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General one-loop reduction in generalized Feynman parametrization form

  • 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 [1, 2]. In this paper, we test the power of Chen's new method using one-loop scalar integrals with propagators of higher power. More explicitly, with the improved version of the method, we can cancel the dimension shift and terms with unwanted power shifting. Thus, the obtained integrating-by-parts relations are significantly simpler and can be solved easily. Using this method, we present explicit examples of a bubble, triangle, box, and pentagon with one doubled propagator. With these results, we complete our previous computations in [3] with the missing tadpole coefficients and show the potential of Chen's method for efficient reduction in higher loop integrals.
  • [1] W. Chen, JHEP 02, 115 (2020), arXiv:1902.10387
    [2] W. Chen, Reduction of Feynman Integrals in the Parametric Representation II: Reduction of Tensor Integrals, arXiv: 1912.08606
    [3] B. Feng and H. Wang, Reduction of one-loop integrals with higher poles by unitarity cut method, arXiv: 2104.00922
    [4] G. Passarino and M. J. G. Veltman, Nucl. Phys. B 160, 151 (1979) doi: 10.1016/0550-3213(79)90234-7
    [5] V. A. Smirnov, Springer Tracts in Modern Physics 211, (2005)
    [6] F. V. Tkachov, Phys. Lett. B 100, 65 (1981) doi: 10.1016/0370-2693(81)90288-4
    [7] K. G. Chetyrkin and F. V. Tkachov, Nucl. Phys. B 192, 159 (1981) doi: 10.1016/0550-3213(81)90199-1
    [8] Z. Bern, L. J. Dixon, D. C. Dunbar et al., Nucl. Phys. B 425, 217 (1994), arXiv:hep-ph/9403226 doi: 10.1016/0550-3213(94)90179-1
    [9] Z. Bern, L. J. Dixon, D. C. Dunbar et al., Nucl. Phys. B 435, 59 (1995), arXiv:hep-ph/9409265 doi: 10.1016/0550-3213(94)00488-Z
    [10] R. Britto, F. Cachazo, and B. Feng, Nucl. Phys. B 725, 275 (2005), arXiv:hep-th/0412103 doi: 10.1016/j.nuclphysb.2005.07.014
    [11] F. Cachazo, P. Svrcek, and E. Witten, JHEP 10, 074 (2004), arXiv:hep-th/0406177
    [12] R. Britto, E. Buchbinder, F. Cachazo et al., Phys. Rev. D 72, 065012 (2005), arXiv:hep-ph/0503132 doi: 10.1103/PhysRevD.72.065012
    [13] C. Anastasiou, R. Britto, B. Feng et al., Phys. Lett. B 645, 213 (2007), arXiv:hep-ph/0609191 doi: 10.1016/j.physletb.2006.12.022
    [14] R. Britto and B. Feng, Phys. Rev. D 75, 105006 (2007), arXiv:hep-ph/0612089 doi: 10.1103/PhysRevD.75.105006
    [15] C. Anastasiou, R. Britto, B. Feng et al., JHEP 03, 111 (2007), arXiv:hep-ph/0612277
    [16] R. Britto, B. Feng, and P. Mastrolia, Phys. Rev. D 73, 105004 (2006), arXiv:hep-ph/0602178 doi: 10.1103/PhysRevD.73.105004
    [17] R. Britto and B. Feng, Phys. Lett. B 681, 376 (2009), arXiv:0904.2766 doi: 10.1016/j.physletb.2009.10.038
    [18] Y. Zhang, An Introduction to Integrals with Uniformally Transcendental Weights, Canonical Differential Equation, Symbols and Polylogarithms, http://staff.ustc.edu.cn/~yzhphy/teaching/summer2021/UT_2021.pdf, 2021
