H Aygun

K Attenborough

A Cummings

PREDICTED EFFECTS OF FLUID LOADING ON THE VIBRATION OF ELASTIC POROUS PLATES

The effects of fluid loading on the vibration of rectangular, clamped, porous elastic plates and on their radiated sound power are considered. This requires the inclusion of an extra term in the equations for the plate vibration, corresponding to the additional external force acting on the plate. As is the case for vibrating non-porous plates, fluid-structure coupling is a very complex phenomenon since the plate modes are coupled by the fluid. The radiation impedance matrix, including direct terms and cross-coupling terms, has been defined and computed. A Gaussian quadrature scheme including twenty terms of the Legendre polynomial has been used to compute the fluid loaded plate deflection for four types of porous and elastic plate and compare the effects of the loading by water and air. The corresponding vibroacoustic indicators including mean square velocity, radiated sound power and radiation efficiency, have been calculated also. For some plate and fluid parameters, the predicted effects of fluid loading are considerable.

INTRODUCTION

The vibrations of fluid loaded, infinite non-porous plates have been studied extensively [START_REF] Maidanik | Influence of fluid loading on the radiation from infinite plates below the critical frequency[END_REF][START_REF] Maidanik | Influence of fluid loading on the radiation from orthotropic plates[END_REF][START_REF] Feit | [END_REF][START_REF] Sandman | Motion of a three-layered elastic-viscoelastic plate under fluid loading[END_REF][START_REF] Sandman | [END_REF]. Sandman [START_REF] Sandman | [END_REF] has solved the governing equations of motion for the normal modes of free vibration of a fluid loaded, simply supported non-porous plate carrying a concentrated mass and has presented examples to illustrate the basic characteristics of the fluid-loaded vibration and radiation from a plate with such a discrete mass discontinuity.

Berry [6] has studied the vibration and sound radiation of fluid-loaded plates with elastic boundary conditions. Berry's approach is based on a variational formulation of the fluidloaded plate motion and a Rayleigh-Ritz method using polynomial displacement functions. The selection of such polynomial functions leads to a new method of solution for the radiation impedance of the plate. Theodorakopoulos and Beskos [START_REF] Theodorakopoulos | Flexural vibration of poroelastic plates[END_REF] have given a rigorous formulation for vibration of a simply supported rectangular plate. Leclaire [START_REF] Leclaire | [END_REF] has developed this formulation to derive a set of equations corresponding to any type of plate (clamped, simply supported and free). Foin [9] has evaluated the acoustic pressure exerted by fluid on a baffled plate by using the Rayleigh integral.

Davies [START_REF] Davies | Low frequency random excitation of water-loaded rectangular plates[END_REF] and Pope and Leibowitz [11] have calculated the radiation impedance matrix for a fluid-loaded simply supported non-porous plate. The coefficients of this matrix express the cross-coupling of in vacuo modes by the fluid. The real part of the radiation impedance matrix is the radiation resistance and expresses the radiation damping of the structure. The imaginary part of the radiation impedance matrix is the radiation reactance and expresses the added mass of fluid on the structure. Berry [6] has calculated the radiation impedance matrix in terms of simple trial displacement functions. The radiation impedance matrix has been rigorously calculated by Foin [9], Sandman [START_REF] Sandman | Motion of a three-layered elastic-viscoelastic plate under fluid loading[END_REF] and Nelisse [12]. Sandman and Nelisse have reduced the resulting quadruple integral into a double integral by using a specific change of variable. They integrated the double integral numerically by means of Gaussian quadrature.

The effects of the fluid loading on the vibration of a rectangular, clamped porous and elastic plate and on the radiated sound power are investigated here. The effect of fluid loading can be incorporated by inserting an extra term in the equations of the plate, corresponding to an additional external force acting on the plate. This leads to the calculation of radiation impedance matrices with non-negligible cross terms. The radiation impedance matrix for a plate has been predicted directly by simply solving the equation of the coefficients of the radiation impedance matrix [9] without interpolation, convergence and reducing the quadruple integral to a double integral with the direct terms and cross-coupling terms [START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF]. A Gaussian quadrature scheme including twenty terms of the Legendre polynomial has been used to compute the fluid loaded plate deflection.

EQUATIONS FOR THE POROUS PLATE MATION

A thin, baffled porous plate of dimension a x b and uniform thickness h, excited by a force density F(x, y, t) is considered. The geometry of a porous plate is shown in Figure 1. The bending vibration of a rectangular porous plate of thickness h is given by equations [START_REF] Leclaire | [END_REF]: .

( ) ∑∑ ∑∑ ∞ = ∞ = ∞ = ∞ = = - ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ - + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + 1 1 6 2 6 2
(

) 1
where D is the bending stiffness of the plate, α and M are Biot's elastic coefficients [START_REF] Biot | The elastic coefficients of the theory of consolidation[END_REF] that can be related to known or measurable elastic constants, is the plate lateral displacement, is the fluid solid relative displacement in the pores, s mn W mn W ρ is the bulk density of the solid- fluid mixture and f ρ is the density of the fluid. The coefficient m(ω ) is a frequency dependent mass [START_REF] Biot | Mechanics of deformation and acoustic propagation in porous media[END_REF] which accounts for the viscous friction between the fluid and the solid. This coefficient depends on the porosity, permeability and tortuosity. The first equation corresponds to the instantaneous elastic response of the plate and the second equation describes the relative motion between the solid and the fluid, including energy losses by viscous friction [START_REF] Biot | Theory of buckling of a porous slab and its thermo elastic analogy[END_REF].

The simple definite integrals (I 1 )-(I 6 ) are given by:

∫ = a m IV m dx x X x X I 0 ) ( 1 ) ( ) ( , , , ∫ = a m m dx x X x X I 0 6 ) ( ) ( ∫ = a m II m dx x X x X I 0 ) ( 3 ) ( ) ( ∫ = b n IV n dy y Y y Y I 0 ) ( 5 ) ( ) ( , , . (2)
∫ = b n n dy y Y y Y I 0 2 ) ( ) ( ∫ = b n II n dy y Y y Y I 0 ) ( 4 ) ( ) (
The exponents (IV) and (II) denote the fourth and second order derivative, respectively. By solving equation ( 1) for m and n modes, the coefficients and can be found. Hence the deflection of a porous plate can be calculated from;

s mn W mn W ) ( ). ( ) , ( 1 1 y Y x X W y x w n m m n s mn s ∑∑ ∞ = ∞ = = , . (3)
) ( ). ( ) , ( 1 1 y Y x X W y x w n m m n mn ∑∑ ∞ = ∞ = =
The beam functions are given by: )

( and ) ( y Y x X n m ) / sin( ) / sinh( ) / cos( ) / cosh( ) ( 4 3 2 1 a x a B a x a B a x a B a x a B x X m m m m m m m m m + + + = , ) / sin( ) / sinh( ) / cos( ) / cosh( ) ( 4 3 2 1 b y b C b y b C b y b C b y b C y Y n n n n n n n n n + + + = . ( 4
)
where and are the frequency parameters corresponding to the m m a n b th and n th roots of a characteristic equation. The constants [START_REF] Szilard | Theory and Analysis of Plates[END_REF] are determined from the boundary conditions at the edges of the plate, and allow for the imposition of any condition involving simply supported, clamped or free edges.

4 3 2 1 , , , n n n n C C C C 4 3 2 1 , , , , m m m m B B B B
For a simply supported plate, the integrals have simple forms with solutions similar to Navier's algebraic solution [START_REF] Szilard | Theory and Analysis of Plates[END_REF] for a porous plate [START_REF] Theodorakopoulos | Flexural vibration of poroelastic plates[END_REF].

The coefficients Q mn and ∆P mn correspond to the external exciting forces. For a load F 0 concentrated at one point (x 0 , y 0 ) of the solid surface (a shaker for example), the forces Q mn. I 2 I 6 and ∆P rn .I 2 I 6 appearing in Eq.( 1) are respectively equal to F 0

) ( ) ( 0 0 y Y x X n m and 0.
Detailed derivations of Q mn and ∆P mn can be found elsewhere [START_REF] Deresiewicz | [END_REF].

DEFLECTION OF FLUID LOADED POROELASTIC PLATES

By using Hamilton's method:

(5)

∫ ∫ + + - = 0 1 0 1 ) ( ) ( t t t t fluid excitation dt W dt W V T w H δ δ
where T is the kinetic energy of the plate, V is the strain energy, is the work done by the excitation force,

A W A W A V A T dt d ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . ( 6
)
The vibrational response of the fluid-loaded porous plate can be obtained by inserting an additional external force (

[ ]{ } pq T mnpq W Z j , ω -
) acting on the plate in Eq.( 1). At a given angular frequencyω , in the frequency domain (i.e. by Fourier transform of Eq.( 6)), the equation of porous plate motion can be written in matrix form. The matrix form of the equation of fluid loaded porous plate motion is given by;

[ ]{ } [ ]{ } [ ] { } ( ) [ ]{ } pq T mnpq mn mn mnpq s mn s mnpq s mn mnpq W Z j q W K W K W M , 1 1 1 2 ω ω - = + + - [ ]{ } [ ]{ } [ ] { } ( ) [ ]{ } pq T mnpq mn mn mnpq s mn s mnpq s mn mnpq W Z j p W K W K W M , 2 2 2 2 ω ω - = + + - . ( 7
)
where [ ] The coefficients of the mass matrix are defined by:

[ ] ( ) ∑ ∑ ∞ ∞ + + ⎟ ⎞ ⎜ ⎛ + 3 2 Mh α 1 mnpq M = = = ⎟ ⎠ ⎜ ⎝ 1 1 6 5 4 3 2 1 2 12 m n I I I I I I D , [ ] 2 mnpq M = . ( 8
)
The coefficients of the complex stiffness matrixes are defined by:

( ) ∑ ∑ ∞ = ∞ = + 1 1 6 4 3 2 m n I I I I Mh α [ ] 1 s mnpq K = = = - 1 1 6 2 m n I I hρω , ∑∑ ∞ ∞ 2 [ ] 1 ∞ ∞ - 2 ω ρ mnpq K = = = 1 1 6 2 m n f I I h , ∑ ∑ [ ] 2 s mnpq K = , ∑∑ ∞ = ∞ = 1 1 6 2 2 m n f I I h ω ρ [ ] 2 mnpq K = . ( 9
)

RADIATION IMPEDANCE MATRIX

edance matrix are given by [START_REF] Sandman | Motion of a three-layered elastic-viscoelastic plate under fluid loading[END_REF]12] (

where is known as the acoustic radiation impedance between the normal modes (m, n)

q), ∑∑ ∞ = ∞ = 1 1 6 2 2 ) ( m n I I hm ω ω
The cross-coupling terms in the radiation imp

∫∫∫∫ ′ ′ ′ ′ ′ = S d
′ ′ = ′ ′ π π ψ
, respectively. The Green function is given by The radiation impedance matrix for a been predicted directly fro Eq.10, without interpolation, convergence and reducing the quadruple integral to a double integral. A Gaussian quadrature scheme with thirteen modal functions in each direction (x, y) has been used. When the variables (x, y) have the values (x′, y′), for which ) 0 , , ; , , (

R R jk y x z y x G π 2 / ) exp( ) 0 , , ; , , ( 0 -
= ′ ′ (11) .
y x z y x G ′ ′
goes to infinity, care should be taken to avoid any singularity. The rigin can be avoided by taking G(0,0) = 0. A point force excitation is assumed at x = 0.08 m, y = 0.07 m assumed properties of the (non-porous) plate are given in Table 1.

