z ∞ 0

e^−th

ln^k+1(t)−(ln(t)−ln(x+ 1))^k+1i

dt=P_k(ln(x+ 1)).

We will remember thatPk a polynomial of degreek+ 1 verifyingPk(0) = 0 and leading coefficient ⁽⁻¹⁾_k+1^k.

2) Calculation of the density associated with ln^k(x+a).

Let’s writeX^k+1=a_0,kP₀(X)+a_1,kP₁(X)+...+a_k−1,kP_k(X), and poseQ_k(X) =a_0,k+a_1,kX...+a_k−1,kX^k. By looking at the leading coefficient we havea_k−1,k= (−1)^k(k+ 1).

By takingX= ln(x+ 1) we have ln^k+1(x) = Z 1

0

(1−t^x)

k−1

X

i=0

ai,k

lnⁱ(ln(1/t)) ln(1/t) . So we have

ln^k+1(x+ 1) = Z 1

0

(1−t^x)Q_k(ln(ln(1/t))) ln(1/t) dt.

Thus, foru_k(x) = ln^k(x+ 1), we haveπ_uk(t) = ^Qk−1(ln(ln(1/t)))

ln(1/t) , that we’ll writeπ_k in the following.

In particularπ_uk(t)∼

0 (−1)^k−1k^lnk−1_ln(1/t)^(ln(1/t)).

3) Equivalent of F_vk at +∞ with v_k(x) = ln^k(x+a).

We so haveuk(x) =− Z

(t^x−1)Q_k−1(ln(ln(1/t))) ln(1/t) dt, and then, forv_k(x) = ln^k(x+a) :

Fv_k(x) = Z 1

0

e^−xS_1,a−1(xt)t^a−1Φ(πk)dt

As Φ(πk)(t)∼πk(t)∼(−1)^k−1k^lnk−1_ln(1/t)^(ln(1/t)), so we haveF_vk(x)∼

Z 1 0

e^−xt(a−1−xt)t^a−1(−1)^k−1kln^k−1(ln(1/t)) ln(1/t) dt, which gives as equivalent

(−1)^k−1k(a−1) Z 1

0

e^−xtt^a−1ln^k−1(ln(1/t))

ln(1/t) dt+ (−1)^kkx Z 1

0

e^−xtt^aln^k−1(ln(1/t)) ln(1/t) dt (provided that the main parts are not simplified)

or, after using the relation∗∗:

F_vk(x)∼ ^(−1)k−1_x^k_aln(x)^{lnk−1(ln(x))}((a−1)Γ(a)−Γ(a+ 1)), which shows thatF_u(x)∼

∞(−1)^kkx^Γ(a)_xa ^lnk−1_ln(x)^(ln(x)). In conclusion :

u(x) = ln^k(x+a) πu(t) =t^a−1Q_k−1(ln(ln(1/t)))

ln(1/t) πu(t)∼

0 (−1)^k−1kt^a−1ln^k−1(ln(1/t)) ln(1/t) Fu(x)∼

∞(−1)^kkxΓ(a) x^a

ln^k−1(ln(x)) ln(x)

7 Principle of superposition of perturbations.

7.1 Preliminaries.

Product of two density sequences.

Proposition :

Let ui two density sequences πi.

The sequence u1u2 admits π=π1? π2 as density withπ(z) = Z 1

z

π1(s)π2(z/s) ds s . Proof :

u1(x)u2(x) = Z 1

0

Z 1 0

t^xs^xπ1(t)π2(s)dsdt.

By putingu=st we obtain Z 1

0

Z t 0

u^x1

tπ1(t)π2(u/t)dt du= Z 1

0

Z 1 u

1

tπ1(t)π2(u/t)dt du= Z 1

0

z^xπ(z)dz, withπ(z) =

Z 1 z

π1(s)π2(z/s) ds s . Properties of the ? operator : 1? is assoiative and commutative.

2 Under conditions of derivability, (π₁? π₂)⁰(z) =−_z¹π₁(z)π₂(1) + Z 1

z

1

s²π₁(s)π₂⁰(z/s)ds.

