Reduction to Master Integrals. V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
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1 Reduction to Master Integrals V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
2 Reduction to Master Integrals IBP (integration by parts) V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
3 Reduction to Master Integrals IBP (integration by parts) Solving IBP relations by hand: simple one-loop examples V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
4 Reduction to Master Integrals IBP (integration by parts) Solving IBP relations by hand: simple one-loop examples Laporta algorithm and its implementations V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
5 Reduction to Master Integrals IBP (integration by parts) Solving IBP relations by hand: simple one-loop examples Laporta algorithm and its implementations Some other approaches to solve IBP relations V.A. Smirnov Atrani, September 30 October 05, 2013 p.1
6 Evaluating a family of Feynman integrals associated with a given graph with general integer powers of the propagators (indices) F Γ (q 1,...,q n ;d;a 1,...,a L ) =... I(q 1,...,q n ;k 1,...,k h ;a 1,...,a L )d d k 1 d d k 2...d d k h I(q 1,...,q n ;k 1,...,k h ;a 1,...,a L ) = 1 (p 2 1 m2 1 )a 1 (p 2 2 m 2 2 )a 2... V.A. Smirnov Atrani, September 30 October 05, 2013 p.2
7 An old straightforward analytical strategy: to evaluate, by some methods, every scalar Feynman integral generated by the given graph. V.A. Smirnov Atrani, September 30 October 05, 2013 p.3
8 The standard modern strategy: to derive, without calculation, and then apply IBP identities between the given family of Feynman integrals as recurrence relations. V.A. Smirnov Atrani, September 30 October 05, 2013 p.4
9 The standard modern strategy: to derive, without calculation, and then apply IBP identities between the given family of Feynman integrals as recurrence relations. Any integral of the given family is expressed as a linear combination of some basic (master) integrals. V.A. Smirnov Atrani, September 30 October 05, 2013 p.4
10 The standard modern strategy: to derive, without calculation, and then apply IBP identities between the given family of Feynman integrals as recurrence relations. Any integral of the given family is expressed as a linear combination of some basic (master) integrals. The whole problem of evaluation constructing a reduction procedure evaluating master integrals V.A. Smirnov Atrani, September 30 October 05, 2013 p.4
11 Integral calculus: b a uv dx = uv b a b a u vdx V.A. Smirnov Atrani, September 30 October 05, 2013 p.5
12 Integral calculus: b a uv dx = uv b a b a u vdx Feynman integral calculus: Use IBP and neglect surface terms [Chetyrkin & Tkachov 81]... [( ) ] 1 q i k j (p 2 1 m2 1 )a 1 (p 2 2 m 2 d d k 2 )a 2 1 d d k 2... = 0;... [ ] 1... k i k j (p 2 1 m2 1 )a 1 (p 2 2 m 2 d d k 2 )a 2 1 d d k 2... = V.A. Smirnov Atrani, September 30 October 05, 2013 p.5
13 An example F(a) = d d k (k 2 m 2 ) a F(a) for integer a 0. We need F(a) for positive integer a. V.A. Smirnov Atrani, September 30 October 05, 2013 p.6
14 An example F(a) = d d k (k 2 m 2 ) a F(a) for integer a 0. We need F(a) for positive integer a. Apply IBP d d k k ( ) 1 k (k 2 m 2 ) a = 0 V.A. Smirnov Atrani, September 30 October 05, 2013 p.6
15 An example F(a) = d d k (k 2 m 2 ) a F(a) for integer a 0. We need F(a) for positive integer a. Apply IBP d d k k ( ) 1 k (k 2 m 2 ) a = 0 Taking derivatives: k k = k µ k µ = d V.A. Smirnov Atrani, September 30 October 05, 2013 p.6
