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The calculation of multi-loop integrals is essential when theoretically predicting the scatting amplitude of a given process. For these calculations, the PV-reduction method [4] is a widely used approach, and one way to implement the reduction method is to use the integrating-by-parts (IBP) relation [5–7]. As one of the most powerful techniques for loop integral reduction, IBP gives a large number of recurrence relations, and the reduction can be represented by a combination of simpler integrals via Gauss elimination. However, as the propagator number and power increase, the IBP method becomes inefficient; hence, more efficient reduction methods must be found.
The unitarity cut method is an alternative reduction method and has been proven to be useful for one-loop integrals [8–17]. In a physical one-loop process, the power of the propagator is just one; however, if the method is complete, it should be able to reduce integrals with higher power propagators. Such a situation is not simply a theoretical curiosity but appears in higher loop diagrams as a sub-diagram. Furthermore, although the scalar basis is natural for one-loop integrals, in general, the choice of basis can be different, depending on the physical input. For example, in the topology of a one-loop bubble, the basis, in which one propagator has a power of two, could be used as part of the UT-basis [18, 19].
In a previous study [3], we successfully obtained an analytical reduction result for one-loop integrals with high power propagators by combining the tricks of differential operators and unitarity cut. We gave coefficients to all bases except the tadpoles'; however, the unitarity method could not be used because the tadpole has only one propagator. To complete this investigation, the missing tadpole coefficients must be found using other efficient methods.
Other than the unitarity cut method, there are proposals to overcome the difficulties of IBP using tricks and other representations of integrals, such as the Baikov representation [20, 21] and Feynman parametrization representation [22, 23] for loop integrals. In recent years, Chen proposed a new representation for loop integrals [1, 2]. His method is based on the generalized Feynman parametrization representation, that is, an extra parameter
xn+1 is introduced to combineU,F in the standard Feynman parametrization representation. Such a generalization will offer several benefits when deriving the IBP recurrence relation, as shown in this paper.As a common feature, the IBP recurrence relation derived using the generalized Feynman parametrization representation will naturally have terms in different spacetime dimensions. Because we are always concerned with reducing a particular dimension D, which is typically set to
4−2ϵ for renormalization, we wish to cancel these terms in different dimensions. In general, this is not easy. In [24], Gluza, Kajda, and Kosower showed how to avoid the change in the power of propagators in standard momentum space. Larsen and Zhang considered the Baikov representation and demonstrated how to eliminate both dimensional shifting and the change in the power of propagators [25–30]. These methods require a solution to the syzygy equations, which is generally not easy. In Chen's second paper [2], he proposed a new technique to simplify the recurrence relation based on non-commutative algebra.Motivated by the above discussion and preparing Chen's method for high-loop computations, in this paper, we use Chen's method to find the missing tadpole coefficients from our previous study. Furthermore, we use the idea of removing terms with dimensional shifting in the derived IBP relation to construct a simpler reduction method, with the analytic results expressed by the elements of the coefficient matrix
ˆA .This paper is organized as follows: In section II, we review and illustrate Chen's new method with a simple example in section II.A. In the example, integrals naturally emerge in different dimensions. We discuss the physical meaning of the boundary terms, which contribute to the sub-topologies. To cancel dimensional shifting in the parametrization form and simplify the IBP relation, a new trick is proposed in section II.B in which free auxiliary parameters are added based on the fact that F in the integrand is a homogeneous function of
xi with degreeL+1 . Using this trick, we successfully cancel dimensional shifting and drop the terms that we are not concerned with. Moreover, we present a simplified IBP relation in which all the integrals are in the particular dimension D and integrals other than the target have a lower total propagator power. The analytic result is presented as a determinant of the cofactor of the matrixˆA , which is entirely determined by a graph. In section III, we calculate a triangleI3(1,1,2) , boxI4(1,1,1,2) , and pentagonI5(1,1,1,1,2) in parametric form using this trick and present the analytic results of all coefficients to the master basis, especially the tadpole parts, to complement our previous study. -
In this section, we introduce a new reduction method proposed by Chen in [1]. The general form of a loop integral is given by
I[N(l)](k)=∫dDl1dDl2⋯dDlLN(l)Dk11Dk22Dk33⋯Dknn,
(1) where, for simplicity, we denote
l=(l1,l2,l3,⋯,lL) andk=(k1,k2,k3,⋯,kn) . Because we consider only scalar integrals withN(l)=1 in this paper, let us labelI(L;λ1+1,⋯,λn+1)=∫dDl1⋯dDlL1Dλ1+11⋯Dλn+1n.
(2) Using the Feynman parametrization procedure,
L∑iαiDi=L∑i,jAijli⋅lj+2L∑i=1Bi⋅li+C,
(3) and thus loop integrals can be found as
∫dDl1⋯dDlLei(∑αiDi)=eiπL(1−D2)/2πLD/2(DetA)−D/2×ei(C−∑A−1ijBi⋅Bj).
(4) Defining
U(α)=DetA andC−∑A−1ijBi⋅Bj≡V(α)/U(α)−∑m2iαi ①, we can see thatU(α) is a homogeneous function ofαi with degree L, whereasV(α) is a homogeneous function ofαi with degreeL+1 . The loop integral becomesI(L;λ1+1,⋯,λn+1)=e−∑((λi+1)/2)iπΠni=1Γ(λi+1)eiπL(1−D/2)/2πLD/2×∫dα1⋯dαnU(α)−D/2ei[V(α)/U(α)−∑m2iαi]αλ11⋯αλnn.
(5) To derive the parametric form suggested by Chen, we perform the following: Using the
α -representation of general propagators,1(l2−m2)λ+1=e−((λ+1)/2)iπΓ(λ+1)∫∞0dαeiα(l2−m2)αλ,Im{l2−m2}>0,
(6) where "
iϵ " is neglected, we obtainI(L;λ1+1,⋯,λn+1)=e−∑ni((λi+1)/2)iπΠni=1Γ(λi+1)∫dDl1⋯dDlL×∫∞0dα1⋯dαnei∑ni=1αiDiαλ11⋯αλnn.
(7) To go further, we change the integral variables to
αi=ηxi . Because there is a total of n independent variables, we must insert another constraint condition. In general, we could let∑i∈S(1,2,3,⋯n)xi=1,
(8) where S is an arbitrary non-trivial subset of
{1,2,3,⋯n} . After carrying out the integration over η, the second line of Eq. (5) becomes(−i)(n+λ−DL/2)Γ(n+λ−DL2)×∫dx1⋯dxnδ(∑j∈Sxj−1)U(x)n+λ−(D/2)(L+1)[−V(x)+U(x)∑m2ixi]n+λ−DL/2xλ11⋯xλnn=(−i)n+λ−DL/2Γ(n+λ−DL2)∫dx1⋯dxnδ(∑j∈Sxj−1)Uλufλfxλ11⋯xλnn, (9) where
U(x)=η−LU(α)=η−LU(ηxi),V(x)=η−L−1V(α)=η−L−1V(ηx),f(x)=−V(x)+U(x)∑m2ixiλ=n∑i=1λi,λu=n+λ−D2(L+1),λf=−n−λ+DL2.
(10) Finally, via Mellin transformation②
Aλ1Bλ2=Γ(−λ1−λ2)Γ(−λ1)Γ(−λ2)∫∞0dx(A+Bx)λ1+λ2x−λ2−1,
(11) we can express (9) as
(−i)n+λ−DL/2Γ(n+λ−DL2)Γ(−λu−λf)Γ(−λu)Γ(−λf)∫dx1⋯dxnδ(∑j∈Sxj−1)∫∞0dxn+1×(Uxn+1+f)λu+λfx−λu−1n+1xλ11⋯xλnn≡(−i)n+λ−DL/2Γ(n+λ−DL/2)Γ(−λu−λf)Γ(−λu)Γ(−λf)∫dΠ(n+1)Fλ0xλ11⋯xλnnxλn+1n+1≡(−i)n+λ−DL/2Γ(n+λ−DL2)Γ(−λu−λf)Γ(−λu)Γ(−λf)iλ0;λ1,⋯λn,
(12) where
dΠ(n+1)=dx1⋯dxn+1δ(∑j∈Sxj−1),F=Uxn+1+f,λ=n∑i=1λi,λ0=λu+λf=−D2,λn+1=−λu−1=D2(L+1)−λ−1−n.
(13) Combined, we finally obtain the parametric form of the scalar loop integrals in (5),
I(L;λ1+1,⋯,λn+1)=(−1)n+λiLπLD/2Γ(−λ0)Πn+1i=1Γ(λi+1)iλ0;λ1,⋯λn.
(14) -
The parametric form of (14) is the starting point of Chen's proposal. The IBP relations in this form are given by③④
∫dΠ(n+1)∂∂xi{Fλ0xλ11⋯xλnnxλn+1+1n+1}+δλi,0∫dΠ(n){Fλ0xλ11⋯xλnnxλn+1+1n+1}|xi=0=0
(15) where
i=1,...,n+1 , anddΠ(n) in the second term isdΠ(n)=dx1⋯^dxi⋯dxndxn+1δ(∑j∈Sxj−1).