3 Equivalent formulas : π1? π2(z) =

Z

√z

z

1

s(π1(s)π2(z/s) +π1(z/s)π2(s)) ds.

z(π₁? π₂)⁰(z) =−π₁(√ z)π₂(√

z) + Z

√z

z

(π₁(z/s)π₂⁰(s) +π₁⁰(s)π₂(z/s)) ds.

Proof :

The first point is left to the reader.

The second is obtained by consideringG(x, y) = Z 1

x

π1(s)π2(y/s) ds s. For the third one :

We consider Z 1

√z

1

s²π₁(s)π₂⁰(z/s)ds

where the variable changeu=z/sis performed, which gives Z

√z

z

π1(z/s)π⁰₂(s)dswhich is integrated by parts and gives the result.

Some examples :

- Forπ₁(t) =t^a etπ₂(t) =t^b :

We haveπ₁? π₂(z) = ^z_b−a^b−za ifa6=b,z^aln(1/z) ifa=b.

- For π₁(t) = ln^a(1/t) et π₂(t) = ln^b(1/t), we have π₁? π₂(z) = B(a+ 1, b+ 1) ln^a+b+1(1/z), B desig- nating the Euler function.

Indeed: The definition gives Z 1

z

ln^a(t/z) ln^b(1/t) dt t

u=ln(1/t)

=

Z ln(1/z) 0

(ln(1/z)−u)^au^b du, the change of variableu=vln(1/z) giving the result.

- Forπ₁(t) =t^a etπ₂(t) =t^bln^c(1/t), we haveπ₁? π₂(z) = Z 1

z

t^b−a−1ln^c(1/t)dt ∼

+∞z^b−a−1ln^c(1/z).

- Forπ(t) =t^a−1 on aπ ?· · ·? π

k fois (t) =t^{a−1 lnk−1}_(k−1)!^(1/t) (by an easy induction), which makes it possible to find the caseu(x) = _(x+a)¹ k.

7.2 Power part of a function.

Definition :

If f is a given function over ]0,1]and under the condition of existence, we will talk about the power part of f, noteda(f) = lim

t→0 tf⁰(t)

f(t) . Remarks :

Iff admits a power part a, then F(x) = Z x

0

f(t) dt admits a power parta+ 1.

Iff and g admit a power part then a(f g) =a(f) +a(g).

Iff admits a power part then we have c₁t^a(f)+ε≤f(t)≤c₂t^a(f)−ε over]0,1].

Proof:

-a(f) =ais equivalent toa(g) = 0 forg(t) =t^−af(t).

- Ifa(f) = 0, thenf(t) =e^α(t)whereα⁰(t) =o(1/t).

IfF(x) = Z x

0

f(t)dt, then Z x

0

e^α(t)dt=h te^α(t)i^x

0− Z x

0

tα⁰(t)e^α(t)dt

| {z }

=o(F(x))

.

Thus ^xF_F(x)^0(x) →1.

The other two points are immediate.

7.3 Problem of overlapping perturbations.

Theorem :

We give u_i with density π_i i= 1,2, so monotonous around 0.

1 Principle of non-superposition of perturbations :

If π2(t) =O t^b

with b > a(π1)

and if π⁰₁ admits a power part, then

π1? π2∼

0 C(π1, π2)π1

and therefore F_u1.u2∼

∞CF_u1 where C(π₁, π₂) =

Z 1 0

t^{−a(π1)−1}π₂(t)dt 2 Principle of superposition of perturbations :

If π1 and π2 admit a power part witha(π1) =a(π2) then

a(π1) =a(π2) =a(π1? π2)

Let’s mention two very important applications before giving the proof : Ifa(π1)< a(π2),· · ·, a(πk) withπ₁⁰ admitting a power part, then for 2≤i≤k, we haveπi(t) =O(t^b) for oneb > a(π1),

soπ2?· · ·? πk(t) =O(t^b), and thus

Fu1···u_k ∼

+∞C(u1, u2?· · ·? πk)Fu1

Ifa(π₁) =a(π₂)< a(π₃),· · ·, a(π_k), meanwhile Fu₁···uk ∼

+∞C(u1? u2, π3?· · ·? πk)Fu₁u₂

Proof : First case :

Leta=a(π1)> b,π1(t) =t^aπ(t).

