Evaluating Feynman integrals using Mellin-Barnes ...prac.us.edu.pl/~gluza/WarsawSem06.pdfJanusz Gluza, University of Silesia Evaluating Feynman integrals using Mellin-Barnes representations

Janusz Gluza, University of Silesia

Evaluating Feynman integrals usingMellin-Barnes representations

Mellin, Robert, Hjalmar, 1854-1933

Barnes, Ernest, William, 1874-1953

MB integrals – 4 December, Warsaw

Outline

Introduction: methods to evaluate FIMellin-Barnes representation, idea

1-loop cases

Multiple MB integrals

Automatizations (very important)Getting MB representations themselves

Analytical continuation in ǫ (and not only!)

Two-, three- and four loop resultsBhabha process in massive QED

Conjectures in SYM theories

PerspectivesMB integrals – 4 December, Warsaw

Methods: analytical, numerical, semi-analytical

For any graph Γ:

(i) reduce all tensorial loop integrals to scalar integrals,

(ii) reduce these to a smaller set of scalar master integrals (MIs),

(iii) evaluate the MIs.

Technically, one has to calculate all the Feynman integrals G(X) related to these

diagrams by a reduction of integrals with irreducible numerators X and denominators

with higher powers νi to a smaller set of scalar master integrals

Graph G with L loops and N propagators can be writen in the following general way:

G(X) =1

(iπD/2)L

∫dDk1 . . . d

DkLX

(q21 − m21)ν1 . . . (q2N − m2N )νN

X = 1, k1α, k1αk2β, . . . stands for tensors in the loop momenta.


General procedure (advanced)

For dimensionally regularized FI we can use Integration By Parts identities (and

always neglect surface terms (!)): ∂∂kµ

i

G(X) = 0, i = 1, ..., L

IBP relations (and LI) defines smaller set of integrals which must be evaluated

(so called Master Integrals)

Procedure:

construction of reduction procedure (Laporta, Remiddi)

evaluating MI

Methods to evaluate MI:

α parametrisation, Feynman parametrisation

Mellin-Barnes representation

Differential equations, difference equations, expansions by regions,

dispersion relations


α-representation, Feynman parametrisation

1

(k2 − m2 + iǫ)ν =i−ν

Γ(ν)

∫ ∞

0

dααν−1 exp iα(k2 − m2 + iǫ)

∫ddk exp[i(Ak2 + 2(pk))] = i

( πiA

)d/2exp

[− ip

2

A

]

1

An11 An22 . . .

=Γ(n1 + . . . + nm)

Γ(n1) . . .Γ(nm)∫ 1

0

dx1 . . .

∫ 1

0

dxmxn1−11 . . . x

nm−1m δ(1 − x1 . . . − xm)

(x1A1 + . . . + xmAm)n1+...nm

see papers by Tarasov (in general, integrals with shifted dimensions),

books by Smirnov, e.g. Feynman Integrals Calculus, Springer 2006


F and U polynomials

Czakon, J.G., Riemann (CGR), PRD71(2005)073009

G(X) =1

(iπD/2)L

∫dDk1 . . . d

DkLX

(q21 − m21)ν1 . . . (q2N − m2N )νN

We can rewrite the above expression in the following form (for X=1):

G(1) = (−1)Nν Γ(Nν −D2 )

Γ(ν1) . . .Γ(νN )

∫ 1

0

N∏

j=1

dxjxνj−1j δ

(1−

N∑

i=1

xi

)U(x)Nν−D2

(L+1)

F (x)Nν−D2

L

where Nν is sum of all propagators:

Nν = ν1 + ν2 . . . νN

for all 1 loop diagrams U(x) = 1


Example: 1-loop box, general case: graphical approach ’50s

Trees contributing to the polynomial U for the square diagram

2 − trees contributing to the polynomial F for the square diagram

x1

x2

x3

x4

p1

p2

p3

p4

U = x1 + x2 + x3 + x4

F = tx1x3 + sx2x4

Cuts of internal lines

such that :

U : (i) every vertex is still connected to every other vertex by a sequence of uncut

lines; (ii) no further cuts without violating (i)

F : (iii) divide the graph into two disjoint parts such that within each part (i) and

(ii) are obeyed and such that at least one external momentum line is connected

to each part;


Singularities in the complex plane: Γ(z) =∫ ∞0

tz−1e−tdt

1

(A + B)λ=

1

Γ(λ)

1

2πi

∫ +i∞

−i∞

dzΓ(λ + z)Γ(−z) Bz

Aλ+z


The Mellin-Barnes relation, exact proof: Whittaker, Watson, Cambridge U.Press, 1965

We can write:

1

(A + B)λ=

1

Aλ1

(1 + B/A)λ≡ 1

Aλ1

(1 + B̃)λ

Taylor series:

LHS =1

(1 + B̃)λ=

∞∑

n=0

(−1)n λ . . . (λ + n − 1)n!