(16) The second term in (15) contributes to a boundary term, which leads to the sub-topologies of the former term.
To illustrate the IBP relation (15), we present the reduction in
I2(1,2) as an example. The general form of one-loop bubble integrals is given byI2(m+1,n+1)=∫dDl(l2−m21)m+1((l−p1)2−m22)n+1,
(17) and the corresponding parametric form is (in this paper, we ignore the former factor
πLD/2 )I2(m+1,n+1)=i(−1)m+n+2×Γ(D2)Γ(m+1)Γ(n+1)Γ(D−2−m−n)×∫dΠ(3)Fλ0xm1xn2xλ33,
(18) where
F=(x1+x2)(m21x1+m22x2+x3)−p21x1x2,
(19) and
iλ0;m,n=∫dΠ(3)Fλ0xm1xn2xλ33,
(20) with
λ0=−D2 , andλ3=−3−m−n−2λ0 . Using Eq. (15), we can obtain three IBP recurrence relations. First, taking∂∂x1 , the first term in (15) givesλ0iλ0−1;m,n+2m21λ0iλ0−1;m+1,n+Δλ0iλ0−1;m,n+1,
(21) where
Δ=m21+m22−p21 . The second term givesδm,0∫dΠ(2)(x3+m22x2)λ0xn+λ02x−2−n−2λ03=δm,0iλ0;−1,n.
(22) Here, the notation
iλ0;−1,n must be explained. From the middle expression of (22), we see that it is the parametric form of the tadpole∫dDl/(l2−m22)n+1 . To emphasize its origin, that is,from a bubble by removingthe first propagator, we extend the definition ofiλ0;λ1,...,λn given in (12) by settingλ1=−1 ⑤. Using the extended notation, we obtain the first IBP relationλ0iλ0−1;m,n+2m21λ0iλ0−1;m+1,n+Δλ0iλ0−1;m,n+1+δm,0iλ0;−1,n=0.
(23) When we set
m=n=0 in (23), this readsλ0iλ0−1;0,0+2m21λ0iλ0−1;1,0+Δλ0iλ0−1;0,1+iλ0;−1,0=0.
(24) Similarly, we can take the differential
∂∂x2 and obtain the second IBP relationλ0iλ0−1;0,0+Δλ0iλ0−1;1,0+2m22λ0iλ0−1;0,1+iλ0;0,−1=0.
(25) We should solve
iλ0;0,1 byiλ0;0,0 from (24) and (25). However, for the bubble part, we haveλ0−1 instead ofλ0 . This could be fixed by rewritingλ0→λ0+1 becauseλ0 is a free parameter. However, the boundary tadpole partiλ0;0,−1 will becomeiλ0+1;0,−1 , that is, it will have the dimensional shifting, which is a common feature in the parametric IBP relation.To deal with this, using the parametric form of tadpoles
iλ0;m,−1=∫dΠ(2)(x1x3+m21x21)λ0xm1x−2−m−2λ03
(26) and taking the
∂∂x1 and∂∂x3 , we can obtain two IBP relations,λ0iλ0−1;m,−1+2m21λ0iλ0−1;m+1,−1+miλ0,m−1,−1=0,λ0iλ0−1;m+1,−1+(−1−m−2λ0)iλ0;m,−1=0,
(27) from which we solve
iλ0;0,−1=−λ02m21(2λ0+1)iλ0−1;0,−1,iλ0;−1,0=−λ02m22(2λ0+1)iλ0−1;−1,0.
(28) Inserting (28) into (24) and (25), we can solve
iλ0−1;0,1 . After shiftingλ0→λ0+1 , we finally getiλ0;0,1=2m21−ΔΔ2−4m21m22iλ0;0,0+−1(2λ0+3)(Δ2−4m21m22)iλ0;0,−1+Δ2m22(2λ0+3)(Δ2−4m21m22)iλ0;−1,0.
(29) Translating back to the scalar basis, we obtain the reduction in
I2(1,2) asI2(1,2)=c2→2I2(1,1)+c2→1ˉ2I2(1,0)+c2→1;ˉ1I2(0,1),
(30) with the coefficients
c2→2=(D−3)(Δ−2m21)Δ2−4m21m22,c2→1;ˉ2=D−2Δ2−4m21m22,c2→1;ˉ1=(D−2)Δ2m22(4m21m22−Δ2).
(31) -
As shown in the previous subsection, the IBP relation given in (15) will contain integrals with dimension shift, which makes the reduction program slightly troublesome. As reviewed in the introduction, there are several references dealing with this or related problems. Based on these studies, an improved version of the IBP relation has been given in [2] (see Eq. (12) and (13)). All these methods require a solution to the syzygy equations, which is not generally an easy task. However, for our one-loop integrals, the function
F(x) is a homogeneous function ofxi with degree of two⑥. This good property simplifies the related syzygy equations, which can then be directly solved⑦. In this paper, we develop a direct algorithm to express the IBP relations without dimension shift and terms with unwanted higher power propagators.In the generalized parametric representation, our improved IBP relation involves multiplying Eq. (15) by a degree zero coefficient
zi , for example,zi=xα1xβ2x−α−β3 . Because the degree of the new integrand does not change, the IBP identity still holds. Summing them together we get⑧n+1∑i=1∫dΠ(n+1)∂∂xi{ziFλ0xλ11xλ22⋯xλn+1+1n+1}+n+1∑i=1δλi,0∫dΠ(n)ziFλ0xλ11⋯xλnnxλn+1n+1|xi=0=0.
(32) Because the second boundary term involves integrals with sub-topologies, we focus on the first term. Expanding it, we get
∫dΠ(n+1)[n+1∑i=1(∂zi∂xi+λ0zi∂F∂xiF+λizixi)+zn+1xn+1]×Fλ0xλi1xλ22⋯xλnnxλn+1+1n+1.
(33) From (13), we can see that the power
λ0 of F is related to dimension. To cancel the dimension shift, we must choose the proper coefficientszi so thatn+1∑i=1zi∂F∂xi is a multiple of the function F, that is,∑n+1i=1zi∂F∂xi+BF=0.
(34) Because the coefficients
zi are not polynomials, (34) is not the "normal syzygy equation," and we cannot directly use the technique developed for the polynomial ring. In [2], Chen developed a method based on the lift and down operators. Here, for the one-loop integrals, we can solve it directly with free auxiliary parameters, as shown later in this paper. When reinserting the solutions to the IBP recurrence relation, we can choose these free parameters to cancel both the dimension shift and unwanted terms with higher power propagators, which leads to a simpler recurrence relation.Now, the idea is explained in detail. Note that in the one loop case, the homogeneous function F is a degree two function of
xi ; therefore, we can write F asF=12Aijxixj,
(35) where A is a symmetric matrix⑨. Thus, we have
fi≡∂F∂xi,ˆf=ˆAˆx,ˆf≡[f1f2⋮fnfn+1],ˆx≡[x1x2⋮xnxn+1],
Solving
ˆx=ˆA−1ˆf , we haveF=12ˆxTAˆx=12ˆfT(ˆA−1)TˆAˆA−1ˆf=12ˆfT(ˆA−1)Tˆf≡ˆfTˆKˆf,K=12A−1,
(36) where the coefficient matrix
ˆK is a real symmetry matrix. In fact, we can go further. Using0=ˆfTˆKAˆf
(37) with any antisymmetric matrix
KA , we can add (37) to (36) to obtain a more general form ⑩F=ˆfTˆKˆf+ˆfTˆKAˆf=ˆfT(ˆK+ˆKA)ˆf≡ˆfTˆRˆf=ˆfTˆRˆAˆx≡ˆfTˆQˆx,ˆQ≡12ˆI+ˆKAˆA.
(38) Note that because the arbitrary matrix
ˆKA is of rankn+1 , there aren(n+1)2 free independent parameters,a1,⋯,a(n(n+1))/2 , in the matrixˆQ in (38).Now, reinserting (38) into (34), we can solve
ˆz asˆfTˆz+BˆfTˆQˆx=0,⟹ˆz=−BˆQˆx.
(39) Note that because z is degree zero, we should ensure B is a homogenous function of degree
−1 . In this study, we chooseB=1/xn+1 . The choice of z given by (39) will guarantee the removal of dimension shift in the IBP relation. Furthermore, by choosing particular values of the free parameters ofˆQ , we may cancel several unwanted terms. Some examples are shown in later computations to illustrate this trick. -
As mentioned in the introduction, one motivation of this study is to complete reduction in the scalar basis with general powers. Using the unitarity cut method in [3], we are able to find reduction coefficients of all bases, except the tadpole. In this section, we will use the improved IBP relation (32) to find the tadpole coefficients as well as other coefficients.