16 k k 1 2k 2 (k 2 m 2 ) a = a (k 2 m 2 ) a+1 [ 1 = 2a (k 2 m 2 ) a + m 2 (k 2 m 2 ) a+1 ] V.A. Smirnov Atrani, September 30 October 05, 2013 p.7
17 k k 1 2k 2 (k 2 m 2 ) a = a (k 2 m 2 ) a+1 [ 1 = 2a (k 2 m 2 ) a + m 2 (k 2 m 2 ) a+1 ] IBP relation (d 2a)F(a) 2am 2 F(a+1) = 0 V.A. Smirnov Atrani, September 30 October 05, 2013 p.7
18 k k 1 2k 2 (k 2 m 2 ) a = a (k 2 m 2 ) a+1 [ 1 = 2a (k 2 m 2 ) a + m 2 (k 2 m 2 ) a+1 ] IBP relation Its solution (d 2a)F(a) 2am 2 F(a+1) = 0 F(a) = d 2a+2 2(a 1)m 2F(a 1) V.A. Smirnov Atrani, September 30 October 05, 2013 p.7
19 Feynman integrals with integer a > 1 can be expressed recursively in terms of one integral F(1) I 1 (master integral). V.A. Smirnov Atrani, September 30 October 05, 2013 p.8
20 Feynman integrals with integer a > 1 can be expressed recursively in terms of one integral F(1) I 1 (master integral). Explicitly, F(a) = ( 1)a (1 d/2) a 1 (a 1)!(m 2 ) a 1 I 1, where (x) a is the Pochhammer symbol V.A. Smirnov Atrani, September 30 October 05, 2013 p.8
21 One more example F Γ (a 1,a 2 ) = d d k (m 2 k 2 ) a 1 ( (q k) 2 ) a 2 V.A. Smirnov Atrani, September 30 October 05, 2013 p.9
22 One more example Apply IBP k k F Γ (a 1,a 2 ) = q k d d k (m 2 k 2 ) a 1 ( (q k) 2 ) a 2 ( ) 1 (m 2 k 2 ) a 1 ( (q k) 2 ) a d d k = 0, 2 ( ) 1 (m 2 k 2 ) a 1 ( (q k) 2 ) a d d k = 0, 2 V.A. Smirnov Atrani, September 30 October 05, 2013 p.9
23 One more example Apply IBP k k F Γ (a 1,a 2 ) = q k d d k (m 2 k 2 ) a 1 ( (q k) 2 ) a 2 ( ) 1 (m 2 k 2 ) a 1 ( (q k) 2 ) a d d k = 0, 2 ( ) 1 (m 2 k 2 ) a 1 ( (q k) 2 ) a d d k = 0, 2 use 2k (k q) (k q) 2 +(k 2 m 2 ) q 2 +m 2 to obtain V.A. Smirnov Atrani, September 30 October 05, 2013 p.9
24 d 2a 1 a 2 2m 2 a a (1 q 2 +m 2 ) = 0 (A) a 2 a 1 a (q 2 +m 2 2 ) a (1 q 2 +m 2 ) = 0 (B) where, e.g., F(a 1,a 2 ) = F(a 1 +1,a 2 1). V.A. Smirnov Atrani, September 30 October 05, 2013 p.10
25 d 2a 1 a 2 2m 2 a a (1 q 2 +m 2 ) = 0 (A) a 2 a 1 a (q 2 +m 2 2 ) a (1 q 2 +m 2 ) = 0 (B) where, e.g., F(a 1,a 2 ) = F(a 1 +1,a 2 1). F Γ (a 1,a 2 ) = 0 for a 1 0 V.A. Smirnov Atrani, September 30 October 05, 2013 p.10
26 d 2a 1 a 2 2m 2 a a (1 q 2 +m 2 ) = 0 (A) a 2 a 1 a (q 2 +m 2 2 ) a (1 q 2 +m 2 ) = 0 (B) where, e.g., F(a 1,a 2 ) = F(a 1 +1,a 2 1). F Γ (a 1,a 2 ) = 0 for a 1 0 A manual solution. V.A. Smirnov Atrani, September 30 October 05, 2013 p.10
27 1. Apply (q 2 +m 2 )(A) 2m 2 (B), (q 2 m 2 ) 2 a = (q 2 m 2 )a (d 2a 1 a 2 )q 2 (d 3a 2 )m 2 +2m 2 a to reduce a 2 to 1 or 0. F[a1_, a2_ /; a2 > 1] := 1/(a2-1)/(qq - mm)ˆ2 ( (a2-1) (qq - mm) F[a1-1, a2] - ((d - 2 a1 - a2 + 1) qq + (d - 3 a2 + 3) mm) F[a1, a2-1] + 2 mm a1 F[a1 + 1, a2-2]); V.A. Smirnov Atrani, September 30 October 05, 2013 p.11
28 2. Suppose that a 2 = 1. Apply (A) (B), i.e. (q 2 m 2 )a = a 1 +2 d+a to reduce a 1 to 1 or a 2 to 0. F[a1_ /; a1 > 1, 1] := 1/(a1-1)/(qq - mm) ((a1-1) F[a1, 0] -(d - a1-1) F[a1-1, 1]); Therefore, any F(a 1,a 2 ) can be reduced to I 1 = F(1,1) and integrals with a 2 0 (which can be evaluated in terms of gamma functions for general d). V.A. Smirnov Atrani, September 30 October 05, 2013 p.12
29 3. Let a 2 0. Apply (A) to reduce a 1 to one. F[a1_ /; a1 > 1, a2_ /; a2 <= 0] := 1/(a1-1)/2/mm ((d - 2 a1 - a2 + 2) F[a1-1, a2] -a2 F[a1-2, a2 + 1] + a2 (qq - mm) F[a1-1, a2 + 1] V.A. Smirnov Atrani, September 30 October 05, 2013 p.13
30 4. Let a 1 = 1. Apply the following corollary of (A) and (B) (d a 2 1)2 = (q 2 m 2 ) 2 a (q 2 +m 2 )(d 2a 2 1) to increase a 2 to zero or one starting from negative values. F[1, a2_ /; a2 < 0] := 1/(d - a2-2) ( (a2 + 1) (qq - mm)ˆ2 F[1, a2 + 2] + (qq + mm) (d - 2 a2-3) F[1, a2 + 1] ); V.A. Smirnov Atrani, September 30 October 05, 2013 p.14