B̃n

On the other hand:

RHS =1

(1 + B̃)λ=

1

Γ(ν)

1

2πi

∫ +i∞

−i∞

dzB̃zΓ(λ + z)Γ(−z)


The Mellin-Barnes relation, exact proof: Whittaker, Watson, Cambridge U.Press, 1965

According to the Cauchy theorem:

∫

C

f(z)dz = 2πi∑

i

reszif

By closing the integration contour to the right and taking a series of residues (with

minus sign) at points z = 0, 1, 2, . . . we obtain:

RHS =1

Γ(λ)

1

2πi

∫ +i∞

−i∞

dz2πi∞∑

n=0

1

(−1)nn!Γ(λ + n)B̃n

By putting Γ(λ + n) = λ . . . (λ + n − 1)Γ(λ), we can see that

LHS = RHS

MB integrals – 4 December, Warsaw –

straightforward generalization

We can expand the M-B relation to the general case:

1

(A1 + . . . + An)λ=

1

Γ(λ)

1

(2πi)n−1

∫ +i∞

−i∞

. . .

∫ +i∞

−i∞

dz1 . . . dzn

n∏

i=2

Azii

A−λ−z2−...−zn1 Γ(λ + z2 + . . . + zn)n∏

i=2

Γ(−zi)


Mellin-Barnes method

Individual integrals: Ussyukina (1975), Davydychev (1989);

Application to dimensionally regularized integrals (systematic solution of the

singularities in ǫ: Smirnov (1999), Tausk (1999)

Massive QED: Actis, Czakon, Kajda, Riemann, J.G.

SYM theories: Czakon, Dixon, Kosower, Smirnov

Conformal symmetries: Smirnov, ....


2-loop Mellin-Barnes representation: construction

B5l2m2:

K1 = k21

K2 = (k2 − p1)2

K3 = (k1 + k2)2 − m2

K4 = (k2 − p3)2

K5 = (k2 − p3 − p4)2 − m2

KB5l2m2 =

∫dDk2

1

Kν11

1

Kν44

1

Kν55Kup

Kup =

∫dDk1

1

Kν11

1

Kν33

K1

K3

K4K2

K5


Two loop example: construction

For Kup we write:

Kup = (−1)ν13Γ(ν13 − D/2)Γ(ν1)Γ(ν3)

∫ 1

0

∏

j=1,3

dxjxνj−1j δ(1 − x1 − x3)

Uν13−Dup

Fν13−D/2up

We find F(x) polynomial (whichever method): Fup = (−k22 + m2)x1x3 + m2x23Next:

use M-B relation for F−(ν13−D/2)up

integrate over x

∫ 1

0

N∏

i

dxjxνj−1j δ

(1 −

N∑

i

xi

)=

Γ(ν1) . . .Γ(νN )

Γ(ν1 + . . . νN )



Result for Kup part:

Kup = (−1)ν131

Γ(ν1)Γ(ν3)Γ(D − ν13)

∫ +i∞

−i∞

dσ13[−k22 + m2]σ13 ×

Γ(−σ13)Γ(ν13 − D/2 + σ13)Γ(ν1 + σ13)Γ(D − 2ν1 − ν3 − σ13)

So now our aim is the following integral:

KB5l2m2 = (−1)ν131

Γ(ν1)Γ(ν3)Γ(D − ν13)∫ +i∞

−i∞

dσ13Γ(−σ13)Γ(ν13 + σ13 − D/2)Γ(ν1 + σ13)

Γ(D − 2ν1 − ν3 − σ13)Kdown

Kdown =

∫dDk2

1

Kν22 Kν44 K

ν55

1

[−k22 + m2]−σ13



For Kdown we perform the same steps as for Kup.

