Asymptotic expansions and Mellin-Barnes representation

Asymptotic expansions and Mellin-Barnes representation David Greynat I.F.A.E. Universitat Autonoma de Barcelona

29 September 2009

in collaboration with Jean-Philippe AGUILAR, Samuel FRIOT and Eduardo de RAFAEL

Conference on ”Approximation and extrapolation of convergent and divergent sequences and series”

Introduction Mellin Transform A physical example MDMT Asymptotic analysis Scheme

Context: Phenomenology in Quantum Field theory

Mellin-Barnes representation may be of use for at least two important questions in the phenomenology of QFT: 1

Perturbative calculations in QFT imply to evaluate numerous Feynman diagrams often with several masses and momenta. How can we evaluate them analytically ?

2

Perturbative expressions are series in powers of the coupling constants. How can we have non-perturbative (i.e. exponentially suppressed) informations ?

D. Greynat

Asymptotic expansions and Mellin-Barnes representation


Why using Mellin-Barnes representation?

The first observation is that the Mellin transform has the following scale property M [f (ax)] (s) = a−s M [f (x)] (s) Renormalization Group solutions lead to consider asymptotic behaviours of the diagrams as aα lnβ a. The Mellin transformation kernel is the most pertinent to obtain this type of asymptotic behaviour. The Mellin-Barnes representation allows expansion in several parameters and it also gives an explicit formula for the remainder that permits a control on perturbative and (in some cases) non-perturbative expansions.

D. Greynat



Mellin-Barnes representation and Feynman diagrams calculation

S. Friot, D. Greynat and E. de Rafael, Phys. Lett. B 628, 73 (2005) J.-Ph.

Aguilar, D. Greynat and E. de Rafael,

D. Greynat

Phys. Rev. D 77, 093010 (2008)


Introduction Mellin Transform A physical example MDMT Asymptotic analysisDefinitions Scheme Converse Mapping theorem

One dimensional Mellin Transform The Mellin transform of a function f and its inverse transform are defined as . M[f (x)](s) =

ˆ

∞

dx x s−1 f (x)

0

If and only if . c = Re s ∈]α, β[

←→

f (x) =

c+i∞ ˆ

ds −s x M[f (x)](s) 2iπ

c−i∞

written

hα, βi

Fundamental strip

It corresponds to the behaviours f (x) = O(x −α )

&

x →0+

(1 + x)−1

←→

(1 + x)−ν

←→

ln(1 + x)

←→ D. Greynat

f (x)

π sin πs Γ(ν − s)Γ(s) Γ(ν) π s sin πs

=

x →+∞

O(x −β )

h0, 1i h0, Re νi h−1, 0i



Idea The singularities in the complex Mellin’s plan govern completely the asymptotic behaviour of the associated function We need to define the singular expansion From the Laurent series of a function ϕ in p ϕ(s) =

A−n A−1 +··· + + A0 + A1 (s − p) + · · · (s − p)n s−p

one can build the formal series by summing all over the poles of ϕ of the singular part: X A−n A−1 + · · · + (s − p)n s−p p this is the singular expansion of ϕ and it is written as ϕ(s)

X p

A−1 A−n + ··· + (s − p)n s−p

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Converse Mapping Theorem

If f satisfies the condition to have a Mellin transform in the fundamental strip hα, βi and M [f ] (s) = O |s|−η for η > 1.

Converse Mapping Theorem M [f (x)]right (s)

M [f (x)]left (s)

X

cp,n (s − p)n

↔

X

dp,n (s + p)n

↔

p>β,n

pβ,n ∼

X (−1)n−1 dn,p p n−1 x ln x x →0 (n − 1)! p 0 We will focus on the vacuum-vacuum transitions 1 Z (0) = √ 2π

ˆ

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+∞

1

dφ e − 2! φ

2

λ φ4 − 4!

−∞


Introduction Mellin Transform A physical example MDMT Asymptotic analysis0-dim Scheme example Perturbative expansion Numerical analysis

Perturbative expansion Taking λ small, one has the perturbative expansion of Z (0) k ∞ k 1 1 X (−1) Γ 2 + 2k 35 2 385 3 λ 1 Z (0) ∼ √ λ − λ +O(λ4 ) = 1− λ+ λ→0 k! 6 8 384 3072 π k =0 Yet

(−1)k Γ 1 + 2k λ k 2 ∀λ, −−−→ ∞ k! 6 k →∞

then this is a divergent series. Let us define SnPert. and Rn

k k n−1 ∞ k k 1 1 1 X (−1) Γ 2 + 2k λ λ 1 X (−1) Γ 2 + 2k +√ Z (0) ∼ √ λ→0 k! 6 k! 6 π k =0 π k =n {z } {z } | | . . =Rn =S Pert. n−1

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Counting diagrams One of the interests of the 0-dimensional theory is that SNPert. counts the number of Feynman diagrams to a given order N in dimension 4. Z (0) = 1 − λ→0

1 35 2 385 3 λ+ λ − λ + O(λ4 ) 8 384 3072

Order λ Z (0) =

(−1) λ 3 1! 4!