31 Any F(a 1,a 2 ) is a linear combination of the two master integrals I 1 = F(1,1) and I 2 = F(1,0). For example, F[3, 2]= (-(((-5 + d) (-3 + d) (-4 mm + d mm - 8 qq + d qq))/( 2 (mm - qq)ˆ4)) I1 + ((-2 + d) (96 mmˆ2-39 d mmˆ2 + 4 dˆ2 mmˆ mm qq - 6 d mm qq - 4 qqˆ2 + d qqˆ2))/ (8 mmˆ2 (mm - qq)ˆ4) I2) V.A. Smirnov Atrani, September 30 October 05, 2013 p.15
32 Triangle rule m 3 = 0 F(a 1,a 2,a 3 ) = d d k [(k +p 1 ) 2 m 2 1 ]a 1 [(k +p2 ) 2 m 2 2 ]a 2 (k 2 ) a 3 V.A. Smirnov Atrani, September 30 October 05, 2013 p.16
33 The IBP identity with the operator ( / k) k 1 = 1 d a 1 a 2 2a 3 [ a 1 1 +( 3 (p 2 1 m 2 1) ) +a 2 2 +( 3 (p 2 2 m 2 2) )] V.A. Smirnov Atrani, September 30 October 05, 2013 p.17
34 A manual solution of IBP relations for massless three-loop propagator diagrams [K.G. Chetyrkin & F.V. Tkachov 81] V.A. Smirnov Atrani, September 30 October 05, 2013 p.18
35 A manual solution of IBP relations for massless three-loop propagator diagrams [K.G. Chetyrkin & F.V. Tkachov 81] MINCER: [S.G. Gorishny, S.A. Larin, L.R. Surguladze & F.V. Tkachov 89] [S.A. Larin, F.V. Tkachov & J. Vermaseren 91] (implemented in FORM) V.A. Smirnov Atrani, September 30 October 05, 2013 p.18
36 Solving IBP relations algorithmically: Laporta s algorithm [Laporta & Remiddi 96; Laporta 00; Gehrmann & Remiddi 01] Use IBP relations written at points (a 1,...,a L ) with ai N and solve them for the Feynman integrals involved. (A Gauss elimination) V.A. Smirnov Atrani, September 30 October 05, 2013 p.19
37 Solving IBP relations algorithmically: Laporta s algorithm [Laporta & Remiddi 96; Laporta 00; Gehrmann & Remiddi 01] Use IBP relations written at points (a 1,...,a L ) with ai N and solve them for the Feynman integrals involved. (A Gauss elimination) Feynman integrals on the right-hand sides of such solutions are master integrals. V.A. Smirnov Atrani, September 30 October 05, 2013 p.19
38 Solving IBP relations algorithmically: Laporta s algorithm [Laporta & Remiddi 96; Laporta 00; Gehrmann & Remiddi 01] Use IBP relations written at points (a 1,...,a L ) with ai N and solve them for the Feynman integrals involved. (A Gauss elimination) Feynman integrals on the right-hand sides of such solutions are master integrals. When N increases, the situation stabilizes, in the sense that the number of the master integrals becomes stable starting from sufficiently large N. V.A. Smirnov Atrani, September 30 October 05, 2013 p.19
39 Experience: the number of master integrals is always finite. V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
40 Experience: the number of master integrals is always finite. Theorem [A. Smirnov & A. Petukhov 10] The number of master integrals is finite V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
41 Experience: the number of master integrals is always finite. Theorem [A. Smirnov & A. Petukhov 10] The number of master integrals is finite (for Feynman integrals with usual propagators) V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
42 Experience: the number of master integrals is always finite. Theorem [A. Smirnov & A. Petukhov 10] The number of master integrals is finite (for Feynman integrals with usual propagators) An algorithm to obtain the number of the master integrals in a given sector [R.N. Lee & A.A. Pomeransky 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