    [19] J. M. Henn, J. Phys. A 48, 153001 (2015), arXiv:1412.2296 doi: 10.1088/1751-8113/48/15/153001
    [20] P. A. Baikov, Phys. Lett. B 385, 404 (1996), arXiv:hep-ph/9603267 doi: 10.1016/0370-2693(96)00835-0
    [21] P. A. Baikov, Explicit solutions of n loop vacuum integral recurrence relations, hep-ph/9604254
    [22] Z. Bern, L. J. Dixon, and D. A. Kosower, Phys. Lett. B 302, 299 (1993), arXiv:hep-ph/9212308 doi: 10.1016/0370-2693(93)90400-C
    [23] Z. Bern, L. J. Dixon, and D. A. Kosower, Nucl. Phys. B 412, 751 (1994), arXiv:hep-ph/9306240 doi: 10.1016/0550-3213(94)90398-0
    [24] J. Gluza, K. Kajda, and D.A. Kosower, Towards a basis for planar two-loop integrals, arXiv: 1009.0472
    [25] K. J. Larsen and Y. Zhang, Integration-by-parts reductions from unitarity cuts and algebraic geometry, arXiv: 1511.01071
    [26] K. J. Larsen and Y. Zhang, Integration-by-parts reductions from the viewpoint of computational algebraic geometry, arXiv: 1606.09447
    [27] K. J. Larsen and Y. Zhang, Phys. Rev. D 93, 041701 (2016), arXiv:1511.01071 doi: 10.1103/PhysRevD.93.041701
    [28] K. J. Larsen and Y. Zhang, PoS LL2016, 029 (2016), arXiv:1606.09447
    [29] Y. Zhang, Lecture Notes on Multi-loop Integral Reduction and Applied Algebraic Geometry, 12, 2016 [1612.02249]
    [30] Y. Jiang and Y. Zhang, JHEP 03, 087 (2018), arXiv:1710.04693
    [31] A.V. Smirnov and F. S. Chuharev, Comput. Phys. Commun. 247, 106877 (2020), arXiv:1901.07808 doi: 10.1016/j.cpc.2019.106877
    [32] R. N. Lee, J. Phys. Conf. Ser. 523, 012059 (2014), arXiv:1310.1145 doi: 10.1088/1742-6596/523/1/012059
  • [1] W. Chen, JHEP 02, 115 (2020), arXiv:1902.10387
    [2] W. Chen, Reduction of Feynman Integrals in the Parametric Representation II: Reduction of Tensor Integrals, arXiv: 1912.08606
    [3] B. Feng and H. Wang, Reduction of one-loop integrals with higher poles by unitarity cut method, arXiv: 2104.00922
    [4] G. Passarino and M. J. G. Veltman, Nucl. Phys. B 160, 151 (1979) doi: 10.1016/0550-3213(79)90234-7
    [5] V. A. Smirnov, Springer Tracts in Modern Physics 211, (2005)
    [6] F. V. Tkachov, Phys. Lett. B 100, 65 (1981) doi: 10.1016/0370-2693(81)90288-4
    [7] K. G. Chetyrkin and F. V. Tkachov, Nucl. Phys. B 192, 159 (1981) doi: 10.1016/0550-3213(81)90199-1
    [8] Z. Bern, L. J. Dixon, D. C. Dunbar et al., Nucl. Phys. B 425, 217 (1994), arXiv:hep-ph/9403226 doi: 10.1016/0550-3213(94)90179-1
    [9] Z. Bern, L. J. Dixon, D. C. Dunbar et al., Nucl. Phys. B 435, 59 (1995), arXiv:hep-ph/9409265 doi: 10.1016/0550-3213(94)00488-Z
    [10] R. Britto, F. Cachazo, and B. Feng, Nucl. Phys. B 725, 275 (2005), arXiv:hep-th/0412103 doi: 10.1016/j.nuclphysb.2005.07.014
    [11] F. Cachazo, P. Svrcek, and E. Witten, JHEP 10, 074 (2004), arXiv:hep-th/0406177
    [12] R. Britto, E. Buchbinder, F. Cachazo et al., Phys. Rev. D 72, 065012 (2005), arXiv:hep-ph/0503132 doi: 10.1103/PhysRevD.72.065012