∫ ∫∫ ∂ ∂ = T S dSdt t t y x w abT V 0 2 2 | ) , , ( 1 .
(

) 12
where

ω π / 2 = T
, and a, b are the coordinates of plates in x and y direction, respectively. S is the surface area of the plate and ω is the angular frequency of the plate.

The radiated sound power expresses the sound energy radiated by the plate. It is obtained from the integration of the intensity over the plate surface. The radiation sound power [9] is given by .

) , , ( ) , 0 , , ( 1 0 ∫
∫∫ ′ ′ ′ ′ ′ - = S f S d y x z y x G y x w z y x P . ) 0 , , ; , , ( ) , ( ) , , ( 2 ω ρ
This is often called the Rayleigh integral. The radiation efficiency expresses the ratio of the vibration energy transformed into sound power. The radiation efficiency of the plate is given [9] by; ρ is the density of air and is the sound speed in air. The radiation efficiency is calculated from the radiated sound power and the mean square velocity of the plate. 0 c

NUMERICAL RESULTS

The coefficients and can be found by solving Eq.1 for the m, n mode. Hence the deflection of porous plates can be calculated by from Eq.3. Then the mean square velocity, the radiated sound power and radiation efficiency can be computed. A Gaussian quadrature scheme with twenty terms of the Legendre polynomial has been used to compute the fluid loaded plate deflection. The effects of the fluid loading on the plates have been studied for three different plates with the properties listed Table 2. The Coustone plate contains flint particles with a mean grain size of about 1mm and an epoxy rubber binder. The G foam and YB10 foam are fabricated from particles of plastic foam obtained from recycled car dashboards. Their grain size varies from a few tens of Figure4: Predicted deflections of the G foam plate Figure5: Predicted deflection of the YB10 foam plate in vacuo and loaded with water [START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].

in vacuo and loaded with either air or water [START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].

The computed spectra of the plate deflections are shown in Figures 456, for G foam, YB10 foam and Coustone, respectively. The magnitudes of the predicted deflection spectra of the plate in vacuo and water-loaded deflection of the plate differ by factors of up to 1000. There is very little difference between the predicted responses of YB10 foam plate in air and in vacuo throughout frequency range. The predicted deflection spectrum of the YB10 foam plate in water is less than the spectra predicted either in vacuo or in air. The Coustone plate is predicted to be relatively little affected by fluid loading. The YB10 plate (see Table 2) was manufactured at the University of Bradford from recycled car dashboards. Its properties have been used to determine the vibroacoustic indicators. A Gaussian quadrature scheme is also used to expand the mean square velocity with sixteen modal functions in each direction (x, y). The plates are assumed to be excited by a point force of 1 N applied at x = 0.1 m, y = 0.09 m from a corner of the plates. The responses of the plates are calculated at the middle of each plate. Figure 7 compares the predicted frequency dependence of the mean square velocity of the YB10 foam plate in water, in air, and in vacuo.

There is no difference between the predicted deflections of air-loaded and in-vacuo YB10 plates. But air loading is predicted to decrease the mean square velocity. The effects of fluid loading may be explained by different patterns of vibration. Although fluid (air) loading does not change the shape of curve, it decreases the mean square velocity by about 15 dB. With water loading the shape of the velocity curve is changed, and also the velocity is decreased by about 20 dB. The differences between the predicted responses in water and in air are very small. So it can be said that fluid loading decreases the mean square velocity of poroelastic materials. YB10 plate in vacuo and loaded with air or water [START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF]. plate loaded with air and water [START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].

The frequency dependence of the radiated sound power of the YB10 foam plate predicted in air and in water is shown in Figure 8. The radiated sound power predicted in vacuo is zero since the sound speed and density in vacuum is zero. Air loading decreases the radiated sound power by about 30 dB in comparison with water loading response. The predicted radiation efficiencies of the YB10 plate loaded with air and water are compared in Figure 9. The predicted radiation efficiency of the plate in air is higher than that predicted in water. With water loading the radiation efficiency of the porous plate is decreased by between 10 dB and 20 dB. The predicted radiation efficiency of the plate in-vacuo response is zero because the radiated sound power in-vacuo is zero.

CONCLUSION

The fluid-loaded vibration of a clamped porous elastic plate has been predicted by the solving the governing equations for flexural vibration. A suitable trial function has been used to calculate the radiation impedance matrix. Calculations have been presented to illustrate the effects of fluid-loading on the vibration and radiation from different porous plates. The deflections of fluid-loaded porous plates have been calculated and compared to the in-vacuo deflection of porous plates for three sets of plate properties and assuming either water or air saturation. The predicted frequency responses of the fluid-loaded plate vibration and radiation have been investigated. Generally speaking, with air and water loading, the predicted radiation efficiency, mean square velocity, and radiated sound power of porous plates are less than the predicted in vacuo values.

ACKNOWLEDGMENT

This investigation has been supported by EPSRC (UK) grant number GR/R43761.

Figure1:

  Figure1:The geometry of a baffled porous plate.

  is the work done by the acoustic pressure exerted by the fluid on the plate, is the displacement of the plate, and t excitation W fluid W w 0 and t 1 are arbitrary times. The formulation of porous plate motion can be written in terms of the Lagrange's equations by using the stationary condition for the Hamiltonian H:

  transverse solid lateral displacement, { } mn W is the magnitude of the transverse fluid-solid relative displacement, { } pq T W , is the total velocity in the vicinity of the plate surface, ( ) mn q is the vector of excitation force, ( ) mn p is the pressure difference in the fluid between two surfaces, and [ ] mnpq Z is the radiation impedance matrix.

,

  and k is the wave num where R is the distance between vectors,

Figure2:

  Figure2: Direct terms in the real part of the radiation impedance matrix for (a) even modes and (b) odd modes.

Figure3:

  Figure3: First two cross coupling terms in the real part of the radiation impedance matrix.

  Three vibroacoustic indicators are in common use. The mean square velocity is defined as a time-space average of the squared vibration velocity of the plate and indicates the global behavior of the plate vibration. It is given by [9];

.

  5 mm. The properties of the saturating fluid have been assumed to be either those of water or those of air.Table2. Characteristics of three types of plate

Figure 6 : 7 :

 67 Figure 6: Predicted deflections of the Coustone plate Figure 7: Predicted mean square velocity of the in vacuo and loaded with water [13].YB10 plate in vacuo and loaded with air or water[START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].

Figure 8 :

 8 Figure 8: Predicted Radiated Sound Power from YB10Figure9: Predicted radiation efficiency of YB10 plate loaded by air and water[START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].plate loaded with air and water[START_REF] Aygun | The design of noise attenuating devices incorporating elastic porous structures[END_REF].


Frank Vega Triangle

Frank Vega
email: vega.frank@gmail.com

Triangle Finding

Keywords: 2012 ACM Subject Classification Theory of computation algorithm, undirected graphs, lower bound, matrix 1 Problem

4

The Algorithm

Given an undirected graph of n vertices represented by a Boolean adjacency matrix, then this should be a valid adjacency matrix for an undirected graph: for the symmetry(equals to its transpose) and the main diagonal is filled only with zeroes. These are the steps that we implicitly follow in the Algorithm 1: 1. First, fill the main diagonal with ones. 2. Next, set to zero all the one values in the entries of the matrix which are above of the main diagonal.

3.

If we find a rectangle whose corners are 1's in the final modified matrix such that the upper right corner is an entry of the main diagonal, then there must be a triangle in the undirected graph input. 4. Otherwise, there is no possible triangle in the undirected graph input.

Theorem 1. After these previous steps there is a triangle in the undirected graph if and only if we find a rectangle whose corners are 1's in the final modified matrix such that the upper right corner is an entry of the main diagonal.

Proof. Since the matrix is symmetric, then we have matrix [c][a] = matrix [a][c] for every pair of nodes a and c. In addition, since we fill the main diagonal with ones, then matrix [a][a] has the value of 1 for every node a (1 is equivalent to the value of true into the Boolean adjacency matrix). Consequently, we have three nodes b < a < c that is a triangle if and only if the entries matrix [a][b], matrix [a][a], matrix [c][b] and matrix [c][a] represent the corners of a rectangle. This rectangle should have all the corners as 1's and the upper right corner must be on the main diagonal. There could be rectangles with corners 1's where the upper right corner is not on the main diagonal, but the algorithm ignores these cases.

Runtime

This takes the running time O(n 2 ) assuming the Binary AND operation could be done in constant time.

Code

This work is implemented into a GitHub Project programmed in Java [3]. We show the Algorithm 1 in pseudo coding.

Programming Techniques

We use the Binary AND Operation between two integers for finding the rectangle whose corners are 1's such that the upper right corner is an entry of the main diagonal.

Highlights

We make the number numbers [j] shorter with the Binary AND Operation (numbers[j] & all) just forcing to have the final result with at most i + 1 binary digits. Hence, the another Binary AND Operation with the number numbers[i] could have the complexity running time as hard as the next comparison lhs > rhs in the following line of coding since the number numbers[i] has exactly i + 1 binary digits. Indeed, we can deduce the comparison operation and the the Binary AND Operation have the same order in their worst cases inside of our Algorithm 1 since they have to check or operate within at most i + 1 binary digits. return "no" 36: end procedure


Luc Abergel

OF SIGNED ENTIERE Luc Abergel Asymptotics

ASYMPTOTICS OF SIGNED ENTIERE SERIES

à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction

The purpose of this work is to study some perturbations in the series e -x = ∞ n=0 (-1) n x n n! .

So we want to give an asymptotic at +∞ of ∞ n=0

(-1) n x n 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 u n gives a perturbation of the sum whose order of magnitude is x n . The case of the e -x series 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 serie u n is not defined for all n, choosing to look at the sum from a certain rank or for the n values for which the terms u n 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 the e -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 + a 1 ) k1 and (n + a 2 ) 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 u n then the sum ∞ n=0

(-1) n x n 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

1 Professor, Lycée Janson de Sailly, PARIS, email : lucabergel@cegetel.net 0 by this method stricto 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)! for a / ∈ -N and B(x, y) = 1 0 t x-1 (1 -t) y-1 dt.

1 Case of a rational fraction.

We're going to study here

∞ n=0 (-1) n x n n! R(n)
where R is a rational fraction. We will note here that if u is polynmial, then F u (x) = O(ρ x ) (this being done by treating the case of the polynomial X(X -1) • • • (X -k + 1)).

1.1 A first case.

We're studying the case u(x) = 1 x+a with (a) > 0. On the one hand, we write ∞ x t a-1 e -t dt = O(x a-1 e -x ), on the other hand

x 0 t a-1 e -t dt = ∞ n=0 (-1) n x n+a n!(n + a)
.

This shows the formula :

∞ n=0 (-1) n x n n!(n + a) = x -a Γ(a) + O(x a-1 e -x ).
The error term is well in O(ρ x ).

1.2 Generalization to the case a / ∈ -N.

Proposition : The case u(x) = 1 (x+a) with a / ∈ -N. We have the following asymptotic :

∞ n=0 (-1) n x n n!(n + a) ∼ ∞ Γ(a)
x a si a / ∈ -N.