43 Experience: the number of master integrals is always finite. Theorem [A. Smirnov & A. Petukhov 10] The number of master integrals is finite (for Feynman integrals with usual propagators) An algorithm to obtain the number of the master integrals in a given sector [R.N. Lee & A.A. Pomeransky 13] A sector σ ν is defined by a subset ν {1,...,N} for which indices are positive: σ ν = {(a 1,...,a N ) : a i > 0 if i ν, a i 0 if i ν}. V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
44 Experience: the number of master integrals is always finite. Theorem [A. Smirnov & A. Petukhov 10] The number of master integrals is finite (for Feynman integrals with usual propagators) An algorithm to obtain the number of the master integrals in a given sector [R.N. Lee & A.A. Pomeransky 13] A sector σ ν is defined by a subset ν {1,...,N} for which indices are positive: σ ν = {(a 1,...,a N ) : a i > 0 if i ν, a i 0 if i ν}. The problem was reduced to evaluating the volume of the Newton polytope associated with the polynomials in the alpha representation. V.A. Smirnov Atrani, September 30 October 05, 2013 p.20
45 The same example d d k F Γ (a 1,a 2 ) = (m 2 k 2 ) a 1 ( (q k) 2 ) a 2 The left-hand sides of the two primary IBP relations: ibp1[a1_, a2_] := (d - 2 a1 - a2) F[a1, a2] - 2 mm a1 F[a1 + 1, a2] - a2 (F[a1-1, a2 + 1] + (mm - qq) F[a1, a2 + 1]); ibp2[a1_, a2_] := (a2 - a1) F[a1, a2] - a1 ((qq + mm) F[a1 + 1, a2] - F[a1 + 1, a2-1]) - a2 (F[a1-1, a2 + 1] + (mm - qq) F[a1, a2 + 1]); V.A. Smirnov Atrani, September 30 October 05, 2013 p.21
46 Let us consider the sector a 1 > 0,a 2 0 Use IBP at various (a 1,a 2 ) with a 1 + a 2 N Solve the corresponding linear system of equation with respect to F(a 1,a 2 ) involved. Increase N. N = 1 Solve[{ibp1[1, 0] == 0, ibp2[1, 0] == 0}, {F[2, 0], F[2, -1]}] {F[2, -1] -> ((-2 qq + d (mm + qq)) F[1, 0])/(2 mm), F[2, 0] -> ((-2 + d) F[1, 0])/(2 mm)} V.A. Smirnov Atrani, September 30 October 05, 2013 p.22
47 N = 2 Solve[{ibp1[1, 0] == 0, ibp2[1, 0] == 0, ibp1[2, 0] == 0, ibp2[2, 0] == 0, ibp1[1, -1] == 0, ibp2[1, -1] == 0 }, {F[2, 0], F[3, 0], F[1, -1], F[2, -1], F[3, -1], F[2, -2]}] {F[2, -2] -> (((2 + d) mmˆ2 + 2 (2 + d) mm qq + (-2 + d) qqˆ2) F[1,0])/(2 mm), F[3, -1] -> ((-2 + d) (-4 qq + d (mm + qq)) F[1, 0])/(8 mmˆ2), F[3, 0] -> ((-4 + d) (-2 + d) F[1, 0])/(8 mmˆ2), F[1, -1] -> (mm + qq) F[1, 0], F[2, -1] -> ((-2 qq + d (mm + qq)) F[1, 0])/(2 mm), F[2, 0] -> ((-2 + d) F[1, 0])/(2 mm)} V.A. Smirnov Atrani, September 30 October 05, 2013 p.23
48 Implementations of the Laporta s algorithm V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
49 Implementations of the Laporta s algorithm Three public versions: V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
50 Implementations of the Laporta s algorithm Three public versions: AIR [Anastasiou & Lazopoulos 04] V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
51 Implementations of the Laporta s algorithm Three public versions: AIR FIRE (in Mathematica; a C++ version is private) [Anastasiou & Lazopoulos 04] [A. Smirnov 08] V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
52 Implementations of the Laporta s algorithm Three public versions: AIR FIRE (in Mathematica; a C++ version is private) Reduze [Anastasiou & Lazopoulos 04] [A. Smirnov 08] [C. Studerus 09, A. von Manteuffel & C. Studerus 12 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
53 Implementations of the Laporta s algorithm Three public versions: AIR FIRE (in Mathematica; a C++ version is private) Reduze Private versions [Anastasiou & Lazopoulos 04] [A. Smirnov 08] [C. Studerus 09, A. von Manteuffel & C. Studerus 12 13] [Gehrmann & Remiddi, Laporta, Czakon, Schröder, Pak, Sturm, Marquard & Seidel, Velizhanin,... ] V.A. Smirnov Atrani, September 30 October 05, 2013 p.24