+ O(λ2 ) = −

1 λ 8

+ O(λ2 )

Order λ2 24 384

λ2

 



 +

3 384

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λ2

+

8 384

λ2



Numerical analysis 1

From now, for all numerical calculations, we fix λ = 13 . Mathematica gives with 8 decimal precision Z (0)|λ= 1 ≈ 0.96556048...

3

5

7

9

11

1

13 1

0.98

0.98 0.96556048

Pert

Sn-1 0.96

0.96

0.94

0.94

3

0.92

0.92 1

3

5

7

9

11

n

Choosing a resummation procedure for the perturbative series leads to the property that |Rn | < |un | and |Rn | < |un−1 |. One then has η−1

Z (0) =

X

uk +

k =0

here

1 1 |uη | ± |uη | , 2 2

Z (0)|λ= 1 = 0.96555187 ± 0.00140990 3

The central value corresponds to the standard Stieltjes approximation which is given, for the rank η defined ∀n, |uη | 6 |un | η−1

Z (0) =

X k =0

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uk +

1 |uη | . 2


13


Z (0) Formal approach

Analytic approach Mellin-Barnes Representation

Mapping

MB - Remainder

Exp. improvement

Perturbative Expansion

Remainder (tail)

Non-perturbative Expansion

Iterative procedure Hyperasymptotic expansions D. Greynat


Hyperasymptotic Theory

Analytic approach

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Initial remarks Construction of the remainder The iterative procedure

Mellin-Barnes Hyperasymptotic Theory Main Idea Construct an exponentially small remainder with its inverse Mellin-Barnes representation (non perturbative in λ). The first step is to obtain an inverse Mellin-Barnes representation of Z (0): −s ˆ λ 4 λ 4 ds φ Γ(s) , e − 4! φ = 2iπ 4! c+iR

with h0, +∞i and | arg λ| < 1 Z (0) = √ π

π . 2

Then we obtain the following results

ˆ

ds 2iπ

c+iR

with 0,

1 4

and | arg λ|
0 and |b0 | < ∞ so that if a0 + b0 |λ|

n=

then Rn is exponentially small in λ (non perturbative). Here, a

|Rn | = O |λ|→0

e

0 − |λ|

2a0 3

a0 |λ|

!

and for a0 = 32 , we have the Optimal Truncation Scheme i.e. the smallest remainder − 3 |Rn | = O e 2|λ| . n→∞

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The iterative procedure

Main Idea It is possible to construct more and more exponentially small remainders order by order in the expansion of Rn .

With the duplication formula 22z−1 1 Γ(2z) = √ Γ(z)Γ z + 2 π we can rewrite the remainder as Rn =

1 √

π 2

ˆ −c+n+iR

dt 2iπ

3 2λ

−t

1 3 π Γ t+4 Γ t+4 sin πt Γ(t + 1)

We introduce now the main tool : the inverse factorial expansion.

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The inverse factorial expansion Barnes’ lemma Γ(s + a)Γ(s + b) = Γ(s + c)

ˆ iR

dt Γ(t + c − a)Γ(t + c − b)Γ(s + ϑ − t)Γ(−t) 2iπ Γ(c − a)Γ(c − b)

with ϑ = a + b − c. Inverse factorial expansion M−1 X (−1)j Γ(s + a)Γ(s + b) = (c − a)j (c − b)j Γ(s + ϑ − j) Γ(s + c) j! j=0 ˆ dt Γ(t + c − a)Γ(t + c − b)Γ(s + ϑ − t)Γ(−t) + 2iπ Γ(c − a)Γ(c − b) M+i R

M > Re (a − c) + δ, M > Re (b − c) + δ, M > Re (s + ϑ) for 0 < δ < 1 and | arg s| < π2 .

Olver 1995

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One has in our example,

m−1 X Γ t + 14 Γ t + 34 (−1)j Aj Γ(t − j) + = Γ(t + 1) j=0

where Aj =

1 j!

