Документ взят из кэша поисковой машины. Адрес оригинального документа : http://theory.sinp.msu.ru/~smirnov/Ex4.pdf Дата изменения: Sat Oct 25 23:02:01 2008 Дата индексирования: Mon Oct 1 19:53:21 2012 Кодировка:

MB 1.1

by Michal Czakon

more info in hep-phЙ0511200 last modified 06 Mar 08 H Example 4a

<< MBЙMB.m

Frules = MBoptimizedRules@F, ep 0, 8<, 8ep
H

F = Gamma@3 Й 2 + ep + zD Gamma@- 1 - 2 ep - zD Gamma@4 ep + zD Gamma@- zD Gamma@1 Й 2 - ep - zD Й Gamma@1 - 2 ep - zD Gamma@1 - 2 ep - zD Strategy 1 2 Gamma@- 1 - 2 ep - zD GammaB L 1 2

L

- ep - zF Gamma@- zD GammaB

3 2

+ ep + zF Gamma@4 ep + zD

MBrules::norules : no rules could be found to regulate this integral

GammaB

H

- Residue@F, 8z, - 1 - 2 ep 3 2

The two residues

- epF GammaB

3 2

F = Gamma@3 Й 2 + ep + zD Gamma@- 1 - 2 ep - z + yD Gamma@4 ep + zD Gamma@- zD Gamma@1 Й 2 - ep - zD Й Gamma@1 - 2 ep - zD Gamma@1 - 2 ep - zD GammaB 1 2 1 - ep - zF Gamma@- 1 - 2 ep + y - zD Gamma@- zD GammaB 7 4 >, :z - 5 4 >> 3 2

H

H plus an integral with the first poles of Gamma@4 ep+zDand Gamma@-1-2 ep-zD of the opposite nature Strategy

- 3 epE Gamma@4 epD Gamma@- 1 + 2 epD GammaA Gamma@1 + 2 epD

+ epF Gamma@- 1 + 2 epD Gamma@1 + 2 epD +
1 2

L

+ 3 epE

2: introduce an auxiliary analytic regularization

L

L

Step1rules = MBoptimizedRules@F, y 0, 8<, 8ep, y
+ ep + zF Gamma@4 ep + zD

MBrules::norules : no rules could be found to regulate this integral


2

Ex4.nb

Level 1 Taking -residue in z = - 1 - 2 ep + y Level 2 Integral 81<

2 integralHsL found ::MBintB MBintB ::ep 1 2

GammaA

3 2

:MBintBGammaB MBintB

exp1 = MBexpand@con1, 1, 8y, 0, 0
GammaB

Gamma@1 - 2 ep - zD 3 2

, y 0>, 8<>F>, 1

+ ep - yE Gamma@1 + 2 ep - yD GammaA GammaB 1 2 Gamma@2 - yD

1 2

- ep + yE Gamma@- 1 + 2 ep + yD

,

+ ep + zF Gamma@4 ep + zD, ::ep

- ep - zF Gamma@- 1 - 2 ep + y - zD Gamma@- zD 1 2 , y 0>, :z - 5 4 >>F>

con2 = Table@ MBcontinue@exp1@@i, 1DD, ep 0, exp1@@i, 2DDD, 8i, Length@exp1D
Gamma@- zD GammaB

+ ep + zF Gamma@4 ep + zD, ::ep

Gamma@- 1 - 2 ep - zD GammaB

+ epF Gamma@- 1 + 2 epD Gamma@1 + 2 epD, ::ep 1 2 - ep - zF 1 2 , y 0>, :z - 5 4

1 2

>>F>

, y 0>, 8<>F,


Ex4.nb

3

Level 1

Level 1

Taking +residue in z = - 1 - 2 ep Taking +residue in z = - 4 ep Taking +residue in z = - 1 - 4 ep Level 2 Integral 81<

1 integralHsL found

4 integralHsL found

Integral 83<

Integral 82<

::MBintBGammaB :MBintB

1 2

::MBintB- GammaB :MBintB- GammaA

- epF GammaB 1 2

3 2

3 2

MBintB

:MBintB

MBmerge@%D GammaA

Gamma@- zD GammaB GammaA
1 2 3 2

Gamma@1 - 2 ep - zD 3 2

GammaA

1 2

- 3 epE Gamma@4 epD Gamma@- 1 + 2 epD GammaA - 3 epE Gamma@2 epD GammaA Gamma@- 1 - 2 ep - zD GammaB Gamma@1 + 2 epD
3 2