54 Solving reduction problems algorithmically in other ways: V.A. Smirnov Atrani, September 30 October 05, 2013 p.25
55 Solving reduction problems algorithmically in other ways: Baikov s method [P.A. Baikov ] [V.A. Smirnov & M. Steinhauser 03] An Ansatz for coefficient functions at master integrals dx1...dx N [... x a xa N P(x ) ] (d h 1)/2 N V.A. Smirnov Atrani, September 30 October 05, 2013 p.25
56 Solving reduction problems algorithmically in other ways: Baikov s method [P.A. Baikov ] [V.A. Smirnov & M. Steinhauser 03] An Ansatz for coefficient functions at master integrals dx1...dx N [... x a xa N P(x ) ] (d h 1)/2 N Gröbner bases [O.V. Tarasov 98] An alternative approach [A. Smirnov & V. Smirnov, 05 08] V.A. Smirnov Atrani, September 30 October 05, 2013 p.25
57 Lee s approach (based on Lie algebras) [R.N. Lee 08 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.26
58 Lee s approach (based on Lie algebras) [R.N. Lee 08 13] the IBP relations have the structure of a Lie algebra. The generators are O ij = k i p j The commutation relations are [O ik,o i j ] = δ ij O i j δ i jo ij. V.A. Smirnov Atrani, September 30 October 05, 2013 p.26
59 Lee s approach (based on Lie algebras) [R.N. Lee 08 13] the IBP relations have the structure of a Lie algebra. The generators are O ij = k i p j The commutation relations are [O ik,o i j ] = δ ij O i j δ i jo ij. It is sufficient to use IBP corresponding to k i k i+1, i = 1,...,h, k h+1 k 1 ; k 1 p j, j = 1,...,n ; h i=1 k i k i. V.A. Smirnov Atrani, September 30 October 05, 2013 p.26
60 Lee s approach (based on Lie algebras) [R.N. Lee 08 13] the IBP relations have the structure of a Lie algebra. The generators are O ij = k i p j The commutation relations are [O ik,o i j ] = δ ij O i j δ i jo ij. It is sufficient to use IBP corresponding to k i k i+1, i = 1,...,h, k h+1 k 1 ; k 1 p j, j = 1,...,n ; h i=1 k i k i. A code to reveal independent IBP relations [P. Kant 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.26
61 LiteRed A package which provides the possibility to find reduction rules in a given sector. It includes various additional tools. [R.N. Lee 12] V.A. Smirnov Atrani, September 30 October 05, 2013 p.27
62 LiteRed A package which provides the possibility to find reduction rules in a given sector. It includes various additional tools. For example, reduction rules for massless four-loop propagator integrals [R.N. Lee 12] M 61 M 62 M 63 have been constructed in all the sectors. A four-loop variant of MINCER! V.A. Smirnov Atrani, September 30 October 05, 2013 p.27
63 The existence of the explicit reduction rules shows that any four-loop massless propagator integral can be represented as a linear combination, with coefficients rational in d, of the twenty eight master integrals revealed by Baikov and Chetyrkin. V.A. Smirnov Atrani, September 30 October 05, 2013 p.28
64 The existence of the explicit reduction rules shows that any four-loop massless propagator integral can be represented as a linear combination, with coefficients rational in d, of the twenty eight master integrals revealed by Baikov and Chetyrkin. This statement has the status of a mathematical theorem. V.A. Smirnov Atrani, September 30 October 05, 2013 p.28