We find Fdown

Fdown = (x13 + x5)2 + [−t]x2x4 + [−s]x13x5

Kup

K4K2

K5

Kdown =(−1)ν245−σ13Γ(νdown)

Γ(ν2)Γ(ν4)Γ(ν5)Γ(−σ13)

∫ 1

0

∏i=2,4,5,13 x

νi−1i δ(1 −

∑xi)

F νdowndown

We use MB-relation, integrate over xi parameters and arrived at the final result


B5l2m2: Final MB-representation, D = 4 − 2ǫ

KB5l2m2 = (−1)ν123451

∏5i=1 Γ(νi)Γ(D − ν13)∫ i∞

−i∞

dσ13dρ1dρ3

(t

s

)ρ3(−s)ν245−D/2−σ13−ρ1

Γ(−σ13)Γ(ν13 − D/2 + σ13)Γ(ν1 + σ13)Γ(D − 2ν1 − ν3 − σ13)Γ(−ρ1)Γ(−ρ3)Γ(ν245 − D/2 − σ13 + ρ1 + ρ3)Γ(−ν24 + D/2 + σ13 − ρ1 − ρ3)Γ(−ν245 + D/2 − ρ1 − ρ3)Γ(−2ν24 − ν5 + D + σ13 − 2ρ3)

Γ(ν2 + ρ3)Γ(ν4 + ρ3)1

Γ(D − ν245 + σ13)1

Γb(ν245 − D/2 − σ13)1

Γa(D − 2ν2 − 2ν4 − ν5 + σ13 − 2ρ1 − 2ρ3)


Loop-by-loop algorithm: planar cases

(i) define kinematic which depends on external legs (invariants);

(ii) make decision about the order in which n 1-loop subloops (n ≥ 1) will beworked out in a sequence;

(iii) construct Feynman integral for the chosen subloop, make manipulations on the

F polynomial to make it the most suitable for using MB representations;

(iv) use the basic MB-relation;

(v) make integration over Feynman parameters;

(vi) go back to the point (iii) and repeat the steps till F in the last n subloop will be

changed to the MB integral.

package AR.m (Kajda, Riemann, J.G.), example: NL 3loop.nb.ps


How to solve a given MB-representation?

means expansion in ǫ


B5l2m2d2 case: fixed indices a1 = a3 = a4 = a5 = 1, a2 = 2

General Tasks

Find a region where integral is regular in the n-fold MB-integral

(FindInstance )

ℜ(α) = 361/384,ℜ(β) = −117/128,ℜ(γ) = −19/32,ℜ(ǫ) = 1/32

Go to the physical region where ǫ → 0 by distorting the integration path step bystep (adding each crossed residuum – per residue this means one integral less

(automatized in MB.m, M.Czakon, hep-ph/0511200)

If we want to get exact analytical result: Take integrals by sums over residua, i.e.

introduce infinite sums

Sum these infinite multiple series into some known functions of a given class,

e.g. Nielsen polylogs, Harmonic polylogs or whatever is appropriate (read:

managable)


Typical situation

B5l2m2 :

Directly from 7-line B7l4m1: After expansion in ǫ we are left with 11

integrals (one 4-dim.)

From our MB-representation: 4 integrals (one 3-dim)

B5l2m2d2 :

From B7m4m1 with help of MB.m: 102 integrals, including 4-dim (!)

From our MB-representation: only one, 3-dim integral,


“Take residues and shift contours”

1. Idea of Tausk automatized by M. Czakon (package in Mathematica), and also by

Anastasiou

2. This procedure defines MB-integrals for ǫn, but not solve resulting MB integrals

after continuation in ǫ (ok, numerically).

3. remark: Smirnov makes it a little bit different way (“glueing”)

I will show how it works “manually”


B5l2m2H* shifting contours *L

In[203]:=

sim = Gamma@-zDOut[203]=

Gamma@-zDIn[227]:=

Sum@-Residue@Gamma@-zD, 8z, n

B5l2m2In[1]:= backGamma =

8G1@x__D -> Gamma@xD, G2@x__D -> Gamma@xD,G3@x__D -> Gamma@xD, G4@x__D -> Gamma@xD,G5@x__D -> Gamma@xD, G6@x__D -> Gamma@xD,G7@x__D -> Gamma@xD, G8@x__D -> Gamma@xD,G9@x__D -> Gamma@xD, G10@x__D -> Gamma@xD,G11@x__D ® Gamma@xD, G12@x__D ® Gamma@xD<

Out[1]= 8G1@x__ D ® Gamma@xD, G2 @x__ D ® Gamma@xD,G3@x__ D ® Gamma@xD, G4 @x__ D ® Gamma@xD,G5@x__ D ® Gamma@xD, G6 @x__ D ® Gamma@xD,G7@x__ D ® Gamma@xD, G8 @x__ D ® Gamma@xD,G9@x__ D ® Gamma@xD, G10 @x__ D ® Gamma@xD,G11@x__ D ® Gamma@xD, G12 @x__ D ® Gamma@xD<