We thus know thata(π) = 0 andt^−aπ2(t) =O(t^α) withα >0.

putf(t) =t^−a−1π2(t) whitch is integrable over ]0,1].

π₁? π₂(z) = =z^a Z 1

z

f(s)π(z/s)ds.

It is therefore sufficient to show Z 1

z

f(s)π(z/s)ds∼

0 π(z) Z 1

0

f(s)ds.

To do this, we integrate by parts by putingF(z) = Z z

0

f(s)ds: Z 1

z

f(s)π(z/s)ds=F(1)π(z)−F(z)π(1) + Z 1

z

z

s²F(s)π⁰(z/s)ds.

a(f)>−1 thusa(F)>0 and so we have an inequality such as F(z)≤cz^ε, which shows thatF(z)π(1) =o(π(z)).

Byz/s=u, the last integral is equal to Z 1

z

F(z/s)π(s)ds, and we have to show that it’s negligible in front ofpi(z).

a(f)>−1 soa(F)>0 and we have an inequality such asF(t)≤Ct^α. The last integral is therefore increased by

Z 1 z

z s

^α

π⁰(s)ds=H(z).

H(z) =z^α Z 1

z

s^−α+1sπ⁰(s)ds= z^α α

−s^−αsπ⁰(s)¹

z+z^α α

Z 1 z

s^−α(sπ⁰(s))⁰ ds

=O(z^α) +1 αzπ⁰(z)

| {z }

o(π(z))

+z^α α

Z 1 z

s^−α(sπ⁰(s))⁰ ds.

But asa(sπ⁰(s)) = 0, (sπ⁰(s))⁰ =o(π⁰(s)) there are two cases.

- If ^π_s^0(s)_α is integrable over ]0,1], then z^α α

Z 1 z

s^−α(sπ⁰(s))⁰ ds=O(z^α).

- Else, z^α α

Z 1 z

s^−α(sπ⁰(s))⁰ ds=o

z^α Z 1

z

π⁰(s) s^α ds

=o(H(z)).

Finally, the relationa(π) = 0 providesz^α=o(π(z)).

Enfin la relationa(π) = 0 fournitz^α=o(π(z)).

Second case :

Let’s start with a lemma : Lemma :

Ifπ1,π2 are locally positive and monotonous around 0, from part to zero power, then π₁(√

z)π₂(√ z) =o

Z

√z

z

π₁(s)π₂(z/s) ds s

! . Proof of the lemma :

Let’s noteπ(z) for the integral. First of all, let’s notice the equality:

Z

√z

z

π1(s)π2(z/s) ds s =

Z 1

√z

π1(z/s)π2(s) ds s

Case π1 decreasing and π2 increasing : Z

√z

z

1

sπ₁(s)π₂(z/s)ds≥π₁(√ z)

Z

√z

z

π₂(z/s) ds

s =π₁(√ z)

Z 1

√z

π₂(s) ds

s ≥π₁(√ z)π₂(√

z) ln(1/√ z).

Casπ1 increasing et π2 decreasing : Z 1

√z

π1(z/s)π2(s) ds

s ≥π2(√ z)

Z 1

√z

π1(z/s) ds

s =π2(√ z)

Z

√z

z

π1(s) ds

s ≥π1(√ z)π2(√

z) ln(1/√ z).

Casπ1 et π2 decreasing :

Ifπ₂(0) = 0, by monotony and positivity we would have the trivial caseπ₂= 0, so

Z 1 0

π2(s)

s dsdiverge.

But becausea(π2) = 0, we haveπ2(1/2)−π2(z) = Z 1/2

z

π₂⁰(s)ds=o Z 1

z

π₂(s) s ds

andπ2(z) =o Z 1

z

π2(s) s ds

. Thusπ2(√

z) =o Z 1

√z

π2(s) s ds

. The inequalityπ(z) =

Z

√z

z

1

sπ₁(s)π₂(z/s)ds≥π₁(√ z)

Z 1

√z

π₂(s)

s dsallows us to conclude.