Proof :

As example, we will only deal with the case of -1

< (a) ≤ 0. Let a = -1 + b with (b) > 0.
We write

∞ n=0 (-1) n x n n!(n + a) = ∞ n=0 (-1) n x n n!(n -1 + b) = 1 -1 + b + ∞ n=0 -(-1) n x n+1 (n + 1)!(n + b) . Then ∞ n=0 (-1) n x n n!(n + a) = 1 b -1 -x ∞ n=0 (-1) n x n n! 1 (n + 1)(n + b) = 1 b -1 -x ∞ n=0 (-1) n x n n! 1 n + 1 - 1 n + b 1 b -1 . Let ∞ n=0 (-1) n x n n!(n + a) = 1 b -1 - x b -1 Γ(1) x -x -b Γ(b) + O(ρ x
) , and we deduce :

∞ n=0 (-1) n x n n!(n + a) = x -b+1 Γ(b) b -1 + O(ρ x ).
This is the formula for 1.1 but in the more general case (a) > -1.

The case a / ∈ -N is also written in the same way.

1.3 Case of a multiple pole.

Case u(x) = 1 (x+a) k with a / ∈ -N and p ∈ N. We have the following asymptotic :

∞ n=0 (-1) n x n n!(n + a) k ∼ ∞ Γ(a) x a ln k-1 (x) (k -1)! si a / ∈ -N.
Proof :

Again, we will only deal with one case, the k = 2 case, the working method becoming generalized without any difficulty for the other values of k.

We star with

∞ n=0 (-1) n x 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 where u is polynomial :

We're dealing with the case

u k (x) = x(x -1)...(x -k + 1) that gives F u k (x) = ∞ n=0 (-1) n x n (n -k)! .
And so

F u k (x) = (-x) k e -x = O(ρ x ).
The case of a rational fraction.

If we give u(x) = P (x) Π 1≤i≤p (x + a i ) pi with P polynomial, with integer exponents p i strictly positive, if we have (a 1 ) < (a i ) pour tout i = 1, and if we finally notice

p = p 1 , a = a 1 so ∞ n=0 (-1) n x n n! u(n) ∼ ∞ A Γ(a) x a ln p-1 (x) (p -1)! . with A = P (-a) Π 1≤i≤p (ai-a) p i .

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

∞ n=p+1 (-1) n x n n!(n -p) k .
We have the asymptotic

∞ n=p+1 (-1) n x n n!(n -p) k ∼ ∞ (-1) k+1 x p p! ln k (x) k! .
Proof :

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

∞ n=p+1 (-1) n x n n!(n -p) k = (-x) p+1 ∞ n=0 (-1) n x n 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

1 (n+1) k+1 : 1 (n+1) k+1 (n+2)•••(n+p+1) = 1 p! 1 (n+1) k+1 + • • • . The equivalent sought is therefore (-x) p+1 1 p! ln k (x) k! Γ(1)
x .

So

∞ n=p+1 (-1) n x n n!(n -p) k ∼ ∞ (-1) p+1 x p p! ln k (x) k! .
It should be noted, given the equivalent obtained, that the result is the same for

n =p (-1) n x n n!(n -p) k .
An example where several poles have even real parts. Let's deal with the case

F (x) = ∞ n=0 (-1) n x n n!(n 2 + 1) .
We write

1 n 2 +1 = i 2 1 x+i -1 x-i . This gives us F (x) = i 2 Γ(i)x -i -Γ(-i)x i + O(ρ x ), or, by writing Γ(i) = ρe iθ : F (x) = +∞ ρ sin(ln(x) -θ) + O(r x ) avec r < 1.
2 Integral form and applications.

Notations and working context :

-We consider functions defined from a certain rank n 0 and we will talk about sequences by looking at the values of such functions over fairly large integers.

-For a given sequence u we note

F u (x) = ∞ n≥n0 (-1) n x n n! u(n).
For a given sequence u :

-If we can write u(x) = 1 0 t x π(t) dt for a function π and for any x ≥ 0, then we'll say that the u is density sequence. It will be said to be associated with π that we will note π u .

-If we can write u(x) =

1 0 t x - p-1 i=0 (x ln(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 assume π monotonous 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, F u (x) = 1 0 e -tx π u (t) dt, it is then the study of this integral that gives an asymptotic of F u in + + ∞. So we'll look for an asymptotic at O(ρ x ) with ρ < 1.

Therefore, sometimes we will write'=' for two terms that differ by a O(ρ x ) (always with ρ < 1).

Finally, let's note that the values of u(x) for x large essentially involve the behavior of π in 1, while those of F u in +∞ that of π in 0.

3 First properties.

-Derivation :

If u is a density sequence associated to π u , then u also is, associated to π u = ln(t)π u (t).

-Primitivation :

If U (x) = x 0 u(t) dt, then U (t) = 1 0 (1 -t x ) π u (t) ln(1/t)
dt. This sequence is therefore deduced from a density sequence.

-Truncation :

The truncated sequence T u(0

) = 0 et T u(n) = u(n) for n ≥ 1 verifies F T u (x) = 1 0 (e -xt -1)π u (t) dt.
If π is integrable in 0 (so that u(0) is defined), noting Π u the primitive of π u which is vanishes in 1,

F T u (x) = x 1 0 e -xt Π 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. If u is a density sequence, then S α (u) is associated to t α π u (t).

And so

F S α (u) (x) = ∞ 0 e -xt t α π u (t) dt.
Proof :

We write S α (u)(x) =

1 0 t x t α π u (t) dt.
This then gives

F S α (u) (x) = ∞ 0 e -xt t α π u (t) dt.

Remark :

For example, if we consider the sequence u(x) = 1 ln(x+2) 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 the u sequence from the rank 1.

It is therefore necessary to consider

F T S -1 (u)(x) = x 1 0 e -xt Π u (t) dt with Π u (t) = - 1 t π u (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 of π, we're interested in the sequence u(x) =

1 0 t x - p-1 i=0 (x ln(t)) i i! π(t)
dt and we want to study F u in +∞. This will require some definitions.

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

-For a function π that can be integrated on ]0.1[, we define the operator Φ by

Φ(π)(t) = 1 t t 0 π(s) ds.
-For a positive integer p, we note Q α,p the polynomial defined by S α,p (X) =

p i=0 b α,i (-X) i i! with b α,i = ∆ i ((X + α) p )(0).
In the case of α = 0, we will simply note b i = b α,i and S p = S 0,p .

Let us note a remark that will be used later :

Note on coefficients b α,i :

p i=0 (-1) i b α,i α + i i = (-1) p .
Proof :

This comes from -α-1 i = (-1) i α+i i and from (X + α) p = p i=0 b α,i X i which is applied in X = -α -1. Let be π an integrable function on ]0, 1[. If u(x) = 1 0 t x - p-1 i=0 (x ln(t)) i i! π(t) dt, then : -u(x) = (-x) p 1 0 t x Φ p (π)(t) dt. -F u (x) = (-1) p 1 0 e -xt S p (xt)Φ p (π)(t) dt.
Proof :

First point :

We integrate by parts

1 0 t x - p-1 i=0 (x ln(t)) i i! π(t) dt.
We then obtain

t x - p-1 i=0 (x ln(t)) i i! tΦ(π)(t) 1 0 - 1 0 x t x-1 - p-1 i=1 (x ln(t)) i-1 t(i -1)! tΦ(π)(t) dt, then (-x) 1 0 t x - p-2 i=0 (x ln(t)) i i! Φ(π)(t) dt,
and we conclude by induction.

Second point :

Note N i (X) = X(X-1)•••(X-i+1) i! = X i .
We write

X p = p i=0 b i N i (X) avec b i = ∆ i (X p )(0). So we have u(x) = (-1) p p i=0 b i N i (x) 1 0 t x Φ p (π)(t) dt. For v i (x) = N i (x) 1 0 t x Φ p (π)(t) dt, we have F vi (x) = n≥0 (-1) n x n n! N i (n) 1 0 t n Φ p (π)(t) dt = n≥i (-1) n x n i!(n -i)! 1 0 t n Φ p (π)(t)
dt, and then

F vi (x) = 1 0 n≥i (-1) n (xt) n i!(n -i)! Φ p (π)(t) dt = 1 0 e -xt (-xt) i i! Φ p (π)(t) dt. So F u (x) = (-1) p 1 0 e -xt p i=0 b i (-xt) i i! Φ p (π)(t) dt = (-1) p 1 0 e -xt S p (xt)Φ p (π)(t) dt.
Application to an integral representation of S α u and of F S α u .

For such a sequence u, per shift we have :

-S α (u)(x) = (-1) p (x + α) p 1 0 t x+α Φ p (π)(t) dt.
-

F S α u (x) = (-1) p 1 0 e -xt S α,p (xt)t α Φ p (π)(t) dt.
Proof :

First point is obvious.

For the second, we must write (X + α

) p = p i=0 a α,i X i with a α,i = ∆ p ((X + α) p )(0).
Thus we obtain the result by the relation S

α,p (X) = p i=0 a α,i (-X) i i! .
4 Some technical results.

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

Si

t α π(t) is integrable on ]0, 1] et si ρ < 1, so ρ x = o (F u (x)). Furthermore, if π 1 ∼ 0 π 2 in 0, then F u1 ∼ +∞ F u2 .
For the first point :

1 0 e -xt π(t) dt = 1 0 t -α e -xt t α π(t) dt ≥ a -α e -xa 1 a t α π(t) dt for a > 0,
this leading term ρ x if we choose a small enough.

For the second :

If a(t) = o(π(t)) in 0, let's check that 1 0 e -xt a(t) dt = o 1 0 e -xt π(t) dt .
We write |a(t)| ≤ επ(t) over ]0, α] and so

1 0 e -xt a(t) dt ≤ ε 1 0 e -xt π(t) dt + e -αx 1 α a(t) dt.
The increase in the first point then makes it possible to conclude.

And finally for a > 0 (voir [2]):

* 1 0 e -xt t a-1 ln b (1/t) dt ∼ ∞ Γ(a) ln b (x) x a = 1 x a p i=0 i b ln b-i (x)(-1) i Γ (i) (a) + o ln b-p x a * * 1/e 0 e -xt t a-1 ln b (1/t) ln c (ln(1/t)) dt ∼ ∞ Γ(a) ln b (x) ln c (ln(x)) x a = 1 x a 0≤i+k≤p and j≤q i b ln b-i-k (x) ln c-j (ln(x))(-1) i Γ (i+k) (a) + o ln b-p ln c-q (ln(x))
x a 5 Practical calculations.

5.1 Case u(x) = 1 (x+a) k with a and k positive.

For u(x) = 1 x+1 we have π u (t) = 1 [0,1]
. We notice π or even π u this function.

By calculating the derivative, we have

(-1) k k! (x + 1) k+1 = 1 0 t x ln k (t) dt. That is, for u k (x) = 1 (x+1) k , π u k (t) = (-1) k-1 ln k-1 (t) (k-1)!
on the interval ]0, 1]. The case of a positive integer exponent k was treated as such.