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We begin with bubble topology. Although this was already done in (30), we redo it using the improved IBP relation (32). The parametric form of bubble is given by (18), (19), and (20). Using our label, we have
ˆf=ˆAˆx,ˆA=[2m21Δ1Δ2m221110],
(40) and
F=ˆfTˆKˆf,ˆK=[14p21−14p21−m21+m22+p214p21−14p2114p21m21−m22+p214p21−m21+m22+p214p21m21−m22+p214p21Δ2−4m21m224p21].
(41) Adding the antisymmetric matrix
KA , we haveˆKA=[0a1a2−a10a3−a2−a30],ˆQ=[1+2a2+2a1m21+2a1m22−2a1p212a2+2a1m22a1a3−2a1m211+2a3−2a1m21−2a1m22+2a1p212−a1−2a2m21−a3(m21+m22−p21)−2a3m22−a2(m21+m22−p21)1−2a2−2a32].
(42) -
Taking
B=−1/x3 in (34), solution (39) giveszi=(Qijxj)/x3 . Expanding (32), we obtain the IBP recurrence relationcm,niλ0;m,n+cm+1,niλ0;m+1,n+cm+1,n−1iλ0;m+1,n−1+cm,n+1iλ0;m,n+1+cm−1,n+1iλ0;m−1,n+1+cm,n−1iλ0;m,n−1+cm−1,niλ0;m−1,n+δ2=0,
(43) where
δ2 is the boundary term, which we will compute later. The other coefficients arecm,n=Q11(1+m)+Q22(1+n)+Q33(1+λ3)+λ0,cm+1,n=Q31λ3=−λ3(a2A11+a3A21),cm+1,n−1=Q21n=−n(a1A11−a3A31),
cm,n+1=Q32λ3=−λ3(a2A12−a3A22),cm−1,n+1=Q12m=m(a1A22+a2A32),cm,n−1=Q23n=−n(a1A13−a3A33),cm−1,n=Q13m=m(a1A32+a2A33).
(44) Because we aim to obtain the reduction in
I2(1,2) , starting fromm=n=0 , we want to eliminate terms with the indices(m+1,n) and(m+1,n−1) while keeping the term with the index(m,n+1) . Thus, we imposecm+1,n=0 andcm+1,n−1=0 , which can be satisfied by choosing the free parameters⑪a2=−a1A21A31=−a1(m21+m22−p21),a3=a1A11A31=2a1m21.
(45) After this choice, the matrix
ˆQ becomesˆQ;r=[12a1A31(A22A31−A21A32)a1A31(A23A31−A21A33)012−a1A31(A12A31−A11A32)a1A31(A11A33−A13A31)0a1A31(A12A21−A11A22)12+a1A31(A13A21−A11A23)], leaving five terms with non-zero coefficients⑫.
cm,n+1=−a1λ3A31(A11A22−A12A21)=−a1λ3A31|˜A33|=a1λ3,cm−1,n+1=−ma1A31(A21A32−A22A31)=−ma1A31|˜A13|=−a1m(m21−m22−p21),cm,n−1=na1A31(A11A33−A13A31)=na1A31|˜A22|=−a1n,cm−1,n=−ma1A31(A21A33−A23A31)=−ma1A31|˜A12|=a1m,cm,n=a1A31((1+n)(A11A32−A12A31)−(λ3+1)(A11A23−A13A21))=a1A31(n−λ3)|˜A23|=a1A31((n−λ3)(m21−m22+p21)).
(46) The boundary
δ2 term: Theδ2 term is given byδ2=∑3i=1δλi,0∫dΠ(2){ziFλ0xm1xn2xλ3+13}|xi=0,
(47) where
λi represents the power ofxi . It is worth emphasizing that becausezi containsxi , the total powerλi ofxi is not equal tom,n,λ3 in general. Expanding it, we get⑬δ2=δλ1,0∫dΠ(2)(Q11Fλ0xm+11xn2xλ33+Q12Fλ0xm1xn+12xλ33+Q13Fλ0xm1xn2xλ3+13)|x1=0+δλ2,0∫dΠ(2)(Q21Fλ0xm+11xn2xλ33+Q22Fλ0xm1xn+12xλ33+Q23Fλ0xm1xn2xλ3+13)|x2=0. (48) Remembering our extended notation explained under (22), we have
∫dΠ(2)F|λ0x1=0xn2≡iλ0;−1,n,∫dΠ(2)F|λ0x2=0xm1≡iλ0;m,−1,
(49) and the
δ2 term can be written asδ2;r=δλ1,0(Q11;riλ0;m+1,n+Q12;riλ0;m,n+1+Q13;riλ0;m,n)+δλ2,0(Q21;riλ0;m+1,n+Q22;riλ0;m,n+1+Q23;riλ0;m,n)=δm,−1Q11;riλ0;−1,n+δm,0Q12;riλ0;,−1,n+1+δm,0Q13;riλ0;−1,n+δn,0Q21;riλ0;m+1,−1+δn,−1Q22;riλ0;m,−1+δn,0Q23;riλ0;m,−1,
(50) where the subscript r in
δ2;r andQij;r indicates thata2 anda3 should be replaced by (45).Because m and n cannot be
−1 , the first and fifth terms are actually zero.Now, we can use (43) and (50) to get our result directly. Setting
m=0 ,n=0 , and all other terms in (43) equal to zero, and we are left with⑭c0,0iλ0;0,0+c0,1iλ0;0,1+δ2;00=0,
(51) with the coefficients
c0,0=−a1(D−3)(m21−m22+p21),c0,1=a1(D−3)(m41+m42p41−2m21p21−2m22p21−2m21m22),δ2;00=Q12;riλ0;−1,1+Q13;riλ0;−1,0+Q21;riλ0;1,−1+Q23;riλ0;0,−1, (52) where
Q21;r=−a1A31(A21A32−A22A31)=−a1A31|˜A13|,Q23;r=−a1A31(A11A33−A13A31)=−a1A31|˜A22|,Q12;r=−a1A31(A21A32−A22A31)=−a1A31|˜A13|,Q13;r=−a1A31(A21A33−A23A31)=−a1A31|˜A12|.
(53) From this, we can directly write the solution as
iλ0;0,1=−c0,0c0,1iλ0;0,0−Q21;rc0,1iλ0;1,−1−Q23;rc0,1iλ0;0,−1−Q12;rc0,1iλ0;−1,1−Q13;rc0,1iλ0;−1,0.
(54) Translating back to scalar integrals, it is
I2(1,2)=c12→11I2(1,1)+c12→10I2(1,0)+c12→20I2(2,0)+c12→01I2(0,1)+c12→02I2(0,2),
(55) with
c12→20=0 andc12→11=−(−3+D)(m21−m22+p21))(m41+(m22−p21)2−2m21(m22+p21),c12→10=D−2−2m21(m22+p21)+m41+(m22−p21)2c12→01=2−D−2m21(m22+p21)+m41+(m22−p21)2,c12→02=−m21+m22+p21−2m21(m22+p21)+m41+(m22−p21)2.
(56) Using
I2(2,0)=D−22m21I2(1,0) ⑮ andI2(0,2)=D−22m22I2(0,1) , we have our final results for the reduction inI2(1,2) ,I2(1,2)=c2→2I2(1,1)+c2→1;ˉ2I2(1,0)+c2→1;ˉ1I2(0,1),
(57) with the coefficients
c2→2=−(D−3)(m21−m22+p21)−2m21(m22+p21)+m41+(m22−p21)2,c2→1;ˉ2=D−2−2m21(m22+p21)+m41+(m22−p21)2,c2→1;ˉ1=−(D−2)(m21+m22−p21)2m22(−2m21(m22+p21)+m41+(m22−p21)2),
(58) which is given in (30).
-
Now, let us consider more complicated examples, that is, bubbles with general higher power propagators. With the choice of (45), we get an IBP recurrence relation (46) and use it to reduce the bubbles
iλ0,m,n+1 to simpler bubbles, which have a lower total propagator power and no higher power inD2 . Similarly, by choosing different values ofa2 anda3 , we can obtain another IBP recurrence relation to reduce the integral to those with no higher power inD1 . The choice isa2=−a1A22A32,a3=a1A12A32,
(59) and the corresponding IBP recurrence is
cm+1,niλ0,m+1,n+cm+1,n−1iλ0,m+1,n−1+cm,n−1iλ0,m,n−1+cm−1,niλ0,m−1,n+cm,niλ0,m,n+δ2;r=0,
(60) with the coefficients
cm+1,n=(|˜A33|)(D−3−m−n),cm+1,n−1=−n|˜A23|,cm,n−1=n|˜A21|,cm−1,n=−m|˜A11|,cm,n=|˜A13|(3+2m+n−D),
(61) and the boundary term
δ2;r′=−δm,0|˜A11|iλ0,m,n+δn,0(−|˜A32|iλ0,m+1,n+|˜A21|iλ0,m,n).