    [13] C. Anastasiou, R. Britto, B. Feng et al., Phys. Lett. B 645, 213 (2007), arXiv:hep-ph/0609191 doi: 10.1016/j.physletb.2006.12.022
    [14] R. Britto and B. Feng, Phys. Rev. D 75, 105006 (2007), arXiv:hep-ph/0612089 doi: 10.1103/PhysRevD.75.105006
    [15] C. Anastasiou, R. Britto, B. Feng et al., JHEP 03, 111 (2007), arXiv:hep-ph/0612277
    [16] R. Britto, B. Feng, and P. Mastrolia, Phys. Rev. D 73, 105004 (2006), arXiv:hep-ph/0602178 doi: 10.1103/PhysRevD.73.105004
    [17] R. Britto and B. Feng, Phys. Lett. B 681, 376 (2009), arXiv:0904.2766 doi: 10.1016/j.physletb.2009.10.038
    [18] Y. Zhang, An Introduction to Integrals with Uniformally Transcendental Weights, Canonical Differential Equation, Symbols and Polylogarithms, http://staff.ustc.edu.cn/~yzhphy/teaching/summer2021/UT_2021.pdf, 2021
    [19] J. M. Henn, J. Phys. A 48, 153001 (2015), arXiv:1412.2296 doi: 10.1088/1751-8113/48/15/153001
    [20] P. A. Baikov, Phys. Lett. B 385, 404 (1996), arXiv:hep-ph/9603267 doi: 10.1016/0370-2693(96)00835-0
    [21] P. A. Baikov, Explicit solutions of n loop vacuum integral recurrence relations, hep-ph/9604254
    [22] Z. Bern, L. J. Dixon, and D. A. Kosower, Phys. Lett. B 302, 299 (1993), arXiv:hep-ph/9212308 doi: 10.1016/0370-2693(93)90400-C
    [23] Z. Bern, L. J. Dixon, and D. A. Kosower, Nucl. Phys. B 412, 751 (1994), arXiv:hep-ph/9306240 doi: 10.1016/0550-3213(94)90398-0
    [24] J. Gluza, K. Kajda, and D.A. Kosower, Towards a basis for planar two-loop integrals, arXiv: 1009.0472
    [25] K. J. Larsen and Y. Zhang, Integration-by-parts reductions from unitarity cuts and algebraic geometry, arXiv: 1511.01071
    [26] K. J. Larsen and Y. Zhang, Integration-by-parts reductions from the viewpoint of computational algebraic geometry, arXiv: 1606.09447
    [27] K. J. Larsen and Y. Zhang, Phys. Rev. D 93, 041701 (2016), arXiv:1511.01071 doi: 10.1103/PhysRevD.93.041701
    [28] K. J. Larsen and Y. Zhang, PoS LL2016, 029 (2016), arXiv:1606.09447
    [29] Y. Zhang, Lecture Notes on Multi-loop Integral Reduction and Applied Algebraic Geometry, 12, 2016 [1612.02249]
    [30] Y. Jiang and Y. Zhang, JHEP 03, 087 (2018), arXiv:1710.04693
    [31] A.V. Smirnov and F. S. Chuharev, Comput. Phys. Commun. 247, 106877 (2020), arXiv:1901.07808 doi: 10.1016/j.cpc.2019.106877
    [32] R. N. Lee, J. Phys. Conf. Ser. 523, 012059 (2014), arXiv:1310.1145 doi: 10.1088/1742-6596/523/1/012059
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Hongbin Wang. General one-loop reduction in generalized Feynman parametrization form[J]. Chinese Physics C. doi: 10.1088/1674-1137/ac7a1c
Hongbin Wang. General one-loop reduction in generalized Feynman parametrization form[J]. Chinese Physics C.  doi: 10.1088/1674-1137/ac7a1c shu
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General one-loop reduction in generalized Feynman parametrization form