In[2]:= fact = HtsL^ro3 H-sL^H-1 - ep + si - ro1L .HtsL^x_ ® H-tL^x H-sL^H-xL

Out[2]= H-sL-1-ep-ro1 -ro3 +si H-t Lro3In[3]:= basic =

1Gamma@2 - 2 epD*G2@a1 + a3 - d2 + siD*G3@a1 +siD*G4@d - 2 a1 - a3 - siD*G5@-ro1D*

G6@-ro3D*G7@a2 + a4 + a5 - d2 - si + ro1 +ro3D*G8@-a2 - a4 + d2 + si - ro1 - ro3D*

G9@-a2 - a4 - a5 + d2 - ro1 - ro3D*G10@-2 a2 - 2 a4 - a5 + d + si - 2 ro3D*G11@a2 + ro3D*G12@a4 + ro3DG@d - a2 - a4 - a5 + siDG@d - 2 a2 - 2 a4 - a5 + si - 2 ro1 - 2 ro3D .

a1 ® 1 . a2 ® 1 . a3 ® 1 . a4 ® 1 .a5 ® 1 . d ® 4 - 2 ep Simplify

Out[3]= HG10@-1 - 2 ep - 2 ro3 + si D G11@1 + ro3 D G12@1 + ro3 DG2@ep si D G3@1 si D G4@1 2 ep si D@ D @ D @ D

1


B5l2m2

In[4]:= ? FindInstance

FindInstance @expr, vars D finds an instanceof vars that makes the statement exprbe True. FindInstance @expr, vars, dom Dfinds an instance over the domain dom.Common choices of dom are Complexes,Reals, Integers and Booleans. FindInstance @expr, vars, dom, n D finds n instances. More¼

In[5]:= FindInstance@Cases@Numerator@basicD . backGamma,Gamma@_D^n_.D . Gamma@x_D^n_. ® x > 0 .

ep ® 0 . si ® -120, 8ro1, ro3 0 .ep ® 110, 8si, ro1, ro3 0 .

ep ® 110 . si ® -120, 8ro1, ro3

B5l2m2

In[10]:= basic . subst . ep ® 110 . si ® -120

Out[10]=1

G@ 34 D G@1D Gamma@ 95 DJG10A 3

8E G11A 3

16E G12A 3

16E G2A 1

20E G3A 19

20E G4A 17

20E

G5A 516E G6A 13

16E G7A 1

40E G8A 39

40E G9A 1

40EN

In[11]:= basic . subst0 . si ® -120

Out[11]=1

G@ 1920 D G@ 65 DJG10A 23

40E G11A 3

16E G12A 3

16E G2A- 1

20E G3A 19

20E

G4A 2120E G5A 5

16E G6A 13

16E G7A- 3

40E G8A 43

40E G9A 1

8EN

In[12]:= constBasic = Normal@Series@fact*basic*Exp@2 ep EulerGammaD .

backGamma . G@x_D ® Gamma@xD, 8ep, 0, 0

B5l2m2

In[19]:= basic

Out[19]= HG10@-1 - 2 ep - 2 ro3 + si D G11@1 + ro3 D G12@1 + ro3 DG2@ep + si D G3@1 + si D G4@1 - 2 ep - si DG5@-ro1 D G6@-ro3 D G7@1 + ep + ro1 + ro3 - si DG8@-ep - ro1 - ro3 + si D G9@-1 - ep - ro1 - ro3 DL

HG@1 - 2 ep + si D G@-1 - 2 ep - 2 ro1 - 2 ro3 + si DGamma@2 - 2 epDL

In[70]:= K7 = Residue@fact*basic . backGamma .G@x_D ® Gamma@xD, 8si, 1+ ep + ro1 + ro3

B5l2m2In[72]:= Normal@

Series@K7*Exp@2 ep EulerGammaD . backGamma .G@x_D ® Gamma@xD, 8ep, 0, 0

B5l2m2

In[80]:= auxK78 = Normal@Series@K78*Exp@2 ep EulerGammaD . backGamma .G@x_D ® Gamma@xD, 8ep, 0, 0