(62) Combining (46) and (60), we can reduce the general bubbles.
-
In the example
I2(1,3) , we simply need to reduceD2 from power3 to1 . The strategy is to use (46) twice. In the first step, by settingm=0 andn=1 in (46), we getI2(1,3)=|˜A23|(D−5)2|˜A33|I2(1,2)+|˜A22|(D−3)2|˜A33|I2(1,1)+−|˜A12|(D−3)2|˜A33|I2(0,2)+|˜A13||˜A33|I2(0,3).
(63) For the first term in (63), setting
m=0 andn=0 in (46) again, we haveI2(1,2)=|˜A23|(D−3)|˜A33|I2(1,1)+|˜A22|(D−2)|˜A33|I2(1,0)+|˜A13||˜A33|I2(0,2)+−|˜A12|(D−2)|˜A33|I2(0,1).
(64) Inserting (64) into (63) and using the reduction in the tadpole⑯, we get
I2(1,3)=c13→11I2(1,1)+c13→10I2(1,0)+c13→01I2(0,1),
(65) with the coefficients
c13→11=(|˜A23||˜A33|+|˜A23|2(D−5))(D−3)2|˜A33|2,c13→10=|˜A22||˜A23|(D−5)(D−2)2|˜A33|2,c13→01=(D−2)8|˜A33|2m42A21(2A32|˜A23|(D−5)m22+A32At33(D−4)−4A33|˜A23|(D−5)m42−2A33|˜A33|(D−3)m22)−A22A31(2|˜A23|(D−5)m22+At33(D−4))+2A23A31m22(2|˜A23|(D−5)m22+|˜A33|(D−3)).
(66) The result is confirmed with FIRE6. In this example, we simply need to solve two equations to reduce the bubble topology.
-
For this example, we must use (60) to lower the power of
D1 and (46) to lower the power ofD2 . Settingm=1 andn=4 in (60), we can reduceI2(3,5) toI2(2,4) ,I2(2,5) ,I2(1,5) , andI2(3,4) .I2(3,5)=|˜A11|(D−7)2|˜A33|I2(1,5)+−|˜A13|(D−9)2|˜A33|I2(2,5)+−|˜A21|(D−7)2|˜A33|I2(2,4)+|˜A23||˜A33|I2(3,4).
(67) Then, setting
m=1 andn=3 in (60), we reduceI2(3,4) toI2(1,4) ,I2(2,3) ,I2(2,4) , andI2(3,3) .I2(3,4)=−|˜A23||˜A33|I2(3,3)+−|˜A13|(D−8)2|˜A33|I2(2,4)+−|˜A21|(D−6)2|˜A33|I2(2,3)+|˜A11|(D−6)2|˜A33|I2(1,4).
(68) Using the same idea, we must solve
14 equations to completely reduceI2(3,5) . The analytic expressions for these14 equations have also been confirmed by FIRE6. -
The triangle
I3(m+1,n+1,q+1) is given byI3(m+1,n+1,q+1)=∫dDl(l2−m21)m+1((l−p1)2−m22)n+1((l+p3)2−m23)q+1.
(69) Its parametric form is
I3(m+1,n+1,q+1)=i(−1)3+m+n+qΓ(−λ0)Γ(m+1)Γ(n+1)Γ(q+1)Γ(λ4+1)iλ0,m,n,q,
(70) where
iλ0;m,n,q=∫dΠ(4)Fλ0xm1xn2xq3xλ44,λ0=−D2,λ4=−4−2λ0−m−n−q=D−4−m−n−q.
(71) Using expression (10), we have
U(x)=x1+x2+x3,V(x)=x1x2p21+x1x3p23+x2x3p22,f(x)=−V+U∑m2ixi=(x1+x2+x3)(x1m21+x2m22+x3m23)−x1x2p21−x2x3p22−x1x3p23,F(x)=U(x)x4+f(x)=(x1+x2+x3)×(m21x1+m22x2+m23x3+x4)−x1x2p21−x2x3p22−x1x3p23.
(72) Thus, we can express the matrices as
ˆA=[2m21m21+m22−p21m21+m23−p231m21+m22−p212m22m22+m23−p221m21+m23−p23m22+m23−p222m2311110],ˆKA=[0a1a2a3−a10a4a5−a2−a40a6−a3−a5−a60],ˆQ=12ˆI+ˆKAˆA.
(73) -
Taking
B=−1x4 in (39), we getzi=Qijxjx4 . Taking this relation into our IBP identities (32), we get4∑i=1∫dΠ(4){ziFλ0xm1xn2xq3xλ4+14}+δ3=0,
(74) for which we deal with the boundary
δ3 term later. After expanding the first term, we getcm,n,qiλ0;m,n,q+cm+1,n,qiλ0;m+1,n,q+cm+1,n,q−1iλ0;m+1,n,q−1+cm+1,n−1,qiλ0;m+1,n−1,q+cm−1,n+1,qiλ0;m−1,q+1,q+cm,n+1,q−1iλ0;m,n+1,q−1+cm,n+1,qiλ0;m,n+1,q+cm,n,q+1iλ0;m,n,q+1+cm,n−1,q+1iλ0;m,n−1,q+1+cm−1,n,q+1iλ0;m−1,n,q+1+cm−1,n,qiλ0;m−1,n,q+cm,n−1,qim,n−1,q+cm,n,q−1iλ0;m,n,q−1+δ3=0,
(75) with the coefficients
cm,n,q=λ0+(m+1)Q11+(n+1)Q22+(q+1)Q33+(λ4+1)Q44,cm+1,n,q=λ4Q41,cm+1,n,q−1=qQ31,cm+1,n−1,q=nQ21,cm,n,q−1=qQ34,cm−1,n+1,q=mQ12,cm,n+1,q−1=qQ32,cm,n+1,q=λ4Q42,cm,n,q+1=λ4Q43,cm,n−1,q+1=nQ23,cm−1,n,q+1=mQ13,cm−1,n,q=mQ14,cm,n−1,q=nQ24.
(76) Now, we can choose particular values for our six parameters
a1 ,a2 ,a3 ,a4 ,a5 , anda6 to let the coefficientscm+1,n,q ,cm+1,n,q−1 ,cm+1,n−1,q ,cm−1,n+1,q ,cm,n+1,q , andcm,n+1,q be zero. The solutions area2=−a1A21A42−A22A41A31A42−A32A41=−a1(−m21+m22+p21)−m21+m22+2(p1⋅p2)+p21,a3=a1(A21A32−A22A31)A31A42−A32A41=−a1(m21−m22−p21)(m22+m23−p22)−m21+m22+2(p1⋅p2)+p21−2a1m22,a4=a1(A11A42−A12A41)A31A42−A32A41=−a1(m21−m22+p21)−m21+m22+2(p1⋅p2)+p21,a5=−a1(A11A32−A12A31)A31A42−A32A41=a1(m21−m22+p21)(m21+m23−2(p1⋅p2)−p21−p22)−m21+m22+2(p1⋅p2)+p21+2a1m21,a6=a1(A11A22−A12A21)A31A42−A32A41=a1(m41−2m21(m22+p21)+(m22−p21)2)−m21+m22+2(p1⋅p2)+p21.
(77) Then, the matrix
ˆQ becomesˆQr=1ΔA[12ΔA0a1|˜A14|a1|˜A13|012ΔA−a1|˜A24|a1|˜A23|0012ΔA+a1|˜A34|a1|˜A33|00−a1|˜A44|12ΔA−a1|˜A43|],ΔA=Det[A31A32A41A42]=A31A42−A32A41.