65 M 61, ε 1 M 62, ε 0 M 63, ε 0 M 51, ε 1 M 41, ε 1 M 42, ε 1 M 44, ε 0 M 45, ε 1 M 34, ε 3 M 35, ε 2 M 36, ε 1 M 52, ε 1 M 43, ε 1 M 32, ε 3 M 33, ε 3 M 21, ε 4 M 22, ε 4 M 26, ε 4 M 27, ε 4 M 23, ε 4 M 24, ε 4 M 25, ε 4 M 11, ε 5 M 12, ε 5 M 13, ε 5 M 14, ε 5 M 01, ε 6 M 31, ε 3 V.A. Smirnov Atrani, September 30 October 05, 2013 p.29
66 FIRE4 can be applied together with LiteRed: [A.V. Smirnov & VS 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.30
67 FIRE4 can be applied together with LiteRed: [A.V. Smirnov & VS 13] to perform a fast reduction when reduction rules have been constructed in all the sectors; V.A. Smirnov Atrani, September 30 October 05, 2013 p.30
68 FIRE4 can be applied together with LiteRed: [A.V. Smirnov & VS 13] to perform a fast reduction when reduction rules have been constructed in all the sectors; to run reduction in the Laporta mode if reduction rules are unknown in some sectors. V.A. Smirnov Atrani, September 30 October 05, 2013 p.30
69 Using symmetries to find extra relations I 11 [S. Laporta & E. Remiddi 96] (a) (b) (at p 2 = m 2 ) is the integral with the numerator is k p where k is the momentum of the massless line, in addition to the corresponding master integral I 10 without numerator (for the three-loop g 2 factor). V.A. Smirnov Atrani, September 30 October 05, 2013 p.31
70 Using symmetries to find extra relations I 11 [S. Laporta & E. Remiddi 96] (a) (b) (at p 2 = m 2 ) is the integral with the numerator is k p where k is the momentum of the massless line, in addition to the corresponding master integral I 10 without numerator (for the three-loop g 2 factor). Indeed, the IBP reduction shows that there are two master integrals in the highest sector. V.A. Smirnov Atrani, September 30 October 05, 2013 p.31
71 G 3 G 4,1 G 4,2 G 4,4 G 5,4 G 4,3 G 4,5 G 4,6 G 5,1 G 5,2 G 7 G 5,3 G 6,1 G 6,2 G 6,3 G 6,4 G 6,5 G 6,6 G 8 I 11 = 2d 5 2(d 2) G 4,4 1 4 G 3. V.A. Smirnov Atrani, September 30 October 05, 2013 p.32
72 G 3 G 4,1 G 4,2 G 4,4 G 5,4 G 4,3 G 4,5 G 4,6 G 5,1 G 5,2 G 7 G 5,3 G 6,1 G 6,2 G 6,3 G 6,4 G 6,5 G 6,6 G 8 I 11 = 2d 5 2(d 2) G 4,4 1 4 G 3. can be obtained from the IBP reduction of F(1,2,1,2,1) = F(1,1,2,2,1) V.A. Smirnov Atrani, September 30 October 05, 2013 p.32
73 The whole problem of evaluating a given family of Feynman integrals constructing a reduction procedure to master integrals (using IBP) evaluating master integrals V.A. Smirnov Atrani, September 30 October 05, 2013 p.33
74 The whole problem of evaluating a given family of Feynman integrals constructing a reduction procedure to master integrals (using IBP) evaluating master integrals Methods of evaluating master integrals based on IBP V.A. Smirnov Atrani, September 30 October 05, 2013 p.33
75 The whole problem of evaluating a given family of Feynman integrals constructing a reduction procedure to master integrals (using IBP) evaluating master integrals Methods of evaluating master integrals based on IBP method of differential equations [A. Kotikov 91, E. Remiddi 97, Gehrmann & Remiddi 00, J. Henn 13] V.A. Smirnov Atrani, September 30 October 05, 2013 p.33
76 The whole problem of evaluating a given family of Feynman integrals constructing a reduction procedure to master integrals (using IBP) evaluating master integrals Methods of evaluating master integrals based on IBP method of differential equations [A. Kotikov 91, E. Remiddi 97, Gehrmann & Remiddi 00, J. Henn 13] R. Lee s method [R. Lee 09, R. Lee & VS 12] based on the use of dimensional recurrence relations [O. Tarasov 96] V.A. Smirnov Atrani, September 30 October 05, 2013 p.33
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