    Corresponding author: Hongbin Wang, 21836003@zju.edu.cn
  • Zhejiang Institute of Modern Physics, Zhejiang University, Hangzhou 310027, China

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 [1, 2]. In this paper, we test the power of Chen's new method using one-loop scalar integrals with propagators of higher power. More explicitly, with the improved version of the method, we can cancel the dimension shift and terms with unwanted power shifting. Thus, the obtained integrating-by-parts relations are significantly simpler and can be solved easily. Using this method, we present explicit examples of a bubble, triangle, box, and pentagon with one doubled propagator. With these results, we complete our previous computations in [3] with the missing tadpole coefficients and show the potential of Chen's method for efficient reduction in higher loop integrals.

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    I.   INTRODUCTION
    • 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 [57]. 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 [817]. 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 combine U,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 42ϵ 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 [2530]. 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 degree L+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 matrix ˆA, which is entirely determined by a graph. In section III, we calculate a triangle I3(1,1,2), box I4(1,1,1,2), and pentagon I5(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.

    II.   CHEN'S REDUCTION METHOD IN PARAMETRIC FORM
    • 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)=dDl1dDl2dDlLN(l)Dk11Dk22Dk33Dknn,

      (1)

      where, for simplicity, we denote l=(l1,l2,l3,,lL) and k=(k1,k2,k3,,kn). Because we consider only scalar integrals with N(l)=1 in this paper, let us label

      I(L;λ1+1,,λn+1)=dDl1dDlL1Dλ1+11Dλn+1n.

      (2)

      Using the Feynman parametrization procedure,

      LiαiDi=Li,jAijlilj+2Li=1Bili+C,

      (3)

      and thus loop integrals can be found as

      dDl1dDlLei(αiDi)=eiπL(1D2)/2πLD/2(DetA)D/2×ei(CA1ijBiBj).

      (4)

      Defining U(α)=DetA and CA1ijBiBjV(α)/U(α)m2iαi , we can see that U(α) is a homogeneous function of αi with degree L, whereas V(α) is a homogeneous function of αi with degree L+1. The loop integral becomes

      I(L;λ1+1,,λn+1)=e((λi+1)/2)iπΠni=1Γ(λi+1)eiπL(1D/2)/2πLD/2×dα1dαnU(α)D/2ei[V(α)/U(α)m2iαi]αλ11αλnn.

      (5)

      To derive the parametric form suggested by Chen, we perform the following: Using the α-representation of general propagators,

      1(l2m2)λ+1=e((λ+1)/2)iπΓ(λ+1)0dαeiα(l2m2)αλ,Im{l2m2}>0,

      (6)

      where "iϵ" is neglected, we obtain

      I(L;λ1+1,,λn+1)=eni((λi+1)/2)iπΠni=1Γ(λi+1)dDl1dDlL×0dα1dαneini=1αiDiαλ11αλnn.

      (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

      iS(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+λDL/2)Γ(n+λDL2)×dx1dxnδ(jSxj1)U(x)n+λ(D/2)(L+1)[V(x)+U(x)m2ixi]n+λDL/2xλ11xλnn=(i)n+λDL/2Γ(n+λDL2)dx1dxnδ(jSxj1)Uλufλfxλ11xλnn,

      (9)

      where

      U(x)=ηLU(α)=ηLU(ηxi),V(x)=ηL1V(α)=ηL1V(ηx),f(x)=V(x)+U(x)m2ixiλ=ni=1λi,λu=n+λD2(L+1),λf=nλ+DL2.

      (10)

      Finally, via Mellin transformation

      Aλ1Bλ2=Γ(λ1λ2)Γ(λ1)Γ(λ2)0dx(A+Bx)λ1+λ2xλ21,

      (11)

      we can express (9) as

      (i)n+λDL/2Γ(n+λDL2)Γ(λuλf)Γ(λu)Γ(λf)dx1dxnδ(jSxj1)0dxn+1×(Uxn+1+f)λu+λfxλu1n+1xλ11xλnn(i)n+λDL/2Γ(n+λDL/2)Γ(λuλf)Γ(λu)Γ(λf)dΠ(n+1)Fλ0xλ11xλnnxλn+1n+1(i)n+λDL/2Γ(n+λDL2)Γ(λuλf)Γ(λu)Γ(λf)iλ0;λ1,λn,

      (12)

      where

      dΠ(n+1)=dx1dxn+1δ(jSxj1),F=Uxn+1+f,λ=ni=1λi,λ0=λu+λf=D2,λn+1=λu1=D2(L+1)λ1n.

      (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)
    • A.   IBP identity in parametric form

    • The parametric form of (14) is the starting point of Chen's proposal. The IBP relations in this form are given by

      dΠ(n+1)xi{Fλ0xλ11xλnnxλn+1+1n+1}+δλi,0dΠ(n){Fλ0xλ11xλnnxλn+1+1n+1}|xi=0=0

      (15)

      where i=1,...,n+1, and dΠ(n) in the second term is

      dΠ(n)=dx1^dxidxndxn+1δ(jSxj1).