B5l2m2

In[86]:= eps1 . x ® x@-12D N

Out[86]= -5.32296

In[47]:= epm2A

Out[47]= -1tHH-t L-ro1 Gamma@-ro1 D3Gamma@1 + ro1 D Gamma@1 + 2 ro1 DL

In[111]:=

Residue@epm2A . t ® -H1 - yL^2 y, 8ro1, 0

B5l2m2

In[132]:=

pol2 =-Normal@Series@Coefficient@eps1, PolyLog@2, xDD*

PolyLog@2, xD, 8x, 0, 10

However,

sometimes happens that this method doesn’t work in the way I’ve just described (an

example B5l2m2d2 for the second planar topology B2 in massive Bhabha QED, after

contraction of appropriate two lines)

Heinrich, Smirnov, PLB598 (2004)

Czakon, J.G., Riemann, PRD (2005)

B7l4m3B7l4m2�2 �4 �3

B7l4m1MB integrals – 4 December, Warsaw –

MB.m, no rules

In[1]:=

no rules found ...

and we need continuation in another parameter ...


MB.m, another analytic continuation

In[16]:= rules = MBoptimizedRules@orig, a1® 0,8a1 < 12 + 1100, ep > -12 - 1100


cont2 = exp1 . MBint@integrand_ , rules_D ¦MBcontinue@integrand, ep® 0, rulesD;

Level 1

Taking -residue in z3 = -3 - 2 ep - z2 - 2 z4

Taking -residue in z3 = -2 - 2 ep - z2 - 2 z4

Level 2

Integral 81<Taking -residue in z4 = -1 -

3 ep

2... ..

Integral 84, 1, 1, 4, 1 <Integral 84, 1, 2, 2, 1 <65 integral HsL foundexp2 = MBexpand@MBmerge@cont2D,

Exp@2 ep EulerGammaD, 8ep, 0, 0


MBintegrate@exp2, 8s ® -2, t ® -3

MB-integrals and IR-singularities

With MB-integrals it is easy to isolate and evaluate analytically the IR-singularities of

simple Feynman integrals.

Let us use as an example the massive QED one-loop box

Box(t, s) =eǫγE

Γ[−2ǫ](−t)(2+ǫ)1

(2πi)2

∫ +i∞

−i∞

dz1

∫ +i∞

−i∞

dz2

(−s)z1(m2)z2(−t)z1+z2 Γ[2 + ǫ + z1 + z2]Γ

2[1 + z1]Γ[−z1]Γ[−z2]

Γ2[−1 − ǫ − z1 − z2]Γ[−2 − 2ǫ − 2z1]

Γ[−2 − 2ǫ − 2z1 − 2z2]

After continuation in ǫ:

Box(t, s) = −1ǫI1 + Log(−s)I1 + ǫ

(1

2

[ζ(2) − Log2(−s)

]I1 − 2I2

).



I1 =eǫγE

st

1

2πi

∫ − 12+i∞

− 12−i∞

dz1

(m2

−t

)z1 Γ3[−z1]Γ[1 + z1]Γ[−2z1]

,

and

I2 =eǫγE

t21

(2πi)2

∫ − 34+i∞

− 34−i∞

dz1

(−s−t

)z1Γ[−z1]Γ[−2(1 + z1)]Γ2[1 + z1]

×∫ − 1

2+i∞

− 12−i∞

dz2

(m2

−t

)z2Γ[−z2]

Γ2[−1 − z1 − z2]Γ[−2(1 + z1 + z2)]

Γ[2 + z1 + z2].

In terms of the conformally mapped variable

y =

√1 − 4m2/t − 1√1 − 4m2/t + 1

,



the first integral I1 can be performed analytically to yield the well known result

I1 =1

m2s

2y

1 − y2 Log(y).

the residua due to poles of Γ(1 + z) in this region have to be taken at:

1 + z = −n, n = 0, 1, · · ·, or equivalently at z = −n, n = 1, 2, · · ·. It is:

Residue[F[z]Gamma[1 + z], {z,−n}] = (−1)n−1

(n − 1)! F[−n]

So the integral becomes:

I =1

ǫ

∞∑

n=0

(−1)n(−s)nΓ(1 + n)3)8n!Γ(−2(−1 − n)) =

1

ǫ

ArcSin(√

s/2)

2√

4 − s√s=

1

ǫ

−x4(1 − x2) ln(x)



Mathematica can do this:

Sum[sˆ(n) Gamma[n + 1]ˆ3/(n!Gamma[2 + 2n]), n, 0, Infinity] =

(4*ArcSin[Sqrt[s]/2])/(Sqrt[4 - s]*Sqrt[s])

Yet another way: binomial sums (Kalmykov, Davydychev)