1 3 4 j 4 j

f (s) =

and

The remainder is then m−1 1 X Rn = − √ (−1)j Aj π 2 j=0

dt 2iπ

ds dt ∧ 2iπ 2iπ

ˆ −c+n+iR

3 2λ

ˆ

ds f (s) Γ(t − s) 2iπ

c+m+i R

Γ(−s)Γ s + 14 Γ s + 43 Γ 14 Γ 34

−t

π Γ (t − j) + Rn,m sin πt

−t

π f (s)Γ(t − s) sin πt

with 1 Rn,m = − √ π 2

ˆ (−c+n+iR) ×(−c+m+iR)

Yet, we have (m 6 n), |Rn,m | = O n→∞

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3 2λ

! −n−c 3 −n n−m m+c− 12 e (n − m) m 2λ Asymptotic expansions and Mellin-Barnes representation



Optimal Truncation Scheme For m =

a1 |λ|

+ b1 , we have

|Rn,m | = O |λ|→0

|Rn | = O |λ|→0

Paris et al.

p e

90’s

a

|λ|e

1 a +ln 3−ln 2 − 0 |λ| a1|λ| (a0

a

0 − |λ|

2a0 3

a0 |λ|

− a1 )

a0 −a1 |λ|

!

We can optimize the remainders in two ways, first satisfying the condition on Rn and then optimizing Rn,m or optimizing directly Rn,m :

OTS 1

3 a0 = 2

OTS 2

a0 = 3

3 a1 = 4 3 a1 = 2

D. Greynat

Rn = O(·) e

3 − 2|λ|

−

p

Rn,m = O(·) |λ|e

3 (1+ln 2) − 2|λ|

p

|λ|e

3 − |λ|




Hyperasymptotic expansion of Z (0) at first hyperasymptotic level

Z (0) =

n−1 X

(−1)k Ak

k =0

3 2λ

k

m−1 1 X − √ (−1)j Aj π 2 j=0

1 − √ π 2

ˆ

dt 2iπ

−c+n+iR

ds dt ∧ 2iπ 2iπ


D. Greynat

3 2λ

3 2λ

−t

−t

π Γ (t − j) sin πt





OTS1 at first hyperasymptotic level h

Z (0) =

i

3 −1 2|λ|

X

(−1)k Ak

k =0

m−1 1 X (−1)j Aj − √ π 2 j=0

+O

p

|λ|e

−

3 2λ

k ˆ

h i 3 −c+ 2|λ| +iR

3(1+ln 2) 2|λ|

dt 2iπ

3 2λ

−t


OTS2 at first hyperasymptotic level Z (0) =

h

i

3 −1 |λ|

X

(−1)k Ak

k =0

1 − √ π 2 +O

h

p

i

3 2λ

k

3 −1 2|λ|

X

j

(−1) Aj

j=0

|λ|e

3 − |λ|

ˆ

h i 3 −c+ |λ| +iR

D. Greynat

dt 2iπ

3 2λ

−t





Iterations... We have shown that it is possible to write the remainder as m−1 1 X Rn = − √ (−1)j Aj π 2 j=0

1 − √ π 2

where f (s) =

ˆ

dt 2iπ

−c+n+iR

3 2λ

ds dt ∧ 2iπ 2iπ


−t


3 2λ

−t


1 3 Γ(−s)Γ s + 14 Γ s + 43 1 Γ s+ 4 Γ s+ 4 π √ = − Γ(s + 1) sin πs Γ 14 Γ 34 π 2

Yet applying the Inverse factorial expansion one has Functional equation  0 m −1 π X 1 (−1)l Al Γ(s − l) + f (s) = − √ π 2 sin πs l=0

ˆ c+m0 −i R

 dt f (t) Γ(s − t) 2iπ

Then using this expression of f in Rn one may get straightforwardly a hyperasymptotic expansions at an arbitrary hyperasymptotic level. D. Greynat


Resummation Resurgence phenomenon Numerics

Formal approach

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The starting point is now the expression of Z (0) as a divergent series k k n−1 ∞ k k 1 1 λ λ 1 X (−1) Γ 2 + 2k 1 X (−1) Γ 2 + 2k Z (0) ∼ √ +√ λ→0 k! 6 k! 6 π k =0 π k =n {z } | {z } | . . =Rn =S Pert. n−1

Using the duplication formula,

Rn =

k ∞ 1 3 1 XΓ k+4 Γ k+4 2λ √ − Γ (k + 1) 3 π 2 k =n

We can apply the Inverse factorial expansion for m−1 X Γ k + 14 Γ k + 34 (−1)j Aj Γ(k − j) + = Γ(k + 1) j=0

1 where Aj = j!