After this, we obtain the reduced IBP relation, where only the propagator
D3=(l+p3)2−m23 has one increasing power,cm,n,qiλ0;m,n,q+cm,n,q+1iλ0;m,n,q+1+cm,n−1,q+1iλ0;m,n−1,q+1+cm−1,n,q+1iλ0;m−1,n,q+1+cm−1,n,qiλ0;m−1,n,q+cm,n−1,qiλ0;m,n−1,q+cm,n,q−1iλ0;m,n,q−1+δ3;r=0,
(78) with the coefficients
cm,n,q=λ0+mQ11;r+nQ22;r+qQ33;r+Q11;r+Q22;r+Q33;r+λ4Q44;r+Q44;r,cm,n,q+1=λ4Q43;r=−a1λ4A31A42−A32A41|˜A44|,cm,n−1;q+1=nQ23;r=−a1nA31A42−A32A41|˜A24|,cm−1,n,q+1=mQ13;r=a1mA31A42−A32A41|˜A14|,cm−1,n,q=mQ14;ra1mA31A42−A32A41|˜A13|,cm,n−1,q=nQ24;r=−a1nA31A42−A32A41|˜A23|,cm,n,q−1=qQ34;r=a1qA31A42−A23A41|˜A33|,
(79) where the subscript r in
δ3;r andQij;r indicates that the parametersa2 toa6 should be replaced by (77).The reduction in the boundary
δ3 part: Similar to the bubble situation, inserting the value ofzi into theδ3 part, we obtainδ3;r=(δm+1,0Q11;r+δm,0Q14;r)iλ0,−1,n,q+δm,0Q12;riλ0,−1,n+1,q+δm,0Q13;riλ0,−1,n,q+1+δn,0Q21;riλ0,m+1,−1,q+(δn+1,0Q22;r+δn,0Q24;r)iλ0,m,−1,q+δn,0Q23;riλ0,m,−1,q+1+δq,0Q31;riλ0,m+1,n,−1+δq,0Q32;riλ0,m,n+1,−1+(δq+1,0Q33;r+δq,0Q34;r)iλ0,m,n,−1,
(80) where
iλ0,m,n,−1 ,iλ0,m,−1,q , andiλ0,−1,n,q contribute to the sub-topology of the triangle, that is, the bubble⑰. -
Now, we apply the complete recurrence relation to the example
I3(1,1,2) . Settingm=n=q=0 in (78), we obtainc0,0,0iλ0,0,0,0+c0,0,1iλ0,0,0,1+δ3;000=0,
(81) with the coefficients
c0,0,1=λ4Q43;r=−1−m21+m22+2(p1⋅p2)+p21×{2a1(D−4)(m41p22−2m21(m22((p1⋅p2)+p22)−m23(p1⋅p2)+p22((p1⋅p2)+p21))+m42(2(p1⋅p2)+p21+p22)+m22(2(p1⋅p2)(2(p1⋅p2)+p21+p22)−2m23((p1⋅p2)+p21))+p21(m43−2m23((p1⋅p2)+p22)+p22(2(p1⋅p2)+p21+p22)))},c0,0,0=−D2+Q11;r+Q22;r+Q33;r+(D−3)Q44;r=−2a1(D−4)(m21(p1⋅p2)−m22((p1⋅p2)+p21)+p21(m23−(p1⋅p2)−p22))−m21+m22+2(p1⋅p2)+p21.
(82) In (81), only two terms of triangle topology remain: one is the scalar basis, and the other is the target we want to reduce. The other five terms in (78) disappear owing to the expression in (79). Thus, there is no need to solve mixed IBP relations. The
δ3 term becomesδp;000≡δp;r|m=0,n=0,q=0=Q14;riλ0,−1,0,0+Q12;riλ0,−1,1,0+Q13;riλ0,−1,0,1+Q21;riλ0,1,−1,0+Q24;riλ0,0,−1,0+Q23;riλ0,0,−1,1+Q31;riλ0,1,0,−1+Q32;riλ0,0,1,−1+Q34;riλ0,0,0,−1.
(83) Translating back to the I form, we obtain the result
I3(1,1,2)=c3→111I3(1,1,1)+c3→110I3(1,1,0)+c3→101I3(1,0,1)+c3→011I3(0,1,1)c3→210I3(2,1,0)+c3→201I3(2,0,1)+c3→120I3(1,2,0)+c3→021I3(0,2,1)c3→102I3(1,0,2)+c3→012I3(0,1,2),
(84) with the coefficients
c3→111=c0,0,0Γ(D−3)c0,0,1Γ(D−4),c3→110=−Q34;rΓ(D−2)c0,0,1Γ(D−4),c3→101=−Q24;rΓ(D−2)c0,0,1Γ(D−4),c3→011=−Q14;rΓ(D−2)c0,0,1Γ(D−4),c3→210=Q31;rΓ(D−3)c0,0,1Γ(D−4),c3→201=Q21;rΓ(D−3)c0,0,1Γ(D−4),c3→021=Q12;rΓ(D−3)c0,0,1Γ(D−4),c3→120=Q32;rΓ(D−3)c0,0,1Γ(D−4),c3→102=Q23;rΓ(D−3)c0,0,1Γ(D−4),c3→012=Q13;rΓ(D−3)c0,0,1Γ(D−4).
(85) The final step is to reduce bubbles that have one propagator with the power two. This problem has been solved in the previous subsection (see (57)). With proper relabeling of the external variables of the last six terms in (84) and collecting all coefficients together, we get
I3(1,1,2)=c3→3I3(1,1,1)+c3→2;ˉ3I3(1,1,0)+c3→2;ˉ2I3(1,0,1)+c3→2;ˉ1I3(0,1,1)+c3→1;ˉ2ˉ3I3(1,0,0)+c3→1;ˉ1ˉ3I3(0,1,0)+c3→1;ˉ1ˉ2I3(0,0,1).
(86) Because the explicit expressions of these coefficients are long, they are provided in the companion Mathematica notebook. The result is confirmed by FIRE6.
-
Similar to the bubble case, with different choices, we can obtain three IBP recurrence relations. In each of these relations, only one term has a propagator with a higher power. For simplicity, we label the IBP recurrence relation
eqi , which shifts the propagatorDi . Now, we can useeqi withi=1,2,3 to calculate the general case for triangles. Let us denoteeq1:(a1+1++a1+3−1+3−+a1+2−1+2−+a3−3−+a2−2−+a1−1−+a0)iλ0,m,n,q+δ3;r,eq1=0,eq2:(b2+2++b2+3−2+3−+b1−2+1−2++b3−3−+b2−2−+b1−1−+b0)iλ0,m,n,q+δ3;r,eq2=0,eq3:(c3+3++c2−3+2−3++c1−3+1−3++c3−3−+c2−2−+c1−1−+c0)iλ0,m,n,q+δ3;r,eq3=0,
(87) where all coefficients have the same form as in (78). Combining these, we can reduce the general triangles. For example, for
I3(2,2,3) , after settingm=0 ,n=1 , andq=2 ineq1 , we can reduceI3(2,2,3) toI3(1,1,3) ,I3(1,2,2) ,I3(1,2,3) ,I3(2,1,3) ,I3(2,2,2) and boundary terms, the general bubbles. Then, settingm=0 ,n=0 , andq=2 ineq1 , we can reduceI3(2,1,3) toI3(1,1,2) ,I3(1,1,3) , andI3(2,1,2) . After 12 steps, we get the result for the reduction in the triangle topology. The boundary terms involve bubbles and tadpoles, which have been dealt with in previous subsections. Finally, we can obtain all coefficients fromI3(2,2,3) to all scalar bases. -
The general form of a box is given by
I4(n1+1,n2+1,n3+1,n4+1)=∫dDlDn1+11Dn2+12Dn3+13Dn4+14,
(88) with
D1=l2−m21,D2=(l−p1)2−m22,D3=(l−p1−p2)2−m23,D4=(l+p4)2−m24.
(89) The parametric form of
I4(n1+1,n2+1,n3+1,n4+1) can be written asI4(n1+1,n2+1,n3+1,n4+1)=i(−1)4+n1+n2+n3+n4Γ(−λ0)Γ(n1+1)Γ(n2+1)Γ(n3+1)Γ(n4+1)Γ(λ5+1)iλ0;n1,n2,n3,n4,
(90) where
iλ0;n1,n2,n3,n4=∫dΠ(5)Fλ0xn11xn22xn33xn44xλ55=∫dΠ(5)(Ux5+f)λ0xn11xn22xn33xn44xλ55dΠ(5)=dx1dx2dx3dx4dx5δ(∑xj−1),λ0=−D2λ5=−5−n1−n2−n3−n4−2λ0=D−5−n1−n2−n3−n4,
(91) and the functions are
U(x)=x1+x2+x3+x4,V(x)=x1x2p21+x1x3(p1+p2)2+x1x4(p1+p2+p3)2+x2x3p22+x2x4(p2+p3)2+x3x4p23f(x)=−V(x)+U(x)∑m2ixi=m21x1+m22x2+m23x3+m24x4+(m21+m22−p21)x1x2+[m21+m23−(p1+p2)2]x1x3+[m21+m24−(p1+p2+p3)2]x1x4,+(m22+m23−p22)x2x3+[m22+m24−(p2+p3)2]x2x4+(m23+m24−p23)x3x4,F(x)=U(x)x5+f(x)=m21x21+m22x22+m23x23+m24x24+(m21+m22−p21)x1x2+[m21+m23−(p1+p2)2]x1x3+[m21+m24−(p1+p2+p3)2]x1x4+[m22+m23−p22]x2x3+[m22+m24−(p2+p3)2]x2x4+[m23+m24−p23]x3x4+x1x5+x2x5+x3x5+x4x5=(x1+x2+x3+x4)(m21x1+m22x2+m23x3+m24x4+x5)−x1x2p21−x1x3(p1+p2)2−x1x4(p1+p2+p3)2−x2x3p22−x2x4(p2+p3)2−x3x4p23.