This is widespread in

: if u k (x) = 1 (x+1) k with k ∈ R + , then π u k (t) = ln k-1 (1/t) (k-1)! 1 ]0,1] .
This relation is easily verified by puting t = e -u in

1 (k -1)! 1 0 t x ln k-1 (1/t) dt, remembering the rating (k -1)! = Γ(k).
This results in the following formula

F u k (x) = 1 (k -1)! 1 0 e -xt ln k-1 (t) dt.
We will now present here the case v k (x) = 1 (x+a) k with a > 0.

We just saw that if

u k (x) = 1 (x+1) k , then π u k (t) = ln k-1 (1/t) (k-1)! 1 ]0,1] . By the shift operator we obtain v k = S a-1 (u k ). So π v k (t) = t a-1 ln k-1 (1/t) (k-1)! 1 ]0,1] ,
and then

F v k (x) = 1 (k -1)! 1 0 e -xt t a-1 ln k-1 (1/t) dt.
For real positive a and k, we have the asymptotic

∞ n=0 (-1) n x n n!(n + a) k = x -a p i=0 k -1 i ln k-1-i (x)(-1) i Γ (i) (a) + o ln k-p (x) x a .
And in particular

∞ n=0 (-1) n x n 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) = 1 (x+a) k with k > -1 and a > 0, we have

π u (t) = t a-1 ln k-1 (1/t) (k -1)! F u (x) ∼ ∞ Γ(a) x a ln k-1 (x) (k -1)! 5.2 Case x -k with k positive.
We will present here the case u

(x) = x -k et u(0) = 0. So we're looking at F u (x) = ∞ n=1 (-1) n x n n!n k .
Here the shift and truncation operators will intervene.

We know that for v(x) = (x + 1) -k we have

-v(x) = 1 0 t x ln k-1 (1/t) (k -1)! dt. -F v (x) = 1 0 e -xt ln k-1 (1/t) (k -1)! dt.
By shifting and then truncation (see page 5), we have

F u (x) = x 1 0 e -xt Π(t) dt with Π(t) = - 1 t ln k-1 (1/s) s(k -1)! ds.
By truncation, we have thus shown

F u (x) = -x 1 0 e -xt ln k (1/t) k! dt.
For k real positive, we have the asymptotic

+∞ n=1 (-1) n x n n!n k = - 1 k! p i=0 k i ln k-i (x)(-1) i Γ (i) (1) + O ln k-p (x) .
And in particular

∞ n=1 (-1) n x n n!n k ∼ ∞ - ln k (x) k! .
This is directly deduced from the integral expression and the relation *

In conclusion :

For u(x) = 1 n k with k real positive and x > 0, we have

F u (x) = -x 1 0 e -xt ln k (1/t) k! dt F u (x) ∼ ∞ - ln k (x) k!
5.3 Case of a negative exponent.

We will study the case u(x) = (x + a) k with k > 0.

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

Integral representation of u(x) = (x + 1) p-k :

If 0 < k < 1 and if p is a positive integer, then

(x + 1) p-k = 1 0 t x - p-1 i=0 (x ln(t)) i i! ln k-p-1 (1/t) (k -p -1)! dt + P k (x).
where P k is polynomial.

Proof :

It should be noted first that t → t x - p-1 i=0 (x ln(t)) i i! ln k-p-1 (1/t) (k -p -1)! is integrable over ]0, 1[. Let us note u p (x) = (x + 1) p-k and v p (x) = 1 0 t x - p-1 i=0 (x ln(t)) i i! ln k-p-1 (1/t) (k -p -1)! dt + Q k (x) with Q k a polynomial whose derivative is P k . A derivation calculation left to the reader shows that u p (x) = (p-k)u p-1 (x) et v p (x) = (p-k)v p-1 (x)+P k (x).
The case p =0 corresponds to u(x) = 1 (x+a) -k with -k > -1 which has been treated in section 5.1 (under the title important remark). It is then sufficient to adjust P k (0) to state that u k = v k . We can therefore conclude by induction that the Proposition is verified for any p ∈ N .

Remark :

We have already reported that for a polynomial u function, F u (x) = O(ρ x ). The appearance of the polynomials P k is therefore of no importance to the asymptotics sought, even if clearly we could calculate these polynomials. The general case u k (x) = (x + a) k .

For k /

∈ N * we have

F u k (x) ∼ ∞ Γ(a) x a ln -k-1 (x) (-k -1)! .
Proof :

We'll actually take 0 < k < 1 and process the case u(x) = (x + a) k-p . Let's pose π k (t) = ln k-p-1 (1/t) (k-p-1)! . In the case a = 1 as u is deduced from a density sequence, we know that

F u (x) = (-1) p 1 0 e -xt S p (-xt)Φ p (π k )(t) dt.
So by the shift operator (see page 4)

F u (x) = (-1) p 1 0 e -xt S a-1,p (-xt)t a-1 Φ p (π k )(t) dt.
We clearly have the implication : If π(t) ∼

0 ln α (1/t) α! then Φ(π)(t) ∼ 0 ln α (1/t) α! . So here, Φ p (π k )(t) ∼ 0 ln k-p-1 (1/t) (k-p-1)! . Let's write S a-1,p (X) = p i=0 b a-1,i (-X) i i! .
As

1 0 e -xt (xt) i t a-1 Φ p (π k )(t) dt ∼ ∞ x i Γ(a + i) x a+i ln k-p-1 (x) (k -p -1)! , we thus obtain F u (x) = (-1) p p i=0 (-1) i b a-1,i Γ(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 i=0 (-1) i b a-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) = 1 0 t x - p-1 i=0 (x ln(t)) i i! ln k-p-1 (1/t) (k -p -1)! dt + Q k (x) with Q k polynomial For k / ∈ N * we have F u (x) ∼ ∞ Γ(a) x a ln -k-1 (x) (-k -1)!
5.4 Case of a negative pole a.

Asymptotic of F u (x) = n≥n0 (-1) n x n n! (n + a) k avec k > 0. Let n 0 -1 < a ≤ n 0 and let u(x) = 1 (x-a) k If n 0 -1 < a < n 0 then F u (x) ∼ ∞ (-1) n0 x a Γ(-a) ln -k-1 (x) (-k-1)! If a = n 0 then F u (x) ∼ ∞ (-1) n0+1 x n0 n 0 ! ln -k (x) (-k)! Proof :
We put a = n 0 -ε with 0 ≤ ε < 1.

If we note n = m + n 0 , we rewrite F u (x) as

F u (x) = (-x) n0 m≥0 (-1) m x m m!(m + 1) • • • (m + n 0 )(m + ε) k . If k is an
integer, by the case of a rational fraction treated at the beginning, we have

F u (x) ∼ ∞ (-x) n0 KF v (x) where v(x) = 1 (x+ε) k and K = [(1 -ε)(2 -ε)) • • • (n 0 -ε)] -1 .
For the other values of k, we will find this result by the principle of no superposition of perturbations in the example paragraph.

-Cas n 0 -1 < a < n 0 :

As F v (x) ∼ ∞ Γ(ε) x ε ln -k-1 (x) (-k-1)! and by the relation Γ(ε) = (ε -1) • • • (ε -n 0 )Γ(ε -n 0 ), we thus obtain the relation F u (x) ∼ ∞ (-1) n0 x a Γ(-a) ln -k-1 (x) (-k-1)! .
-Case a = n 0 :

In this case v(x) = 1 x k and F v (x) ∼ ∞ - ln -k (x) (-k)! . So F u (x) ∼ ∞ (-1) n0+1 x n 0 n0! ln -k (x) (-k)! .
It should be noted that, as in the case of rational fractions, we have the same equivalent if we study

F u (x) = n =n0 (-1) n x n n! u(n).

Conclusion.

We consider u(x) = 1 (x+a) k with x and k real We define n 0 = 0 if a > 0 and n

0 -1 < a ≤ n 0 if a ≤ 0 F u (x) ∼ ∞ (-1) n0 Γ(a) x a ln k-1 (x) (k -1)! si a / ∈ -N F u (x) = O(ρ x ) si a ∈ -N
6 Logarithmic cases.

6.1 Cas ln(x + a).

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

Case u(x) = ln(x + a).

If u(x) = ln(x + a), then

F u (x) ∼ ∞ - Γ(a) x a ln(x) Proof : Let u(x) = ln(x + 1).
Clearly, by primitization, u(x) =

1 0 t x -1 ln(t) dt.
Let Li(t) = t 0 ds ln(s) .

Because u is deduced from a density sequence, we then have

F u (x) = - 1 0 e -xt S 1 (xt)Φ(Li)(t) dt.
We have the asymptotic Li(t)

∼ 0 t ln(t) ,
and then Φ(Li)(t) ∼ 1 ln(t) , so,

1 0 e -xt Li(t) dt ∼ ∞ x 1 0 e -xt t ln(1/t) dt (Remember S 1 (X) = -X).
Thus, by the relation * ,

F u (x) ∼ ∞ -1 x ln(x) .
Let us treat the case a = 1 in the same way :

For v(x) = ln(x + a), by shifting u by S a-1 ,

F v (x) = - 1 0 e -tx t a-1 S 1,a-1 (xt)Φ(Li)(t) dt,
and then

F v (x) = - 1 0 e -xt (a -1 -xt)Li(t)t a-1 dt,
and finally, after calculations left to the reader,

F v (x) ∼ ∞ -Γ(a)
x a ln(x) .

In conclusion :

u(x) = ln(x + a) F u (x) = - 1 0 e -xt (a -1 -xt)Li(t)t a-1 dt F u (x) ∼ ∞ - Γ(a) x a ln(x) 6.2 Case ln k (x + a) with k integer. Case u(x) = ln k (x + a) with k positive integer. If v k (x) = ln k (x + a) for a positive integer k, then F v k (x) ∼ ∞ (-1) k k a! x a ln k-1 (ln(x)) ln(x) .
Proof :

We're going to proceed in 3 steps.

1) Study of I k (x) = 1 0 (1 -t x ) ln k (ln(1/t)) ln(1/t) dt.
In this paragraph we will note

c k = Γ (k) (1) = ∞ 0 ln k (t)e -t dt.
By changing the variable e -t = u we get Γ (k) (1) = 1 0 ln k (ln(1/u)) du.

We clearly have

I k (x) = 1 0 t x ln k (ln(1/t)) dt.
We write I k (x) =

x 0 t x-1 t ln k (ln(1/t)) dt that we integrate by parts :

I k (x) = (t x -1) x t ln k (ln(1/t)) 1 0 0 - 1 0 (t x -1) x ln k (ln(1/t)) -kt ln k-1 (ln(1/t)) -1/t ln(1/t) ln k (ln(1/t))+k ln k-1 (ln(1/t)) ln(1/t) dt. Thus xI k (x) = -I k (x) + c k -kI k-1 (x).
We then obtain the relation (x + 1)

I k (x) = c k -kI k-1 (x).
We then define P k by I k (x) = P k (ln(x + 1)).

The recurrence relation is written as P k = c k -kP k-1 with P k (0) = 0 and also P 0 = X since I 0 (x) = ln(x+1).

Let's check by induction the relation

(k + 1)P k (X) = ∞ 0 e -t ln k+1 (t) -(ln(t) -X) k+1 dt : Note J k (X) the integral 1 k + 1 ∞ 0 e -t ln k+1 (t) -(ln(t) -X) k+1 dt.
For k = 0 we clearly have J 0 (X) = X.