I1 ∝∞∑

n=0

sn 2n

n

(2n + 1)

and solving this binomial we get the same result,

it is more general approach,

Mathematica will not always solve directly :-(((


MI B1 B2 B3 B4 B5 B6 solved:

B7l4m1 + – – – – – Smirnov:2001,CGR:2006

B7l4m1N + – – – – – Smirnov,Heinrich:2004,CGR:2006

B7l4m2 – + – – – – Smirnov,Heinrich:2004†,CGR:2006

B7l4m2[d1--d3] – + – – – – CGR:2006

B7l4m3 – – + – – – NP:Smirnov,Heinrich:2004†

B7l4m3[d1--d2] – – + – – – NP

B6l3m1 + – + – – – CGR:2006

B6l3m1d + – + – – – CGR:2006

B6l3m2 – + – + – – CGR:2006

B6l3m2d – + – + – – CGR:2006

B6l3m3 – – + – – – NP

B6l3m3[d1--d5] – – + – – – NP

B5l2m1 + – + – – – CGR:2004

B5l2m2 – + – + – + CGR:2006

B5l2m2[d1--d2] – + – + – + CGR:2006

B5l2m3 + – + – – – CGR:2006

B5l2m3[d1--d3] + – + – – – CGR:2006

B5l3m – + + + – – CGR:2006

B5l3m[d1--d3] – + + + – – CGR:2006

B5l4m – + + + + – Bonciani, Mastrolia, Remiddi:2002

B5l4md – + + + + – CGR:2004


Second planar, Smirnov, Heinrich, PLB598(2004)

BPL,2(1, . . . , 1, 0; s, t, m2; ǫ) =

(iπd/2ǫ−γEǫ

)2 [− xy

s2(−t)1+2ǫ

×(

c22(x, y)

ǫ2+

c21(x, y)

ǫ+ c201(x, y) + c

202(x, y)

)+c203(s, t, m

2) + O(ǫ)]

,

c203(s, t, m2) = − 4

s√−t

∫ 1

0

∫ 1

0

dx1dx2

√x1√

1 − x1

×Arsh[(√−t√x1

√1 − x2)/(2m

√x1 + x2 − x1x2)

]

(4m2 − sx1)x2√

(4m2 − t)x1(1 − x2) + 4m2x2×(ln(−s/m2) + 2 lnx1 − ln

[4(1 − x1)x2 − sx1(x1 + x2 − x1x2)/m

+1

s

∫ 1

0

∫ 1

0

dx1dx2ln(x1 + x2 − x1x2)√

1 − x1(4m2 − sx1)√

1 − x2(4m2 − tx2)×

(4 ln 2 + 2 ln(−4m2/s + x1) + 2 ln(1 − x1)

−2 lnx1 + 2 ln(1 − x2) − ln(x1 + x2 − x1x2)) . MB integrals – 4 December, Warsaw –

Approximations

e.g. (one initial 3-dim integral for B5l2m2d2)

B5l2m2d2 = m−4−2ǫ(−m2/s

)ǫ+α(−x)β

× Γ[1 − ǫ − α − β]Γ[−β]Γ[1 + β]× Γ[1 + ǫ + β]Γ[ǫ + α + β]Γ[−2ǫ − γ]Γ[1 − α − γ].......

ℜ(α) = 361/384,ℜ(β) = −117/128,ℜ(γ) = −19/32

m2/s

Final structure for approximated MIs

In all planar cases, after expansions, we arrived at 1-dimensional integrals of the kind

(B5l2m2d2 example)(−m2/s

)n=2(−x)βΓ[...β...]

or(−m2/s

)...β...(−x)...β...Γ[...β...], or ...Γ[...2β...]...,

which can be solved analytically using XSummer (Uwer, Moch), or PSLQ (Ferguson,

Bailey) algorithms

Finding iteratively residues must be automatized (sometimes hundreds of terms must

be considered).

Finally, we get

I ∼ snF(

ln

(−m

2

s

),t

s

)+ O

(m2

), n =

1

2dim I + 2ǫ,

This is requirement we impose for the structure of MIs


Approximations: chosen MIs for B5l2m2 system

B5l2m2 = − 1ǫ

1

6t

[−3 L2 + 6 L ln(x) − 3 ln2(x) − 24 ζ2

]

− 16t

{const

}

B5l2m2d2 = +1

6st

{−12 L3 + 9 L2 ln(x) − 36 L ζ2

− 12 ζ3 − 12 ζ2 ln(x) − ln3(x) + 18 ζ2 ln(1 + x)