(92) Now, the matrices are given by
ˆA=[2m21m21+m22−p21m21+m23−p212m21+m24−p2131m21+m22−p212m22m22+m23−p22m22+m24−p2231m21+m23−p212m22+m23−p222m23m23+m24−p231m21+m24−p213m22+m24−p223m23+m24−p232m24111110],KA=[0a1a2a3a4−a10a5a6a7−a2−a50a8a9−a3−a6−a80a10−a4−a7−a9−a100],
(93) where
pij≡pi+pi+1⋯pj . -
Taking
B=−1x5 in (39), we get{cn1+1,n2,n2,n41++cn1+1,n2,n3,n4−1‘+4−+cn1+1,n2,n3−1,n41+3−+cn1+1,n2−1,n3,n41+2−+cn1,n2+1,n3,n42++cn1,n2+1,n3,n4−12+4−+cn1,n2+1,n3−1,n42+3−+cn1−1,n2+1,n3,n42+1−+cn1,n2,n3+1,n43++cn1,n2,n3+1,n4−13+4−+cn1,n2−1,n3+1,n43+2−+cn1−1,n2,n3+1,n43+1−+cn1,n2,n3,n4+14++cn1,n2,n3−1,n4+14+3−+cn1,n2−1,n3,n4+14+2−+cn1−1,n2,n3,n4+14+1−+cn1,n2,n3,n4−14−+cn1,n2,n3−1,n43−+cn1,n2−1,n3,n42−+cn1−1,n2,n3,n41−+cn1,n2,n3,n4}in1,n2,n3,n4+δ4=0,
(94) where
j+in1⋯nj⋯nk=in1⋯nj+1⋯nk,j−in1⋯nj⋯nk=in1⋯nj−1⋯nk.
(95) Similarly, we can choose particular values of the parameters
a2 toa10 with a freea1 to ensure the coefficients of the terms in the first three lines of (94) equal zero. The analytic solution is provided in the companion Mathematica notebook. Here, we can express the solution for the parameters using the matrix elements ofˆA .a2=−a1ΔBox|˜A13,45|,a3=a1ΔBox|˜A14,45|,a4=−a1ΔBox|˜A15,45|,a5=a1ΔBox|˜A23,45|,a6=−a1ΔBox|˜A24,45|,a7=a1ΔBox|˜A25,45|,a8=a1ΔBox|˜A34,45|,a9=−a1ΔBox|˜A35,45|,a10=a1ΔBox|˜A45,45|,ΔBox=|A31A32A33A41A42A43A51A52A53|,
where
|˜Aij,kl| represents the determinant of the matrix A after we removed thei,j th rows andk,l th columns. Then, the matrixˆQ becomesˆQr=1ΔBox[12ΔBox00−a1|˜A15|−a1|˜A14|012ΔBox0a1|˜A25|a1|˜A24|0012ΔBox−a1|˜A35|−a1|˜A34|00012ΔBox+a1|˜A45|a1|˜A44|000−a1|˜A55|12ΔBox−a1|˜A54|].
We then obtain the simplified recurrence relation
cn1,n2,n3,n4+1in1,n2,n3,n4+1+cn1,n2,n3−1,n4+1in1,n2,n3−1,n4+1+cn1,n2−1,n3,n4+1in1,n2−1,n3,n4+1+cn1−1,n2,n3,n4in1−1,n2,n3,n4+cn1,n2,n3,n4−1in1,n2,n3,n4−1+cn1,n2,n3−1,n4in1,n2,n3−1,n4+cn1,n2−1,n3,n4in1,n2−1,n3,n4+cn1−1,n2,n3,n4in1−1,n2,n3,n4+cn1,n2,n3,n4in1,n2,n3,n4+δ4;r=0.
(96) Now, we must calculate the
δ4 term.The reduction in the boundary
δ4 term: Similar to the former case, we can expand theδ4 term and take the values of the parametersa2 toa10 into theδ4 part. Subsequently, we getδ4;r=δn1+1,0Q11;ri−1,n2,n3,n4+δn1,0Q12;ri−1,n2+1,n3,n4+δn1,0Q13;ri−1,n2,n3+1,n4+δn1,0Q14;ri−1,n2,n3,n4+1+δn1,0Q15;ri−1,n2,n3,n4+δn2,0Q21;rin1+1,−1,n3,n4+δn2+1,0Q22;rin1,−1,n3,n4+δn2,0Q23;rin1,−1,n3+1,n4+δn2,0Q24;rin1,−1,n3,n4+1+δn2,0Q25;rin1,−1,n3,n4+δn3,0Q31;rin1+1,n2,−1,n4+δn3,0Q32;rin1,n2+1,−1,n4+δn3+1,0Q33;rin1,n2,−1,n4+δn3,0Q34;rin1,n2,−1,n4+1+δn3,0Q35;rin1,n2,−1,n4+δn4,0Q41;rin1+1,n2,n3,−1+δn4,0Q42;rin1,n2+1,n3,−1+δn4,0Q43;rin1,n2,n3+1,−1+δn4+1,0Q44;rin1,n2,n3,−1+δn4,0Q45;rin1,n2,n3,−1,
(97) where the subscript "r" represents the value of the parameter Q after we set
a2 toa10 . -
Now, we can use recurrence relation (96) to calculate the example
I4(1,1,1,2) . Lettingn1=n2=n3=n4=0 , we get (the coefficients of the other terms are all zero)c0,0,0,0i0,0,0,0+c0,0,0,1i0,0,0,1+δ4;0000=0,
(98) where
δ4;0000≡δ4;r|n1=n2=n3=n4=0 . Translating to I, we obtain the following result:I4(1,1,1,2)=c4→1111I4(1,1,1,1)+c4→1110I4(1,1,1,0)+c4→1101I4(1,1,0,1)+c4→1011I4(1,0,1,1)+c4→0111I4(0,1,1,1)+c4→2110I4(2,1,1,0)+c4→2101I4(2,1,0,1)+c4→2011I4(2,0,1,1)+c4→1210I4(1,2,1,0)+c4→1201I4(1,2,0,1)+c4→0211I4(0,2,1,1)+c4→1120I4(1,1,2,0)+c4→1021I4(1,0,2,1)+c4→0121I4(0,1,2,1)+c4→1102I4(1,1,0,2)+c4→1012I4(1,0,1,2)+c4→0112I4(0,1,1,2),
(99) with the coefficients
c4→1111=c0,0,0,0c0,0,0,1(D−5)=TrˆQij;r+(D−5)Q55;r−D2Q54;r,c4→0111=−Q15;rΓ(D−3)c0,0,0,1Γ(D−5),c4→1011=−Q25;rΓ(D−3)c0,0,0,1Γ(D−5),c4→1101=−Q35;rΓ(D−3)c0,0,0,1Γ(D−5),c4→1110=−Q45;rΓ(D−3)c0,0,0,1Γ(D−5),c4→0211=Q12;rΓ(D−4)c0,0,0,1Γ(D−5),c4→0121=Q13;rΓ(D−4)c0,0,0,1Γ(D−5),c4→0112=Q14;rΓ(D−4)c0,0,0,1Γ(D−5),c4→2011=Q21;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1021=Q23;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1012=Q24;rΓ(D−4)c0,0,0,1Γ(D−5),c4→2101=Q31;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1201=Q32;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1102=Q34;rΓ(D−4)c0,0,0,1Γ(D−5),c4→2110=Q41;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1210=Q42;rΓ(D−4)c0,0,0,1Γ(D−5),c4→1120=Q43;rΓ(D−4)c0,0,0,1Γ(D−5).
(100) Next, we must use the reduction in triangles with one double propagator given in (86). Inserting them into (99), we obtain the complete reduction in the box
I4(1,1,1,2) .I4(1,1,1,2)=c4→4I4(1,1,1,1)+c4→3;ˉ1I4(0,1,1,1)+c4→3;ˉ2I4(1,0,1,1)+c4→3;ˉ3I4(1,1,0,1)+c4→3;ˉ4I4(1,1,1,0)+c4→2;ˉ1ˉ2I4(0,0,1,1)+c4→2;ˉ1ˉ3I4(0,1,0,1)+c4→2;ˉ1ˉ4I4(0,1,1,0)+c4→2;ˉ2ˉ3I4(1,0,0,1)+c4→2;ˉ2ˉ4I4(1,0,1,0)+c4→2;ˉ3ˉ4I4(1,1,0,0)+c4→1;D1I4(1,0,0,0)+c4→1;D2I4(0,1,0,0)+c4→1;D3I4(0,0,1,0)+c4→1;D4I4(0,0,0,1),
(101) the long coefficient expressions of which are given in the companion Mathematica notebook. The result is confirmed by FIRE6.
-
The general form of a pentagon is given by
I5(n1+1,n2+1,n3+1,n4+1,n5+1)=∫dDlDn1+11Dn2+12Dn3+13Dn4+14Dn5+15
(102) with
D1=l2−m21,D2=(l−p1)2−m22,D3=(l−p1−p2)2−m23,D4=(l−p1−p2−p3)2−m24,D5=(l+p5)2−m25.