- epF GammaB

+ epF Gamma@- 1 + 2 epD Gamma@1 + 2 epD, 88ep 0, y 0<, 8<, 3 2
1 2

+ epF Gamma@- 1 + 2 epD Gamma@1 + 2 epD, 88ep 0, y 0<, 8<, + 3 epE Gamma@1 + 4 epD 1 2 - ep - zF + 3 epE , 88ep 0, y 0<, 8<, 5 4 >>F>> , 88ep 0, y 0<, 8<,

1

MBintB

exp2 = MBexpand@%, Exp@2 ep EulerGammaD, 8ep, 0, 0
GammaB

Gamma@1 - 2 ep - zD 3 2

- 3 epE Gamma@2 epD GammaA 1

- 3 epE Gamma@4 epD Gamma@- 1 + 2 epD GammaA Gamma@2 + 2 epD Gamma@- 1 - 2 ep - zD GammaB Gamma@1 + 2 epD
3 2

+ ep + zF Gamma@4 ep + zD, :8ep 0, y 0<, :z -
1 2

Gamma@2 + 2 epD

+ 3 epE Gamma@1 + 4 epD 1 2

+ 3 epE

-

+ ep + zF Gamma@4 ep + zD, :8ep 0, y 0<, :z -

- ep - zF Gamma@- zD 5 4 >>F>

, 88ep 0, y 0<, 8<

4

Ex4.nb

:MBintB-

MBmerge@%D

1 ep2

ep2 12 EulerGamma2 + 35 2 - 36 EulerGamma 2 + PolyGammaB0, 3 - 44 + 9 PolyGammaB0, 6 PolyGammaB0, 1 2
1 2

96

6 - 6 ep - 6 + 2 EulerGamma - 3 PolyGammaB0, 1 2 F + 12 PolyGammaB0,
2

1 2

F + 3 PolyGammaB0, 1 2 3 2
2

3 2

3 2

MBintB

:8ep 0, y 0<, :z - 5 4 >>F>

Gamma@- 1 - zD GammaA

- zE Gamma@- zD Gamma@zD GammaA Gamma@1 - zD

F 2 + 3 PolyGammaB0,

3 2

F

F + 9 PolyGammaB0,
3 2

F - PolyGammaB0, F-

F+

3 2

F+

, 88ep 0, y 0<, 8<
In[26]:=

H

Out[26]=

In[27]:=

H Strategy 1: there are no poles. Expand the integrand in epsilon. However, the contour cannot be a straight line. L H Strategy 2 GammaB 3 2 - ep - zF Gamma@- zD GammaB- L 1 2 + ep + zF Gamma@1 + ep + zD

F = Gamma@- 1 Й 2 + ep + zD Gamma@1 + ep + zD Gamma@3 Й 2 - ep - zD Gamma@- zD GammaB 3 2 - ep - zF Gamma@- zD GammaB- 1 2 + ep + zF Gamma@1 + ep + zD

Example

4b

L

Out[27]=

F = Gamma@- 1 Й 2 + ep + zD Gamma@1 + ep + zD Gamma@3 Й 2 - ep - zD Gamma@- zD Frules = MBoptimizedRules@F, ep 0, 8<, 8ep
In[28]:=

MBresidues::contour : contour starts andЙor ends on a pole of Gamma@1 + ep + zD
Out[28]=

MBresidues::contour : contour starts andЙor ends on a pole of Gamma@1 + ep + zD
$Aborted

MBresidues::contour : contour starts andЙor ends on a pole of Gamma@1 + ep + zD

General::stop : Further output of MBresidues::contour will be suppressed during this calculation. Ю
HThe integral of Gamma@a+sD Gamma@b+sD Gamma@c-sD Gamma@dd-sDL Mel40@- 1 Й 2 + ep, 1 + ep, 3 Й 2 - ep, 0D 3 GammaA- 4 Gamma@2 + epD
1 2

In[29]:= In[30]:=

Mel40@a_, b_, c_, d_D := Gamma@a + cD Gamma@a + dD Gamma@b + cD Gamma@b + dDЙ Gamma@a + b + c + dD; + epE Gamma@1 + epD

Out[30]=