As J k (0) = 0, to check the relation to the k rank by induction, just check that J k satisfies the relation

J k (X) = c k -kJ k-1 (X) : J k (X) = ∞ 0 e -t (ln(t) -X) k dt = -kJ k-1 (X) + k ∞ 0 e -t ln k (t) dt = -kJ k-1 + c k-1 . Thus 1 0 (1 -t x ) ln k (ln(1/t)) ln(1/t) dt = 1 k + 1 ∞ 0 e -t ln k+1 (t) -(ln(t) -ln(x + 1))
k+1 dt = P k (ln(x + 1)).

We will remember that P k a polynomial of degree k + 1 verifying P k (0) = 0 and leading coefficient (-1) k k+1 .

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

Let's write X k+1 = a 0,k P 0 (X)+a 1,k P 1 (X)+...+a k-1,k P k (X), and pose

Q k (X) = a 0,k +a 1,k X...+a k-1,k X k .
By looking at the leading coefficient we have a k-1,k = (-1) k (k + 1).

By taking X = ln(x + 1) we have ln

k+1 (x) = 1 0 (1 -t x ) k-1 i=0 a i,k ln i (ln(1/t)) ln(1/t) .
So we have

ln k+1 (x + 1) = 1 0 (1 -t x ) Q k (ln(ln(1/t))) ln(1/t) dt.
Thus, for u k (x) = ln k (x + 1), we have

π u k (t) = Q k-1 (ln(ln(1/t))) ln(1/t)
, that we'll write π k in the following.

In particular

π u k (t) ∼ 0 (-1) k-1 k ln k-1 (ln(1/t)) ln(1/t)
.

3) Equivalent of F v k at +∞ with v k (x) = ln k (x + a).

We so have

u k (x) = -(t x -1) Q k-1 (ln(ln(1/t))) ln(1/t) dt,
and then, for v k (x) = ln k (x + a) :

F v k (x) = 1 0 e -x S 1,a-1 (xt)t a-1 Φ(π k ) dt As Φ(π k )(t) ∼ π k (t) ∼ (-1) k-1 k ln k-1 (ln(1/t)) ln(1/t) , so we have F v k (x) ∼ 1 0 e -xt (a -1 -xt)t a-1 (-1) k-1 k ln k-1 (ln(1/t)) ln(1/t) dt,
which gives as equivalent

(-1) k-1 k(a -1) 1 0 e -xt t a-1 ln k-1 (ln(1/t)) ln(1/t) dt + (-1) k kx 1 0 e -xt t a ln k-1 (ln(1/t)) ln(1/t) dt
(provided that the main parts are not simplified) or, after using the relation * * :

F v k (x) ∼ (-1) k-1 k ln k-1 (ln(x))
x a ln(x)

((a -1)Γ(a) -Γ(a + 1)), which shows that F u (x) ∼ ∞ (-1) k kx Γ(a)
x a ln k-1 (ln(x)) ln(x)

.

In conclusion :

u(x) = ln k (x + a) π u (t) = t a-1 Q k-1 (ln(ln(1/t))) ln(1/t) π u (t) ∼ 0 (-1) k-1 kt a-1 ln k-1 (ln(1/t)) ln(1/t) F u (x) ∼ ∞ (-1) k kx Γ(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 u i two density sequences π i .

The sequence u 1 u 2 admits π = π 1 π 2 as density with π(z)

= 1 z π 1 (s)π 2 (z/s) ds s .
Proof :

1 (x)u 2 (x) = 1 0 1 0 t x s x π 1 (t)π 2 (s) dsdt.
By puting u = st we obtain

1 0 t 0 u x 1 t π 1 (t)π 2 (u/t) dt du = 1 0 1 u 1 t π 1 (t)π 2 (u/t) dt du = 1 0 z x π(z) dz, with π(z) = 1 z π 1 (s)π 2 (z/s) ds s .
Properties of the operator :

1 is assoiative and commutative.

2 Under conditions of derivability, (π

1 π 2 ) (z) = -1 z π 1 (z)π 2 (1) + 1 z 1 s 2 π 1 (s)π 2 (z/s) ds.
3 Equivalent formulas :

π 1 π 2 (z) = √ z z 1 s (π 1 (s)π 2 (z/s) + π 1 (z/s)π 2 (s)) ds. z (π 1 π 2 ) (z) = -π 1 ( √ z)π 2 ( √ z) + √ z z (π 1 (z/s)π 2 (s) + π 1 (s)π 2 (z/s)) ds.
Proof :

The first point is left to the reader.

The second is obtained by considering G(x, y) =

1 x π 1 (s)π 2 (y/s) ds s .
For the third one :

We consider

1 √ z 1 s 2 π 1 (s)π 2 (z/s
) ds where the variable change u = z/s is performed, which gives

√ z z π 1 (z/s)π 2 (s)
ds which is integrated by parts and gives the result.

Some examples :

-For π

1 (t) = t a et π 2 (t) = t b : We have π 1 π 2 (z) = z b -z a b-a if a = b, z a ln(1/z) if a = b. -For π 1 (t) = ln a (1/t) et π 2 (t) = ln b (1/t), we have π 1 π 2 (z) = B(a + 1, b + 1) ln a+b+1
(1/z), B designating the Euler function.

Indeed: The definition gives

1 z ln a (t/z) ln b (1/t) dt t u=ln(1/t) = ln(1/z) 0 (ln(1/z) -u) a u b du,
the change of variable u = v ln(1/z) giving the result.

-For

π 1 (t) = t a et π 2 (t) = b ln c (1/t), we have π 1 π 2 (z) = 1 z t b-a-1 ln c (1/t) dt ∼ +∞ z b-a-1 ln c (1/z). -For π(t) = t a-1 on a π • • • π k fois (t) = t a-1 ln k-1 (1/t) (k-1)!
(by an easy induction), which makes it possible to find the case u(x) = 1 (x+a) k .

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 , noted a(f

) = lim t→0 tf (t) f (t) .
Remarks :

If f admits a power part a, then F (x) =

x 0 f (t) dt admits a power part a + 1.

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

If f admits a power part then we have

c 1 t a(f )+ε ≤ f (t) ≤ c 2 t a(f )-ε over ]0, 1].
Proof:

-a(f ) = a is equivalent to a(g) = 0 for g(t) = t -a f (t).

-If a(f ) = 0, then f (t) = e α(t) where α (t) = o(1/t).

If F (x) = x 0 f (t) dt, then x 0 e α(t) dt = te α(t) x 0 - x 0 tα (t)e α(t) dt =o(F (x))
.

Thus xF (x) F (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 π 2 (t) = O t b with b > a(π 1 )
and if π 1 admits a power part, then

π 1 π 2 ∼ 0 C(π 1 , π 2 )π 1
and therefore

F ∼ ∞ CF u1
where

C(π 1 , π 2 ) = 1 0 t -a(π1)-1 π 2 (t) dt
2 Principle of superposition of perturbations :

If π 1 and π 2 admit a power part with a(π

1 ) = a(π 2 ) then a(π 1 ) = a(π 2 ) = a(π 1 π 2 )
Let's mention two very important applications before giving the proof :

If a(π 1 ) < a(π 2 ), • • • , a(π k ) with π 1 admitting a power part, then for 2 ≤ i ≤ k, we have π i (t) = O(t b ) for one b > a(π 1 ), so π 2 • • • π k (t) = O(t b ),
and thus

F u1•••u k ∼ +∞ C(u 1 , u 2 • • • π k )F u1 If a(π 1 ) = a(π 2 ) < a(π 3 ), • • • , a(π k ), meanwhile F u1•••u k ∼ +∞ C(u 1 u 2 , π 3 • • • π k )F u1u2 Proof : First case : Let a = a(π 1 ) > b, π 1 (t) = t a π(t).
We thus know that a(π) = 0 and t -a π 2 (t) = O(t α ) with α > 0. put f (t) = t -a-1 π 2 (t) whitch is integrable over ]0, 1].

π 1 π 2 (z) = = z a 1 z f (s)π(z/s) ds.
It is therefore sufficient to show

1 z f (s)π(z/s) ds ∼ 0 π(z) 1 0 f (s) ds.
To do this, we integrate by parts by puting F (z) = z 0 f (s) ds :

1 z f (s)π(z/s) ds = F (1)π(z) -F (z)π(1) + 1 z z
s 2 F (s)π (z/s) ds. a(f ) > -1 thus a(F ) > 0 and so we have an inequality such as F (z) ≤ cz ε , which shows that F (z)π(1) = o(π(z)).

By z/s = u, the last integral is equal to

1 z F (z/s)π(s) ds,
and we have to show that it's negligible in front of pi(z). a(f ) > -1 so a(F ) > 0 and we have an inequality such as F (t) ≤ Ct α .

The last integral is therefore increased by

1 z z s α π (s) ds = H(z) = z α 1 z s -α+1 sπ (s) ds = z α α -s -α sπ (s) 1 z + z α α 1 z s -α (sπ (s)) ds = O(z α ) + 1 α zπ (z) o(π(z)) + z α α 1 z s -α (sπ (s)) ds.
But as a(sπ (s)) = 0, (sπ (s)) = o(π (s)) there are two cases.

-If π (s) s α is integrable over ]0, 1], then

z α α 1 z s -α (sπ (s)) ds = O(z α ).
-Else,

z α α 1 z s -α (sπ (s)) ds = o z α 1 z π (s) s α ds = o(H(z)). Finally, the relation a(π) = 0 provides z α = o(π(z)).
Enfin la relation a(π) = 0 fournit z α = o(π(z)).

Second case :

Let's start with a lemma :

Lemma :

If π 1 , π 2 are locally positive and monotonous around 0, from part to zero power, then

π 1 ( √ z)π 2 ( √ z) = o √ z z π 1 (s)π 2 (z/s) ds s .
Proof of the lemma :

Let's note π(z) for the integral. First of all, let's notice the equality:

√ z z π 1 (s)π 2 (z/s) ds s = 1 √ z π 1 (z/s)π 2 (s) ds s
Case π 1 decreasing and π 2 increasing :

√ z z 1 s π 1 (s)π 2 (z/s) ds ≥ π 1 ( √ z) √ z z π 2 (z/s) ds s = π 1 ( √ z) 1 √ z π 2 (s) ds s ≥ π 1 ( √ z)π 2 ( √ z) ln(1/ √ z).
Cas π 1 increasing et π 2 decreasing :

1 √ z π 1 (z/s)π 2 (s) ds s ≥ π 2 ( √ z) 1 √ z π 1 (z/s) ds s = π 2 ( √ z) √ z z π 1 (s) ds s ≥ π 1 ( √ z)π 2 ( √ z) ln(1/ √ z).
Cas π 1 et π 2 decreasing :

If π 2 (0) = 0, by monotony and positivity we would have the trivial case π 2 = 0, so

1 0 π 2 (s) s ds diverge. But because a(π 2 ) = 0, we have π 2 (1/2) -π 2 (z) = 1/2 z π 2 (s) ds = o 1 z π 2 (s) s ds and π 2 (z) = o 1 z π 2 (s) s ds . Thus π 2 ( √ z) = o 1 √ z π 2 (s) s ds . The inequality π(z) = √ z z 1 s π 1 (s)π 2 (z/s) ds ≥ π 1 ( √ z) 1 √ z π 2 (s)
s ds allows us to conclude.