+ 3 ln2(x) ln(1 + x) + 6 ln(x) Li2(−x) − 6 Li3(−x)}

B5l2m2(k2 · p3) = +1

ǫ

s

12t

[3 L2 + L (12 + 6 x − 6 ln(x))

+ 24 ζ2 − 12 ln(x) − 6 x ln(x) + 3 ln2(x)]

+s

12t

{const

}


Solutions: The planar box masters for m2/s

B5l2m3 (Musashi), x = t/s

B5l2m3 = +1

12u

{−6 L2 (6 ζ2 + ln2(x))

− 6 L (−4 ζ3 + 4 ζ2 ln(x) − 12 ζ2 ln(1 + x)− 2 ln2(x) ln(1 + x) − 4 ln(x) Li2(−x) + 4 Li3(−x)) + 312 ζ4+ 72 ζ3 ln(x) + 36 ζ2 ln

2(x) + ln4(x) − 24 ζ3 ln(1 + x)+ 24 ζ2 ln(x) ln(1 + x) − 36 ζ2 ln2(1 + x)− 6 ln2(x) ln2(1 + x) − 24 ln(x) S1,2(−x) + 12 (8 ζ2 + ln2(x)− 2 ln(x) ln(1 + x)) Li2(−x) − 48 ln(x) Li3(−x)

+ 24 ln(1 + x) Li3(−x) + 72 Li4(−x) + 24 S2,2(−x)}


B5l2m3, transcendentality: 4

Higher orders in ǫ may be determined:

B5l2m3d2 = · · ·

+ ǫ1

3st

{{−5 L4 + 111 L2 ζ2 + 10 L3 ln(x) + L (104 ζ3

− 126 ζ2 ln(x) − 6 ln3(x) + 108 ζ2 ln(1 + x) + 18 ln2(x) ln(1 + x)+ 36 ln(x) Li2(−x) − 36 Li3(−x)) − 372 ζ4 − 78 ζ3 ln(x) + 30 ζ2 ln2(x) ++ ...

− 6 (4 ζ2 + 3 ln2(x) − 2 ln(x) ln(1 + x)) Li2(−x) + 24 ln(x) Li3(−x)

− 12 ln(1 + x) Li3(−x) − 12 Li4(−x) − 12 S2,2(−x)}}


Analytical, numerical checks

some results can be checked by comparing with different methods, e.g. DEqs,

Czakon, J.G. , Kajda, Riemann, NPB (2005);

MB-DEqs - complementary methods;

Multidimensional MB-integrals checked with MB.m (by Czakon)

Comparison with sector decomposition (Binoth, Heinrich)

Expansions (analytical form)

m2/s = 1/1000, t/s = 1/4

B5l2m2d2 = 0.00252823

B5l2m2d2expanded = 0.0025307


The Nf > 1 contributions

The 2-box-diagrams repre-

sent a three-scale problem:

s/m2, t/m2, M2/m2

SE3l2M1mV4l2M1m

V4l2M2m B5l2M2m

2

3

4

1

3

5

1

3

4

3

21 2

4

2

1


Results for the Finite Parts

[B5l2M2m]fin =1

st

{−2 ln2 (ms) ln (Ms) + ζ2 ln (ms)

+[2 ln2 (ms) + 2 ln (ms) ln (Ms)

]ln

(t

s

)

− 2 ln (ms) ln2(

t

s

)

+[3ζ2 +

1

2ln2

(t

s

)]ln

(1 +

t

s

)+ ln

(t

s

)Li2

(− t

s

)

− Li3(− t

s

)}

ms ≡ −m2

sMs ≡ −

M2

s


Numerics for Nf = 2

soon, this year


Magic relations, Drummond, Henn, Smirnov, Sokatchev’06

For off-shell Feynman diagrams (p2i 6= 0) exactly at d=4 we have:

Tennis court = s × Triple box


MSYM, AdS/CFT, MHV, four-loops and MB

Bern, Czakon, Dixon, Kosower, Smirnov

“Heuristically, the Maldacena duality conjecture hints that even quantities

unprotected by supersymmetry should have perturbative series that can be

resummed in closed form.”

Higher-loop on-shell scattering amplitudes of colour non-singlet gluons

“There is now significant evidence of a very simple structure in the planar limit.

In particular, the planar contributions to the two-loop and three-loop four-gluon

amplitudes have been shown to obey iterative relations ...”