(103) The parametric form of
I5(n1+1,n2+1,n3+1,n4+1,n5+1) can be written asI5(n1+1,n2+1,n3+1,n4+1,n5+1)=i(−1)5+n1+n2+n3+n4+n5Γ(−λ0)5∑i=1Γ(ni+1)Γ(λ6+1)iλ0;n1,n2,n3,n4,n5,
(104) where
iλ0;n1,n2,n3,n4,n5=∫dΠ(6)Fλ0xn11xn22xn33xn44xn55xλ6+16,dΠ(5)=dx1dx2dx3dx4dx5dx6δ(∑xj−1),λ0=−D2,λ6=(D−6)−n1−n2−n3−n4−n5,
(105) and the function
U(x)=x1+x2+x3+x4+x5,V(x)=x1x2p21+x1x3p212+x1x4p213+x1x5p214+x2x3p22+x2x4p223+x2x5p224+x3x4p23+x3x5p234+x4x5p24,f(x)=(x1+x2+x3+x4+x5)(m21x1+m22x2+m23x3+m24x4+m25x5)−x1x2p21−x1x3p212−x1x4p213−x1x5p214−x2x3p22−x2x4p223−x2x5p224−x3x4p23−x3x5p234−x4x5p24,F(x)=(x1+x2+x3+x4+x5)(m21x1+m22x2+m23x3+m24x4+m25x5+x6)−x1x2p21−x1x3p212−x1x4p213−x1x5p214−x2x3p22−x2x4p223−x2x5p224−x3x4p23−x3x5p234−x4x5p24,
(106) where
pij≡pi+pi+1+⋯pj−1+pj . Now the matrix are given byˆA=[2m21m21+m22−p21m21+m23−p212m21+m24−p213m21+m25−p211m21+m22−p212m22m22+m23−p22m22+m24−p223m22+m25−p2241m21+m23−p212m22+m23−p2232m23m23+m24−p23m23+m25−p2341m21+m24−p213m22+m24−p223m23+m24−p232m24m24+m25−p241m21+m25−p21m22+m25−p224m23+m25−p234m24+m25−p242m251111110],
(107) ˆKA=[0a1a2a3a4a5−a10a6a7a8a9−a2−a60a10a11a12−a3−a7−a100a13a14−a4−a8−a11−a130a15−a5−a9−a12−a14−a150].
(107) Taking
B=−1/x6 , and insertingzi into the IBP identities,6∑i=1∫∂∂xi{ziFλ0xn11xn22xn33xn44xn55xλ6+16}+δ5=0,
(108) where
δ5 is given byδ5=5∑i=1δλi,0∫dΠ(5){ziFλ0xn11xn22xn33xn44xn55xλ6+16}|xi=0.
(109) -
Similar to the previous subsections, by expanding the IBP relation, we get
{cn1+1,n2,n3,n4,n51++cn1+1,n2−1,n3,n4,n51+2−+cn1+1,n2,n3−1,n4,n51+3−+cn1+1,n2,n3,n4−1,n51+4−+cn1+1,n2,n3,n4,n5−11+5−+cn1,n2+1,n3,n4,n52++cn1−1,n2+1,n3,n4,n51−2++cn1,n2+1,n3−1,n4,n52+3−+cn1,n2+1,n3,n4−1,n52+4−+cn1,n2+1,n3,n4,n5−12+5−+cn1,n2,n3+1,n4,n53++cn1−1,n2,n3+1,n4,n51−3++cn1,n2−1,n3+1,n4,n52−3++cn1,n2,n3+1,n4−1,n53+4−+cn1,n2,n3+1,n4,n5−13+5−+cn1,n2,n3,n4+1,n54++cn1−1,n2,n3,n4+1,n51−4++cn1,n2−1,n3,n4+1,n52−4++cn1,n2,n3−1,n4+1,n53−4++cn1,n2,n3,n4+1,n5−14+5−+cn1,n2,n3,n4,n5+15++cn1−1,n2,n3,n4,n5+11−5++cn1,n2−1,n3,n4,n5+12−5++cn1,n2,n3−1,n4,n5+13−5++cn1,n2,n3,n4−1,n5+14−5++cn1−1,n2,n3,n4,n51−+cn1,n2−1,n3,n4,n52−+cn1,n2,n3−1,n4,n53−+cn1,n2,n3,n4−1,n54−+cn1,n2,n3,n4,n5−15−+cn1,n2,n3,n4,n5}in1,n2,n3,n4,n5+δ5=0.
(110) We can choose particular values for parameters
a2 toa15 to ensure the coefficients of the first three line of (110) equal zero. The solution isa2=−a1Δpen|˜A13,56|,a3=a1Δpen|˜A14,56|,a4=−a1Δpen|˜A15,56|,a5=a1Δpen|˜A16,56|,a6=a1Δpen|˜A23,56|,a7=−a1Δpen|˜A24,56|,a8=a1Δpen|˜A25,56|,a9=−a1Δpen|˜A26,56|,a10=a1Δpen|˜A34,56|,a11=−a1Δpen|˜A35,56|,a12=a1Δpen|˜A36,56|,a13=a1Δpen|˜A45,56|,a14=−a1Δpen|˜A46,56|,a15=a1Δpen|˜A56,56|,
(111) where
Δpen=|A31A32A33A34A41A42A43A44A51A52A53A54A61A62A63A64|.
Subsequently, we get
{cn1,n2,n3,n4,n5+1;r5++cn1−1,n2,n3,n4,n5+1;r1−5++cn1,n2−1,n3,n4,n5+1;r2−5++cn1,n2,n3−1,n4,n5+1;r3−5++cn1,n2,n3,n4−1,n5+1;r4−5+cn1−1,n2,n3,n4,n5;r1−+cn1,n2−1,n3,n4,n5;r2−+cn1,n2,n3−1,n4,n5;r3−+cn1,n2,n3,n4−1,n5;r4−+cn1,n2,n3,n4,n5−1;r5−+cn1,n2,n3,n4,n5;r}iλ0;n1,n2,n3,n4,n5+δ5;r=0,
(112) where we define
i+iλ0;n1,n2,n3,n4,n5≡iλ0;n1,⋯ni+1,⋯n5,i−iλ0;n1,n2,n3,n4,n5≡iλ0;n1,⋯ni−1,⋯n5,
(113) with the coefficients
c0,0,0,0,1=Q65;rλ6,c−1,0,0,0,1=n1Q15;r,c0,−1,0,0,1=n2Q25;r,c0,0,−1,0,1=n3Q35;r,c0,0,0,−1,1=n4Q45;r,c−1,0,0,0,0=n1Q16;r,c0,−1,0,0,0=n2Q26;r,c0,0,−1,0,0=n3Q36;r,c0,0,0,−1,0=n4Q46;r,c0,0,0,0,−1=n5Q56;r,c00000;r=TrˆQij;r+((D−6))Q66;rr−D2+n1Q11;r+n2Q22;r+n3Q33;r+n4Q44;r+n5Q55;r,
(114) while the matrix
ˆQ becomesˆQr=1Δpen[12Δpen000a1|˜A1,6|a1|˜A1,5|012Δpen00−a1|˜A2,6|−a1|˜A2,5|0012Δpen0a1|˜A3,6|a1|˜A3,5|00012Δpen−a1|˜A4,6|−a1|˜A4,5|000012Δpen+a1|˜A5,6|a1|˜A5,5|0000−a1|˜A6,6|12Δpen−a1|˜A6,5|].
-
Similar to the former situation, the
δ6;r term is given byδ5;r=Q11;rδn1,−1i−1,n2,n3,n4,n5+Q12;rδn1,0i−1,n2+1,n3,n4,n5+Q13;rδn1,0i−1,n2,n3+1,n4,n5+Q14;rδn1,0i−1,n2,n3,n4+1,n5+Q15;rδn1,0i−1,n2,n3,n4,n5+1+Q16;rδn1,0i−1,n2,n3,n4,n5+Q21;rδn2,0in1+1,−1,n3,n4,n5+Q22;rδn2,−1in1,−1,n2,n3,n4,n5+Q23;rδn2,0in1,−1,n3+1,n4,n5+Q24;rδn2,0in1,−1,n3,n4+1,n5+Q25;rδn2,0in1,−1,n3,n4,n5+1+Q26;rδn2,0in1,−1,n3,n4,n5+Q31;rδn3,0in1+1,n2,−1,n4,n5Q32;rδn3,0in1,n2+1,−1,n4,n5+Q33;rδn3,−1in1,n2,−1,n4,n5+Q34;rδn3,0in1,n2,−1,n4+1,n5+Q35;rδn3,0in1,n2,−1,n4,n5+1+Q36;rδn3,0in1,n2,−1,n4,n5Q41;rδn4,0in1+1,n2,n3,−1,n5+Q42;rδn4,0in1,n2+1,n3,−1,n5+Q43;rδn4,0in1,n2,n3+1,−1,n5+Q44;rδn4,−1in1,n2,n3,−1,n5+Q45;rδn4,0in1,n2,n3,−1,n5+1+Q46;rδn4,0in1,n2,n3,−1,n5+Q51;rδn5,0in1+1,n2,n3,n4,−1+Q52;rδn5,0in1,n2+1,n3,n4,−1+Q53;rδn5,0in1,n2,n3+1,n4,−1+Q54;rδn5,0in1,n2,n3,n4+1,−1+Q55;rδn5,−1in1,n2,n3,n4,−1+Q56;rδn5,0in1,n2,n3,n4,n5.