Cas π 1 et π 2 increasing :

π 2 ( √ z) -π 2 (z) = √ z z π 2 (s) ds = o √ z z π 2 (s) s ds . Futhermore, √ z z π 2 (s) s ds ≥ π 2 (z) [ln(t)] √ z z = π 2 (z) ln(1/z), so π 2 (z) = o √ z z π 2 (s) s ds , puis 2 ( √ z) = o √ z z π 2 (s) s ds . Finally π(z) ≥ 1 √ z π 1 (s)π 2 (z/s) ds s ≥ π 1 ( √ z) 1 √ z π 2 (z/s) s ds = π 1 ( √ z) √ z z π 2 (s)
s ds, hence the result.

Remark :

A proof of this lemma not using monotony would be welcome.

Proof of the second case :

If a(π 1 ) = a(π 2 ) = a : By puting π i (s) = t a π * i (s), and by π 1 π 2 (z) = z a π * 1 π * 2 (z), we'll come back in case a = 0. We will therefore assume a(π i ) = 0 and note π = π 1 π 2 . We have

zπ (z) = -π 1 ( √ z)π 2 ( √ z) + √ z z (π 1 (z/s)π 2 (s) + π 1 (s)π 2 (z/s)) ds
The relations π i (z) = o(π i (z)/z) as well as the nature of the integration interval then show that the integral is negligeable in front of pi(z), and so it remains to show π

1 ( √ z)π 2 ( √ z) = o(π(z))
, which is ensured by the lemma.

7.4 Exemples.

Fisrt example

As promised, we will give an equivalent of m≥0 (-1) m x m (m + n 0 )!(m + ε) si 0 ≤ ε < 1 thanks to the principle of non-superposition of perturbations.

We write (m + 1)

• • • (m + n 0 )(m + ε) k -1 = 1 (m + ε) k n0 i=1 a i m + i . For v(x) = 1 (x+ε) k we have π v (t) = t ε-1 ln k-1 (1/t) (k-1)! and for u i (x) = 1 x+i we have π i (t) = t i-1 . As a(π v ) = ε -1 < a(π i ) = i -1, by non-superposition, we have π v π i ∼ C i π v avec C i = 1 0 t -(ε-1)-1 t i-1 dt = 1 i -ε . Thus for u(x) = (x + 1) • • • (x + n 0 )(x + ε) k -1 , we have π u ∼ π v n0 i=1 a i i -ε = 1 (1 -ε) • • • (n 0 -ε) , whitch gives F u (x) ∼ ∞ (-x) n0 KF v (x) ou v(x) = 1 (x+ε) k and K = [(1 -ε)(2 -ε)) • • • (n 0 -ε)]
-1 as wished.

Second example

We're going to focus on w n = ln(n+a) (n+b) k . We want an equivalent at +∞ of

F w (x) = +∞ n=0 (-1) n x n ln(n + a) (n + b) k
It should be noted that this is not the product of two density sequences.

We therefore pose w(x) = ln(x+a) (x+b) k .

We know that ln(x + a) = x 1 0 t x t a-1 Li(t) dt.

So we're going to pose u(x) = 1 0 t x t a-1 Li(t) dt, and v

(x) = 1 (x+b) k = 1 0 t x t b-1 ln k-1 (1/t) (k -1)! dt, let π u (t) = t a-1 Li(t) ∼ t a ln(t) and π v (t) = t b-1 k-1 (1/t) (k-1)! , and finally π = π 1 π 2 . We know that F w (x) = 1 0 e -xt S 1 (xt)Φ(π)(t) dt with S 1 (X) = -X, a(π u ) = a and a(π v ) = b -1.
There are therefore three cases.

First case a < b -1 :

π u.v ∼ C(π u , π v )π u with C = 1 0 t -a-1 t b-1 ln k-1 (1/t) (k -1)! dt, Φ(π)(t) ∼ C t a (a+1) ln(t) . Thus F w (x) ∼ -C 1 0 e -xt xt t a (a + 1) ln(t) dt, or F w (x) ∼ -CΓ(a+2)x (a+1)x a+2 ln(x) , or even F w (x) ∼ +∞ - CΓ(a + 1) x a+1 ln(x) Second case a > b -1 : This time, C = C(π v , π u ) = 1 0 t -(b-1)-1 ta -1Li(t) dt, and π u.v ∼ Cπ v , Φ(π v )(t) ∼ bt b-1 ln k-1 (1/t) (k-1)! Thus F w (x) ∼ -Cbx 1 0 e -xt t b ln k-1 (1/t) (k -1)! dt and finally F w (x) ∼ - CbΓ(b + 1) ln k-1 (x) (k -1)!x b Third case a = b -1 : π(z) = 1 z t a-1 Li(t)(z/t) b-1 ln k-1 (t/z) (k -1)! 1 t dt. This gives π(z) ∼ z a ln k-1 (1/z) (k -1)! 1 z Li(t) t 2 dt ∼ -z a ln k-1 (1/z) (k -1)! ln(ln(1/z))
(we will not justify here the first equivalent), puis φ(π) ∼ 1 a+1 π.

In this way

F w (x) ∼ x 1 0 e -xt t t a a + 1 ln k-1 (1/t) (k -1)! ln(ln(1/t)) dt,
and at last

F w (x) ∼ +∞ Γ(a + 1) ln k-1 (x) ln(ln(x)) (a + 1)x a (k -1)!

Third example

We're going to treat the example u(x) = 1 (x α -a α ) k with 0 < a < 1, α > 0 and k ∈ N.

First method with the principle of non-superposition of perturbations :

Put u(x) = 1 (x α -a α ) k , v(x) = 1 (x-a) k and w(x) =

x-a

x α -a α . We assume that v, and therefore w, are deduced from density sequences admitting a power part.

By truncation, F v (x) = x 1 0 e -xt π V (t) dt avec π V (t) = -1 t π v (s) ds, so π v = -π V , and the same is true for w.

We have the relations a(π V ) = a(π v ) + 1, a(Φ 2 (π V )) = a(π V ). In addition, by practical calculations, π V (t) = t -a-1 ln k-1 (1/t) (k-1)! .

Since W is defined for some x < a, by holomorphy, f (t) = t a-ε π W (t) is can be integrated in 0.

Having a power part, tf (t) is also and by an integration by parts, it is easy to see that f (t) = O(1/t), whitch gives π W (t) = O(t -1-a+ε ).

So a(π W ) > a(π V ) and by non-superposition, π V π W ∼ Cπ V .

Thus u(x) ∼ Cx 2 1 0 t x π V (t) dt, and then F u (x)

∼ 1 0 e -xt S 2 (xt)Φ 2 (π V )(t) dt.
S 2 (X) = X 2 + X and F u (x) ∼ 1 0 e -xt (xt + x 2 t 2 )t -a-1 ln k-1 (1/t) dt.

This gives F u (x) ∼ C ln k-1 (x)(x Γ(-a+1)

x -a+1 + x 2 Γ(-a+2)

x -a+2 ) (because main parts don't simplifie). After simplification we find

F u (x) ∼ +∞ Kx a ln k-1 (x)
Second method, by direct work :

It's more delicate.

If we give 0 < a < 1 et α > 0 and if we put u(x) = 1 (x α -a α ) k , then k+i) .

F u (x) ∼ ∞ α -k a k(1-α) Γ(-a)x a ln k-1 (x) (k -1)! Proof : We write u(x) = ∞ i=0 -k i (-1) i a αi x α(
The case v(x) = x -p gives F v (x) = -x 1 0 e -xt ln k (1/t) k! dt.

By this formula, we obtain here Second example :

We'll shamefully add up equivalents (but we'll say where the crimes take place), by studying u(x) = 1 x+a , which will validate the calculations on the logarithms.

We write :

u(x) = k≥0 (-1) k k! ln k (x + a) u k .
We know that π u k (t) ∼ (-1) k-1 kt a-1 ln k-1 (ln(1/t))

ln(1/t) . So we're going to assume that we have (it's here !) :

π u (t) ∼ k≥0 (-1) k
k! (-1) k-1 kt a-1 ln k-1 (ln(1/t)) ln(1/t) , whitch gives π u (t) ∼ -t a-1 ln(1/t) k≥1 ln k-1 (ln(1/t))

ln(1/t) = t a-1 .

Furthermore, F u k (x) ∼ (-1) k k Γ(a) ln k-1 (ln(x))

x a ln(x)

. We'll also assume (it's also here !):

F u (x) ∼ k≥1 (-1) k k Γ(a) ln k-1 (ln(x))
x a ln(x) , whitch gives F u (x) ∼ k≥0 Γ(a) ln k-1 (ln(x)) (k -1)!x a ln(x)) .

We find well F u (x) ∼ +∞ Γ(a)

x a .

Finally, it would remain to treat the case ln k (x + a) with k and a real.

Références :

[1] https://hal.archives-ouvertes.fr/hal-00747720v4/document

[2] https://hal.archives-ouvertes.fr/hal-00644237/document August 30th 2019 2

2 Luc Abergel, Lycée Janson de Sailly, 106 rue de la pompe 75116 PARIS FRANCE, email : lucabergel@cegetel.net

F

  i (a ln(1/t)) αi (α(k + i))! . Let's note c i = -k i (-1) i (αi)! (α(k+i))! = k+i-1 i (αi)! (α(k+i))! , so that ϕ(t) = ∞ i=0 c i (a ln(1/t)) αi (αi

Because of the equivalent for c i , we have ϕ(t) ∼ 0 K i≥0 i b (a ln(1/t)) αi (αi)! .

But we have

And so ϕ(t)

Here again, after simplification, we have

8 Generalizations.

There are two possible generalizations.

For example, if c is a regular (and therefore positive) sequence, that means

Example :

and

This requires an asymptotic of

in 0 to get one of F u . This can be done, again in the case where c(k) is a positive sequence, with the techniques proposed in this article.

First example :

We use

and so π u (t) ∼ t a-1 ln α (ln(1/t))

α! ln(1/t) ,


Jacques Amblard

Pascal

Dusapin's Third Style in Faustus, the Last Night

Born in 1555, Pascal Dusapin is now probably France's most famous living composer. His opera Faustus (2006) was composed during a transitional period, when his second and third styles coexisted. His second style, which we could call « intonationnism » (close to speaking), was mostly typical of the 1990s. His third style came in the early 21st century. The musical language was then more romantic, more devoted to strings. Another details firmly enshrine this opera in the key topics of the new millennium : postmodernity, humour, regression.

cultivate some kind of auctoritas for a "Faustian" musician who, at this point in his career, was forging his world reputation (in this regard, the choice of a libretto directly written in English might not be innocent). Composer of the music but also author of the libretto, like the master of Bayreuth (see Chapter 6), Dusapin obviously wished to conceive an ambitious Gesamtkunstwerk (total dramatic artwork). The purpose of this chapter is not, however, to follow the creator in each of his particular, complex intentions. Rather, trying to show what makes the work lyrical ("operatic"), original, and/ or typical of its time may prove more interesting.