M(2)4 (ρ; ǫ) =

1

2

[M

(1)4 (ρ; ǫ)

]2

− (ζ2 + ζ3ǫ + ζ4ǫ2) M (1)4 (ρ; 2ǫ) −1

2ζ22 + O(ǫ)



M(3)4 (ρ; ǫ) = −

1

3

[M

(1)4 (ρ; ǫ)

]3+ M

(1)4 (ρ; ǫ) M

(2)4 (ρ; ǫ)

+ f (3)(ǫ) M(1)4 (ρ; 3 ǫ)

+C(3) + O(ǫ)

f (3)(ǫ) =11

2ζ4 + ǫ(6ζ5 + 5ζ2ζ3) + ǫ

2(c1ζ6 + c2ζ23 ),

C(3) =

(341

216+

2

9c1

)ζ6 +

(−17

9+

2

9c2

)ζ23 .

c1 and c2 are expected to be rational, they cancel out on the RHS.



M(4)4 (ρ; ǫ) =

1

4

[M

(1)4 (ρ; ǫ)

]4−

[M

(1)4 (ρ; ǫ)

]2M

(2)4 (ρ; ǫ) + M

(1)4 (ρ; ǫ)M

(3)4 (ρ;

+1

2

[M

(2)4 (ρ; ǫ)

]2+ f (4)(ǫ) M

(1)4 (ρ; 4 ǫ) + C

(4) + O(ǫ)

11

(d)

9

8

10 21 3

7

65

413

12

113)

2l + l 1(

14

2s

(e)

9

8

10

1

12

76

5s)2

1315

23 4

1311

14

3)2

l + l 19

3

76 8

11

10

5

1 12

13

2 4s14

8)2

l + l (4)

2l + l 2(

15

(f)

s 32 3

41

12 7

3

6

45

10

89

13

11129

(a) (b)

10)4

l + l 8(t 123

76

54

109

8

1213 14

(c)

2s 3)2

l + l 1(11

12

14

13

8

10

9 21 3

7

65

4

12(l + lx x(

6


Perspectives

A lot to do with summations (Xsummer by Moch, Uwer generalized to the

massive cases);

Nonplanar topologies;

Limits: more legs?

...


{it scriptsize {white { Janusz Gluza, University of Silesia}}}{scriptsize it Outline}{scriptsize it Methods: analytical, numerical, semi-analytical}{scriptsize it General procedure (advanced)}{scriptsize it $alpha $-representation, Feynman parametrisation}{scriptsize it $F$ and $U$ polynomials}{scriptsize it Example: 1-loop box, general case: graphical approach '50s }{scriptsize it Singularities in the complex plane: $ Gamma (z)=int_{0}^{infty } t^{z-1} e^{-t}dt $ }{scriptsize it The Mellin-Barnes relation, exact proof: Whittaker, Watson, Cambridge U.Press, 1965}{scriptsize it The Mellin-Barnes relation, exact proof: Whittaker, Watson, Cambridge U.Press, 1965}{scriptsize it straightforward generalization}{scriptsize it Mellin-Barnes method}scriptsize it it 2-loop Mellin-Barnes representation: construction{scriptsize it Two loop example: construction}{scriptsize it Two loop example: construction}{scriptsize it Two loop example: construction}{scriptsize it B5l2m2: Final MB-representation, $D=4-2 epsilon $}{scriptsize it Loop-by-loop algorithm: planar cases}scriptsize it How to solve a given MB-representation?scriptsize it B5l2m2d2 case: fixed indices $a_1=a_3=a_4=a_5=1$, $a_2=2$scriptsize it Typical situation{scriptsize it ``Take residues and shift contours''}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}{scriptsize it B5l2m2}scriptsize it However,{scriptsize it MB.m, no rules}scriptsize it no rules found ...{scriptsize it MB.m, another analytic continuation}{scriptsize it MB.m, another analytic continuation}{scriptsize it MB.m, another analytic continuation}scriptsize it MB-integrals and IR-singularitiesscriptsize it MB-integrals and IR-singularitiesscriptsize it MB-integrals and IR-singularitiesscriptsize it MB-integrals and IR-singularities{scriptsize it }scriptsize it Second planar, Smirnov, Heinrich, PLB598(2004)scriptsize it Approximationsscriptsize it Final structure for approximated MIsscriptsize it Approximations: chosen MIs for B5l2m2 systemscriptsize it Solutions: The planar box masters for $m^2/s

Documents

Evaluating Feynman integrals using Mellin-Barnes ...prac.us.edu.pl/~gluza/WarsawSem06.pdfJanusz Gluza, University of Silesia Evaluating Feynman integrals using Mellin-Barnes representations