(115) -
Setting
n1=n2=n3=n4=n5=0 , we get the IBP recurrence relation (other coefficients are all zero)c0,0,0,0,1iλ0;0,0,0,0,1+c0,0,0,0,0iλ0;0,0,0,0,0+δ5;00000=0,
(116) where
δ5;00000≡δ5;r|n1=n2=n3=n4=n5=0 .Comparing them with our scalar basis, we have the result
I5(1,1,1,1,2)=c5→5I5(1,1,1,1,1)+c5→01111I4(0,1,1,1,1)+c5→10111I5(1,0,1,1,1)+c5→11011I5(1,1,0,1,1)+c5→11101I5(1,1,1,0,1)+c5→11110I5(1,1,1,1,0)+c5→20111I5(2,0,1,1,1)+c5→21011I5(2,1,0,1,1)+c5→21101I5(2,1,1,0,1)
+c5→21110I5(2,1,1,1,0)+c5→02111I5(0,2,1,1,1)+c5→12011I5(1,2,0,1,1)+c5→12101I5(1,2,1,0,1)+c5→12110I5(1,2,1,1,0)+c5→01211I5(0,1,2,1,1)+c5→10211I5(1,0,2,1,1)+c5→11201I5(1,1,2,0,1)+c5→11210I5(1,1,2,1,0)+c5→01121I5(0,1,1,2,1)+c5→10121I5(1,0,1,2,1)+c5→11021I5(1,1,0,2,1)+c5→11120I5(1,1,1,2,0)+c5→01112I5(0,1,1,1,2)+c5→10112I5(1,0,1,1,2)+c5→11012I5(1,1,0,1,2)+c5→11102I5(1,1,1,0,2),
(117) with the coefficients
c5→5=(D−6)c0,0,0,0,0c0,0,0,0,1,c5→01111=(D−6)(5−D)Q16;rc0,0,0,0,1,c5→4;10111=(D−6)(5−D)Q26;rc0,0,0,0,1,c5→4;11011=(D−6)(5−D)Q36;rc0,0,0,0,1,c5→4;11101=(D−6)(5−D)Q46;rc0,0,0,0,1,c5→4;11110=(D−6)(5−D)Q56;rc0,0,0,0,1,c5→20111=(D−6)Q21;rc0,0,0,0,1,c5→21011=(D−6)Q31;rc0,0,0,0,1,c5→21101=(D−6)Q41;rc0,0,0,0,1,c5→21110=(D−6)Q51;rc0,0,0,0,1,c5→02111=(D−6)Q12;rc0,0,0,0,1,c5→12011=(D−6)Q32;rc0,0,0,0,1,c5→12101=(D−6)Q42;rc0,0,0,0,1,c5→12110=(D−6)Q52;rc0,0,0,0,1,c5→01211=(D−6)Q13;rc0,0,0,0,1,c5→10211=(D−6)Q23;rc0,0,0,0,1,c5→11201=(D−6)Q43;rc0,0,0,0,1,c5→11210=(D−6)Q53;rc0,0,0,0,1,c5→01121=(D−6)Q14;rc0,0,0,0,1,c5→10121=(D−6)Q24;rc0,0,0,0,1,c5→11021=(D−6)Q34;rc0,0,0,0,1,c5→11120=(D−6)Q54;rc0,0,0,0,1,c5→01112=(D−6)Q15;rc0,0,0,0,1,c5→10112=(D−6)Q25;rc0,0,0,0,1,c5→11012=(D−6)Q35;rc0,0,0,0,1,c5→11102=(D−6)Q45;rc0,0,0,0,1.
(118) The final step is to reduce the coefficients of the general boxes to the scalar basis.
After this reduction, we obtain the final solution.
\begin{aligned}[b] I_{5}(1,1,1,1,2) = &c_{5\to5}I_5(1,1,1,1,1)+c_{5\to4;\bar1}I_5(0,1,1,1,1)+c_{5\to4;\bar2}I_5(1,0,1,1,1)+c_{5\to4;\bar3}I_5(1,1,0,1,1) \\ &+c_{5\to4;\bar4}I_5(1,1,1,0,1)+c_{5\to4;\bar5}I_5(1,1,1,1,0)+c_{5\to3;\bar1\bar2}I_5(0,0,1,1,1)+c_{5\to3;\bar1\bar3}I_5(0,1,0,1,1) \\ &+c_{5\to3;\bar1\bar4}I_5(0,1,1,0,1)+c_{5\to3;\bar1\bar5}I_5(0,1,1,1,0)+c_{5\to3;\bar2\bar3}I_5(1,0,0,1,1)+c_{5\to3;\bar2\bar4}I_5(1,0,1,0,1) \\ &+c_{5\to3;\bar2\bar5}I_5(1,0,1,1,0)c_{5\to3;\bar3\bar4}I_5(1,1,0,0,1)+c_{5\to3;\bar3\bar5}I_5(1,1,0,1,0)+c_{5\to3;\bar4\bar5}I_5(1,1,1,0,0) \\ &+c_{5\to2;D_1D_2}I_5(1,1,0,0,0)+c_{5\to2;D_1D_3}I_5(1,0,1,0,0)+c_{5\to2;D_1D_4}I_5(1,0,0,1,0)+c_{5\to2;D_1D_5}I_5(1,0,0,0,1) \\ &+c_{5\to2;D_2D_3}I_5(0,1,1,0,0)+c_{5\to2;D_2D_4}I_5(0,1,0,1,0)+c_{5\to2;D_2D_5}I_5(0,1,0,0,1)+c_{5\to2;D_3D_4}I_5(0,0,1,1,0) \\ &+c_{5\to2;D_3D_5}I_5(0,0,1,0,1)+c_{5\to2;D_4D_5}I_5(0,0,0,1,1)+c_{5\to1;D_1}I_5(1,0,0,0,0)+c_{5\to1;D_2}I_5(0,1,0,0,0) \\ &+c_{5\to1;D_3}I_5(0,0,1,0,0)+c_{5\to1;D_4}I_5(0,0,0,1,0)+c_{5\to1;D_5}I_5(0,0,0,0,1), \end{aligned}
(119) with the coefficients given in the attached Mathematica notebook. Now, all coefficients are complete.
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The analytic results are provided in the Mathematica notebooks, which are publicly available at https://github.com/Wanghongbin123/oneloop_parametric.
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In this paper, we consider one-loop scalar integrals in the parametric representation given by Chen. However, in the recurrence relation, there are typically several terms that we do not want as well as terms with dimensional shifting in general, which makes calculations difficult and inefficient. In Chen's later paper [2], he used a method based on non-commutative algebra to cancel the dimension shift. Unlike other methods, the one-loop case involves a straightforward method in which linear equation systems are solved to simplify the IBP recurrence relation in the parametric representation. Benefiting from the fact that F is a homogeneous function of
x_i with a degree of two in the one-loop situation, we can solvex_i using{{{\partial }} F}/{{{\partial }} x_i} with several free parameters. Then, combining all the IBP identities with a particular factorz_i and choosing particular values for the free parameters, we succeed in canceling the dimension shift and terms with higher total power. As a complement to the tadpole coefficients of the reduction explored within our previous paper, we calculate several examples and provide an analytic result of the reduction.For further research, there are several factors to be considered. In our calculations, the constructed coefficients
z_i are not polynomial since they have a denominator with the formx_{n+1}^{{\gamma}} ; therefore, we cannot directly use the technique of syzygy. Moreover, the application of Chen's method to a higher loop is definitely another future research direction. For this case, the homogeneous functionF(x) is of degreeL+1 , where L is the number of loops. For the high loop case, we should consider how to construct the coefficientsz_i efficiently and find a relation similar to (37) to cancel the terms we do not need. Finally, the sub-topologies are entirely decided by the boundary term in the parametric representation, which may lead to simplification of calculation. -
I would like to thank Bo Feng for the inspiring discussion and guidance.
General one-loop reduction in generalized Feynman parametrization form
- Received Date: 2022-04-19
- Available Online: 2022-10-15
Abstract: The search for an effective reduction method is one of the main topics in higher loop computation. Recently, an alternative reduction method was proposed by Chen in [