One hallmark of Dusapin's artistry is the theater, and even more specifically, conversation. He has always been interested in fundamental sonic links: between two people, two animals, a mother and her baby, two lovers, or two Neanderthal men trying to communicate yet ignoring any language but their intonations (their "music"). More precisely, as one sought to explain elsewhere, the "music of language" is what might speak to any listener of Dusapin's works.3 This "speech" of the music seems familiar, as if it called back the music of speech. Hanno Müller-Brachmann, the first performer of Dusapin's Mephistopheles, observed, "In the beginning, when we [the singers] opened the score, it was total panic. Thousands of notes, thousands of very complicated rhythms, and we thought that it was really impossible to produce so much information with one voice… [But] [o] ne day, I understood that it was something very close to a discussion, to speech."4 Speech was the key.

At the very beginning of Faustus, the Last Night, a very peculiar detail sounds "new," seems unheard. Faustus asks Mephistopheles, "Let me breathe a while!" and Mephistopheles laughs and repeats with a ridiculous intonation, "let me breathe!" The two "let me breathe" lines are identical in the libretto. On the stage, however, they are opposite: the first, tragic; the second, comic. Thus, right from the start of the piece, one understands that for Dusapin, the significant information may be less the text itself (even one written by the composer) than its prosodic lines. The subtle comparison of two melodic processes of speech, and then their translation into melodic lines of the score, forms the French composer's specific techne, which he has refined over more than two decades.

Through such means, Dusapin can recreate very simplyand with his own distinctive processesthe two poles of theater, tragedy and comedy, in which the tragic is immediately mocked by the comic. In fact, this is specifically facilitated by atonal aesthetics, which represents a considerable success for the latter. The atonal language, not always appreciated by the public during the twentieth century, is suddenly humanized. And reciprocally, the musical model of intonation, which emerged in the Baroque era with the invention of opera, is much more precisely imitated with an atonal technique because prosody is also atonal music. Finally, the model of intonation may have been one of the few, or rare, efficient ideas of late twentieth-century modernism.

The New Lyricism

The first lyrical trump card of Faustus, the Last Night might still be this well-known secret of Dusapin. While his first style, which could be considered reflective of youth, still influenced by Iannis Xenakis and micro intervals, was more related to the 1980s, his second style, typical of the 1990s, may draw upon the common source of the intonation of speech, the spirit of tragedy, and thus the roots of opera itself, and he may have reached a third style since the beginning of the new century. This powerful framework had already permitted the resounding success of Medeamaterial (1991),5 Dusapin's second opera, which definitively installed him as a credible "opera composer" at a time when Pierre Boulez, for example, had long since buried the dramatic genre as obsolete. Dusapin used to transcribe speech intonations into the form of a score, or instrumental music. That led him to what one could call an efficient intonationism. Thus, in the works of the 1990s, one could invariably hear those typical repetitive melodic lines, written in a restricted ambitus and in "restricted scales of tones." However, this second style, even if it belongs to the past of the composer, reappears in Faustus, the Last Night as a process inextricably linked to opera. Here, it is related specifically to the comical figure of Sly, invariably arising with the jester. Indeed, intonationism might be popular because intonation may be the affective code understood by anyone. Intonation is an "intra-cultural consensus," according to the Japanese linguist Kawada Junzō.6 Thus, intonationism may be associated at least with comical figures, since they are assumed to be the most popular characters.

In Faustus, Dusapin keeps his subtlest processes of instrumental intonationism for the entourage of a single character and for rare moments, because for him they might just be the heritage of earlier times, aged and a priori artificial in the context of his new language. He seems to be interested in other processes, at this moment of his career, although not averse to older ones that have shown their efficiency. It is even possible that the character of Sly itself would have been invented just to allow the return of intonationism. One of the greatest successes of instrumental intonationism was the one permitted by the trombone. This was the very purpose of the trombone concerto Watt (1994), one of Dusapin's masterpieces. [START_REF] Amblard | Even if he's probably now France's most famous living composer, Dusapin was always a much disputed figure of References[END_REF] The trombone can slide. Its register is close to that of the human voice. It can almost "pronounce" consonants like the French "rrr" when it uses Flatterzunge. Thus, the trombone is the best "instrumental double" of the popular Dusapin. When Sly "wakes up" (at the end of number 7, just after the Angel sings "but your soul has been killed by the Light"), the solo trombone (in a trio on stage: a so-called musique de scene) "pronounces" mocking sounds using glissando and Flatterzunge, repeating a slide between two notes separated by a simple tone.

One hears, then, a sonic border: the limit that any musical instrument reaches when it tries to imitate a human voice and, more precisely, a communication effort. One hears a very peculiar mocking prosody, which incidentally seems to summarize the very spirit of comedy (and thus largely of theater itself). This spirit of irreverence is confirmed by the clownish elephant-like timbre of the slide trombone.

Dusapin is drawn to Shakespeare's tragicomic spirit. Sly yells and laughs while the Angel cries. The trombone lows when the orchestra sings in a peculiar, new, particularly lyrical way. The ubiquitous strings, heard throughout the opera, assume the role of the coryphaeus in Greek tragedy. For Dusapin, as mentioned, everything is reflexive upon the theater. As explained elsewhere, the composer, in his best scores, recovers, in Nietzsche's terms, the "birth of tragedy."8 He might cure our corrupted times of thei disenchantment or a very ancient poison, in Nietzsche's terms: a "newly born daemon called Socrates." 9By building characters with instruments or groups of instruments, Dusapin literally tries to "re-enchant" our world and, more specifically, contemporary atonal music. That is probably why he wrote a series of true symphonies, since Go (1992), which he called "soli for orchestra." The orchestra is imagined as a human being, crossed by contradictory emotions figured by the different sonic strata. In Faustus, the Last Night, the stringswhich are, in a neo-Romantic way, the heart of the orchestracomment on the drama in the form of a permanent and deep sob, a black melancholia.

Plus encore que l'ange, l'orchestre, c'est le personnage qui est traversé par une grande variété d'émotions. Quelquefois il commente, quelquefois il amplifie certaines actions musicales de la scène, parfois il anticipe ce qui va se passer. Dans ce sens, il n'est pas très différent de celui que j'ai utilisé dans Perela ou dans d'autres partitions. Mais celuilà, il est aussi marqué par une très grande mélancolie.

(More than the Angel the orchestra is a character crossed by a great variety of emotions.

Sometimes it comments, sometimes it amplifies some musical actions of the scene, sometimes it anticipates what is going to happen. In this way, it is not very different from the orchestra I used in Perela or in other scores. But this precise orchestra is also marked by a very deep melancholia).10

The strings, better than "anyone" else, better than Faustus, are aware of his tragic fate. They articulate what Jung could call the "collective unconscious" of the characters and double the role of the Angel, in a more subterranean way.

The importance of the strings in Faustus, the Last Night recalls the composer's latest "solo for orchestra" (Reverso, 2005-6), first performed on July 1, 2007, by the prestigious Berliner Philharmoniker. We recognize here a new, third style. It is no longer a question of restricted scales and ambitus. In recent years, Dusapin seems weary of restricting his modes in only four degrees11 and his lines in this severe fifth, already imagined by Denys of Halicarnasse during the first century bce.12 Here, he obviously wants to free the expressive capacities of the instruments, especially those of the strings (heralds of Romantic expression par excellence) after ten years of self-imposed frustration. Thus, the ambitus of the strings explodes, notably up to the high pitches of the violin, and the notes take their time: they do not have to "pronounce" fast and precise "messages" (prosodic lines). This results in an impression of Romantic material. The chords and melodic lines obviously do not borrow anything from the nineteenth century. But, from this hypnotic time period, the complacent resonance of the violins comes back.

This return, at the beginning of the twenty-first century, is quite logical. Takemitsu Tōru had already noticed this "Romantic" feeling about the 1990s. 13 Boulez, in Antheme II (1997), composed a piece for violin and electronics whose stunning effects of echo chambers celebrated the resonance of the instrument. The lovers of the first ascetic Boulez (Le marteau sans maître) (1954) were surprised, sometimes even disappointed, by this strange and "guilty" softness they heard. Also the late Messiaen (Éclairs sur l'au-delà) (1991) had markedly come back to the strings.14 This Romantic tendency could not be explained as just an "end of century" spirit, because it remains today (unless the twenty-first century might not have started yet, since the twentieth century is often considered to have begun with World War I). It might also betray our outmoded celebration of some eternal "Y2K" (Year 2000), a mythical date which, in a way, never came. The resonance, the echo, is like a cosmic call,15 which reminds one of the earlier optimism of science fiction. In the case of Dusapin and of his generation of relatively young composers, the echo is also the sign of a synthesis, or perhaps a compromisetypical of our timebetween high modernism (here simply represented by the atonal language of Faustus, the Last Night) and the postmodernism of the 1980s, which inaugurated this propensity for bowed instruments (but caricaturing tonal language this time). The postmodernist Henryk Górecki, in this very way, drew his Third Symphony (specially dedicated to strings) to outstanding fame in 1993.16

More precisely Romantic, in Dusapin's third style, is the composer's continued use of very conjunct, often chromatic motions, as in his second style, but in this case imitating the human voice with the instruments. However, the result is not exactly Wagnerian, even though the German composer used that kind of ultra-chromatic line in Tristan und Isolde. More precisely, the conjunct motions, associated with atonal aggregates, give the impression of recovering the less tonal work of Bartók (which remained conjunct), as in Zene húros hangszerekre, ütökre és cselesztára (Music for Strings, Percussion, and Celesta); however, Dusapin evokes a Bartók who would have frozen his tempi, in a slower and perhaps more complacent way, while exclusively conceiving his works in what Boulez called a "temps lisse" (smooth time).17

As a matter of fact, even the strings, emphasized in the third style, are driven by a spirit of theater. They often organize "conversations" between one another, entering into deep dialectic relationships. Some violent oppositions cross them, and hence they become "human." The simplest and most contrasted process that Dusapin likes to use is to reduce the entire orchestra to only two strata (or voices) of strings. The effect of this is to generate strange, massive, perhaps even monstrous, duets.18 One of the two strata is very low-pitched (occupied by the cellos and/ or the double basses), and the second is as high-pitched as possible, in the very high zones of violins, the upper echelons of the E string. The "dirty result" is an example of bad (and even absurd) orchestration, because the two strata excavate a large lacuna between them (the totally unoccupied medium register), which recalls the comic duets for piccolo and double bass composed during the 1970s by the Russian composer Galina Ustvolskaya.