1 3 4 j 4 j

and

D. Greynat

ˆ

ds f (s) Γ(k − s) 2iπ

c+m+i R

Γ(−s)Γ s + 14 Γ s + 43 f (s) = Γ 14 Γ 34



Resummation and terminant function The remainder is then Rn =

k ∞ m−1 X −2λ 1 X √ Γ(k − j) (−1)j Aj 3 π 2 j=0 k =n k ˆ ∞ X ds −2λ 1 Γ(k − s) + √ f (s) 2iπ 3 π 2 c+m+i R

k =n

One can now perform a Borel resummation on the two infinite sums in Rn , "∞ k # n X 2λ 3 −2λ = Γ(n − j) − Λn−j−1 Γ(k − j) B 3 3 2λ k =n

The function Λ` is known as a terminant function, here Λ` (x) = x

`+1

ˆ

∞

where Γ(a, x) =

Dingle 1973

x

e Γ(−`, x)

dy y a−1 e −y

x

is the incomplete Gamma function. D. Greynat



Therefore performing the summation, one has the expansion (m 6 n) k n−1 k 1 1 X (−1) Γ 2 + 2k λ Z (0) = √ k! 6 π k =0 (−1)n + √ e π 2 +

3 2λ

(−1)n √ e π 2

m−1 X j=0

3 2λ

j

(−1) Aj Γ(n − j) ˆ

c+m+i R

ds 2iπ

2 3λ

3 Γ −n + j + 1, 2λ 3 f (s) Γ(n − s) Γ −n + s + 1, 2λ

2λ 3

−s

j

where the sub-dominant terms appear explicitly through the expression of the tail of the series. Using the MB representation of the incomplete Gamma function we have exactly the hyperasymptotic expansion of Z (0) at first level obtained in the analytic approach.

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Resurgence phenomenon If we consider a few terms in the perturbative expansion of Z (0) Z (0) = 1 −

1 35 2 385 3 λ+ λ − λ + ··· 8 384 3072

we have the same coefficient in the non-perturbative expansion of Rn as (−1)n 3 Rn = − √ e 2λ π 2 3 1 3 × 1 Γ(5)Γ −4, − λ Γ(4)Γ −3, 2λ 8 2λ 35 2 3 385 3 3 + λ Γ(3)Γ −2, − λ Γ(2)Γ −1, +··· 384 2λ 3072 2λ This resurgence phenomenon also appears in higher order hyperasymptotic levels.

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Numerical analysis

OTS 1

OTS 2

Pertur. expa.

OTS 2

m

m0

m00

1 2 3

3 ]=4 [ 2|λ| 3 [ 2|λ| ]=4 3 [ 2|λ| ]=4

3 ]=2 [ 4|λ| 3 [ 4|λ| ]=2 3 [ 4|λ| ]=2

3 [ 8|λ| ]=1 3 [ 8|λ| ]=1

3 [ 8|λ| ]=0

1 2 3

3 [ |λ| ]=9 9 [ 2|λ| ] = 13 6 [ |λ| ] = 18

3 [ 2|λ| ]=4 3 [ |λ| ] = 9 9 [ 2|λ| ] = 13

3 [ 2|λ| ]=4 3 [ |λ| ]=9

3 [ 2|λ| ]=4

Z (0) 0.965560481.. 0.96555187

Mathematica

OTS 1

n

Sn

Sm

0.9638 0.9638 0.9638 0.9696 1.0573 -27.696

0.0017 0.0017 0.0017 -0.0040 -0.0917 28.662

Sm 0

Sm00

0.000041 0.000041

0

±0.001410990

1 2 3 1 2 3

0.96552297 0.96556492 0.96556492 0.965562911 0.965560477 0.965560486

D. Greynat

0.0000061 -0.0001292

9 × 10−9



• Numerical stability

Olver et al.

’95

It has been proven that it is possible to obtain numerical stability by choosing another OTS, taking n+1−j aj = n+1 For the third hyperasymptotic level we have n=8

m0 = 4

m=6

m00 = 2

and Z (0) = 0.965560480 • The non-Borel summable sector For λ = − 13 , we have Mathematica Stieljes S5 +

S2NP

1.05995021 − 0.00758472 i

1.06098837 + 0 i ± 0.00140990 1.05990083 − 0.00752794 i D. Greynat


C ONCLUSIONS The Mellin Barnes representation is a perfect tool to obtain asymptotic expansions of Feynman diagrams containing one or several scales. It assures the calculation at any order and under certain conditions the exact representation. The Mellin Barnes representation is also a powerful instrument to deal with perturbative divergent series and their resummation. It allows also through the hyperasymptotic method to get non perturbative contributions directly from the perturbative expansion itself.

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