The tragicomic effect may appeal to murky, but perhaps fundamental, feelings in the listener's mind, evoking freaky myths, as in, for instance, "La belle et la bête," "Der Froschkönig oder der eiserne Heinrich," or in every other fable that compares innocence (violins) and monstrosity (double basses). The tragicomic end of Watt organizes such a dialogue between the solo trombone in its low register and the piccolo. It is clear that the trombone personifies a comic character, typically Beckettian (recalling the tramp hero of Watt) or Shakespearean (any Bottom comparable to Sly in The Taming of the Shrew or Faustus, the Last Night). The piccolo plays the role of his guardian angel, or of his "innocence" (according to Dusapin).19

The paradigm of those procedures may be to model the fundamental links that build every society: very different beings (symbolized, dramatized, by the very low-and the very highpitched) finding common points between each other and finally discovering that they are products of the same essence. Faustus, the Last Night ends in the same way, with a dialogue between the violins (in the high pitches) and the double basses, an interaction that already occurs twice (each time when the opera seems to find some kind of simplification, a particularly emblematic spirit, or when the libretto has to be underlined), and even in Reverso, as if it would constitute a real musical tic of the third style. At the end of Faustus, long after Togod has sung -into Faustus's earthe last repetitions of the opera ("Repeat after me… There is… Nothing!.. Nothing… nothing… Nothing… That's it… That's right… That's the way it is…"), these two very contrasted instrumental voices, like two pure consciousnesses (easily identifiable because of their great difference of register), have not understood or admitted that the die has been cast, they still agonize, imitating each other20 during a long decrescendo smorzando, around a few conjunct notes (A♯, B, C♯, and D), which form one of those usual "restricted scales"21 in a tiny ambitus (a diminished fourth).22 Finally, the two voices stop on the minor third B-D, but each note in its own register: in the end, the vacuum that separates the two instrumental "characters" (or "ghosts") reaches more than five octaves (plus a minor third)probably a tragic symbol of the fatal distance that separates them, or perhaps not. The work seems to have found, at least, a compromise between innocence and sin because of the common scale that the two strata use (this restricted A♯-B-C♯-D scale like a common language, in fact a common prosody).

The compromise could be between innocence and pride: Faustus's pride, obviously,23 which led him to his loss, perhaps a relative loss, precisely because of this final balance between God and the devil, right and wrong, innocence and pride, although this balance is precisely possible only when Faustus is finally damned. In the end, the consequence for the character of Faustus himself is of an absolute loss, but not so for listeners, because we are only partially Faustus. The latter is a scapegoat -"absolute pride," according to Dusapin24and his damnation is a good thing for us: it allows this final balance.

Faustus is one of those anti-heroes, rare, but at the same time, crucial in the history of opera. He remains a hero because he is the main character, which probably means that we are supposed to identify with him. Faustus is more than a banal evil character (which is always secondary). Rather than being our adversary, he occupies the same unusual and subtle position as Da Ponte's character, Don Giovanni, immortalized by Mozart and, according to Kierkegaard, the character par excellence, who accomplished "[t]he immediate stages of the erotic, or musical erotic":25 music, across opera, seems to take advantage of amoral (not only immoral) subject matter because it may rank beyond every dichotomy.

Nietzsche, at the start of Die Geburt der Tragödie aus dem Geiste der Musik, saw in the essence of all music either an Apollonian or a Dionysian principle, never a moral one. Then, a possible interpretation, if one wished, could link Don Giovanni to Dionysus and Faustus to Apollo. In both cases, as in Carmen,26 the operas (for once) found a real musical subject, that is to say, true musical heroes, beyond every parasitic morality. Even if the destiny of the characters is moral (the end, for Faustus, as for Don Giovanni, is punishment), the existence itself of those characters in a musical work may be the important thing. In a way, Don Giovanni is Dionysus himself, as Faustus is Apollo (the god of beauty but also of knowledge, Faustus's ultimate search)both damned by Christianity.

A Twenty-First-Century Faustus

In the end, how do we recognize Faustus, the Last Night as an opera of our time? In other words, how does it differ from a work of the twentieth century? In the first analysis, this Faustus may not appear so different from a work by Berg: the characters mostly express an excessive pain in an atonal language. Faustus and Mephistopheles (the first abused by the second), both shavenheaded and clad in Andrea Schmidt-Futterer's sober costumes, and the first staging by Peter Mussbach, recall Wozzeck manipulated by the doctor or by his captain. Both Faustus and the Angel, having become mad, seem to yell in a typically expressionistic way. But here is the eternal anathema of modern and contemporary operas: they always risk reminding us of Berg, whenever they look for excessiveness (or only because they employ disjunct and dissonant intervals), which is oftenif not alwaysthe case.

However, subtle details give us information. First, the choice of subject itself takes on a postmodern color. There is no question here of the postmodernist aesthetics typical of the 1980s, to which Dusapin never adhered. However, as Judith Lochhead has noted, the term "postmodern" is increasingly used in an extended sense to characterize the present period.27 Even atonal composers (against whom postmodernists reacted) choose more and more popular librettos for their operas. In the emerging dramatic works, one notes the taste for supernatural, for fairy tale, for the world traditionally associated with childhood, as in, for instance, Das Mädchen mit den Schwefelhölzern, conceived by the very "serious" and, in any case, hyper-modernist German composer Helmut Lachenmann in 1996. Even if the political and subversive aspects of the tale by Hans Christian Andersen are well known, the choice remained surprising in its simplicity.

More recently, the Portuguese musician Emmanuel Nunes, also a hyper-modernist, gave us Das Märchen (2006-8), based on the poem by Goethe; the Belgian Philippe Boesmans, Wintermärchen (1999), based on The Winter's Tale by Shakespeare; and the Korean composer Unsuk Chin, Alice in Wonderland (2004-7). Those tales are admittedly high cultural references, yet they still summon the child in every listener.

In the case of Faustus, the Last Night, the myth is darker and seems "for adults." At first sight, what child would sell his soul to the devil? However, the fable stays popular, and its treatment by Dusapin takes comic aspects and even regressive, childlike, typicalof-our-time colors27-bis that play in a strange and sickly way, perhaps in order to forget what Time called the "decade from hell,"28 marked by September 11, 2001, and then by various severe economic recessions. Here is the greatest distance from Berg, who did not put a drop of humor in his two operas.29 Or humor, in Berg's case, may be closer to dark sarcasm (which is supposed to be politically useful and then "serious") than the regressive, free, lightweight, almost kitsch games held in Faustus, the Last Night.

Beyond the comic character of Sly, or the childishand popularmockeries of Mephistopheles, the original staging by Mussbach adds greatly to the overall atmosphere of kitsch. As Mephistopheles reappears in the middle of the work, he is disguised as a monstrous (ridiculous) white bunny, recalling Puppy (1992), the enormous work by Jeff Koons.30 Similarly, the Angel wears an outdated costume, whose feathered white wings are anything but discrete. One is reminded, also, of the latest opera by Péter Eötvös, Angels in America . The Hungarian composer remarked, "what I like in this story from Kushner is its unreality."31 What the public could like, indeed, might be its supernatural statement. This may also be the degeneration, or, at least, the simplified lecture of the myth of Faustus in our time: not a moral lecture, but an infantile one. What is delectable in it, for us, may be simply its underlying magic in general. Mephistopheles is no longer frightening, if we are already in hell. He is a magician when our own hell does not even contain magic and we, Harry Potter readers (children and adults), are probably thirsty for a reenchantment of the world. That is perhaps the new context that André Malraux glimpsed as he famously prophesied, "The 21st century will be spiritual or will not be." Notes contemporary French music, notably because of his apparent (and in fact assumed) clumsiness of the self-taught composer. However, even the famous and much respected French composer Henri Dutilleux recognized at least that he liked Watt.

8. See Amblard, Pascal Dusapin, 51-53. 9. Friedrich Nietzsche, The Birth of Tragedy, trans. Douglas Smith (Oxford: Oxford University Press, 2000), 68. 10. Dusapin, "À propos d'une histoire de Faustus," in Faustus, the Last Night, 137.

11. Those four degrees confirm the analyses of modern phonology, which often modelizes four "intonation contour points" in several languages, including French and American. See Kenneth Lee Pike, The Intonation of American English (Ann Arbor: University of Michigan Press, 1945), 27, or for the prosody of French, Mary-Annick Morel, Grammaire de l'intonation l'exemple du français (Paris: Ophrys, 1998), 12, or Pierre R. Léon, Phonétisme et prononciations du français (Paris: Nathan, 1992), 125.

12. The Greek grammarian had already analyzed that the intonation of speech (of ancient Greek in this case) was "contained in a fifth." ("Dans le discours, le chant de la voix se renferme dans l'intervalle de la quinte ou à peu près.") Denys d'Halicarnasse, Traité de l'arrangement des mots, trans. Charles Batteux (Paris: Nyon l'aîné et fils, 1788), 59.

13. Takemitsu Tōru, Chosakushū (Tokyo: Shinchōsha, 2000), 5:162.

14. The last movement ("Le Christ, lumière du paradis"), like the fifth ("Demeurer dans l'amour"), is entirely dedicated to them.

15. The echo, since Beethoven emphasized it for the first time in his famous Sonata no. 14 (op. 27, no. 2), is traditionally related to the idea of night. Thus, the sonata is subtitled "Moonlight." This is probably because the echo (the new resonance of the pianoforte which was allowed by the new pedal) reminds one of the depth of a cavern, but also of the human soul, of course, or of a sky that reveals its stars only at night. The cosmos is then the echo, as a sign of depth (as the sound of the universe).

16. The second recording of the London Sinfonietta had remained on the Billboard Classical Charts for 134 weeks and even appeared in the British Pop Album Charts (number six, before Madonna and the pop group R.E.M.). 17. Pierre Boulez, Points de repere (Paris: Christian Bourgois, 1981), 81.

18. These monstrous (at least clumsy) effects may be those of a self-taught composer, in a way, but are also fully assumed by the musician because they generate a vivid effect of improvisation, of "truth," as if, each time, something "deeply human" would happen.

19. Dusapin, interview by the author, June 1998, Paris. 20. This process often occurs in the counterpoint-like manner of Dusapin. However, those imitations, here, are driven at a slower pace than in the prosodic style (second style), which is typical of the third style.

21. Those restricted scales form the main vocabulary of the second style. But here, they obviously do not intend to make the instruments "speak" because of the slow tempo (which is not the one of speech).

22. However, when in the second style, each instrumental stratum should have stayed in this strict ambitus; in the present third style, the four notes sometimes change their register (octave), in order for the expression to be deployed. 23. Faustus, as Dusapin represents him, is not only a careless scientist, in business with the wrong man, but a monster of pretentiousness who verges on madness, capable of yelling, "I! Myself! I! But I, myself, my… Iiiiiii. Without myself?" (number 9, near the end, when Faustus is alone with Togod and the Angel). 24. After he wrote an opera "about innocence" (Perela), Dusapin wanted to compose another about the opposite, the "absolute evil." Interview by the author, 2005, Paris. 25. This is the title of an essay that became a chapter of Either/ Or. Søren Kierkegaard, Either/Or, trans. Howard V. Hong and Edna H. Hong (Princeton, NJ: Princeton University Press, 1987). 28. See Andy Serwer, "The '00s: Goodbye (at Last) to the Decade from Hell," Time, November 24, 2009. Der Spiegel also painted a grim picture, featuring "Das verlorene Jahrzehnt" as a cover headline in December 2009.

29. Unless one sees the use of Wozzeck as guinea pig (by the doctor) as bitter irony. 30. This is a gigantic flower sculpture, first assembled in 1992 and notably remade in 1995, in front of the Guggenheim museum in Bilbao.

31. See "Great Performers 2009-2010: Peter Eotvos Talks about His Opera Angels in America," http:// www.youtube.com/ watch?v=9cp2gnobFzQ.

  26. It is possible to think that the success of Carmen also finds deep reasons in similar arguments. 27. Judith Lochhead, introduction to Postmodern Music/ Postmodern Thought, ed. Judith Lochhead and Joseph Auner (New York: Routledge, 2002), 4. 27-bis. See Jacques Amblard & Emmanuelle Aymès, Micromusique et ludismes régressifs depuis 2000 (Aix-en-Provence: PUP, 2017).