      (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 by

      I2(m+1,n+1)=dDl(l2m21)m+1((lp1)2m22)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(D2mn)×dΠ(3)Fλ0xm1xn2xλ33,

      (18)

      where

      F=(x1+x2)(m21x1+m22x2+x3)p21x1x2,

      (19)

      and

      iλ0;m,n=dΠ(3)Fλ0xm1xn2xλ33,

      (20)

      with λ0=D2, and λ3=3mn2λ0. Using Eq. (15), we can obtain three IBP recurrence relations. First, taking x1, the first term in (15) gives

      λ0iλ01;m,n+2m21λ0iλ01;m+1,n+Δλ0iλ01;m,n+1,

      (21)

      where Δ=m21+m22p21. The second term gives

      δm,0dΠ(2)(x3+m22x2)λ0xn+λ02x2n2λ03=δm,0iλ0;1,n.

      (22)

      Here, the notation iλ0;1,nmust be explained. From the middle expression of (22), we see that it is the parametric form of the tadpole dDl/(l2m22)n+1. To emphasize its origin, that is,from a bubble by removingthe first propagator, we extend the definition of iλ0;λ1,...,λn given in (12) by setting λ1=1 . Using the extended notation, we obtain the first IBP relation

      λ0iλ01;m,n+2m21λ0iλ01;m+1,n+Δλ0iλ01;m,n+1+δm,0iλ0;1,n=0.

      (23)

      When we set m=n=0 in (23), this reads

      λ0iλ01;0,0+2m21λ0iλ01;1,0+Δλ0iλ01;0,1+iλ0;1,0=0.

      (24)

      Similarly, we can take the differential x2 and obtain the second IBP relation

      λ0iλ01;0,0+Δλ0iλ01;1,0+2m22λ0iλ01;0,1+iλ0;0,1=0.

      (25)

      We should solve iλ0;0,1 by iλ0;0,0 from (24) and (25). However, for the bubble part, we have λ01 instead of λ0. This could be fixed by rewriting λ0λ0+1 because λ0 is a free parameter. However, the boundary tadpole part iλ0;0,1 will become iλ0+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)λ0xm1x2m2λ03

      (26)

      and taking the x1 and x3, we can obtain two IBP relations,

      λ0iλ01;m,1+2m21λ0iλ01;m+1,1+miλ0,m1,1=0,λ0iλ01;m+1,1+(1m2λ0)iλ0;m,1=0,

      (27)

      from which we solve

      iλ0;0,1=λ02m21(2λ0+1)iλ01;0,1,iλ0;1,0=λ02m22(2λ0+1)iλ01;1,0.

      (28)

      Inserting (28) into (24) and (25), we can solve iλ01;0,1. After shifting λ0λ0+1, we finally get

      iλ0;0,1=2m21ΔΔ24m21m22iλ0;0,0+1(2λ0+3)(Δ24m21m22)iλ0;0,1+Δ2m22(2λ0+3)(Δ24m21m22)iλ0;1,0.

      (29)

      Translating back to the scalar basis, we obtain the reduction in I2(1,2) as

      I2(1,2)=c22I2(1,1)+c21ˉ2I2(1,0)+c21;ˉ1I2(0,1),

      (30)

      with the coefficients

      c22=(D3)(Δ2m21)Δ24m21m22,c21;ˉ2=D2Δ24m21m22,c21;ˉ1=(D2)Δ2m22(4m21m22Δ2).

      (31)

      This result is confirmed with FIRE6 [31, 32].

    • B.   Improvement of parametric IBP

    • 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 of xi with degree of two. This good property simplifies the related syzygy equations, which can then be directly solved. 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=xα1xβ2xαβ3. Because the degree of the new integrand does not change, the IBP identity still holds. Summing them together we get

      n+1i=1dΠ(n+1)xi{ziFλ0xλ11xλ22xλn+1+1n+1}+n+1i=1δλi,0dΠ(n)ziFλ0xλ11xλnnxλn+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+1i=1(zixi+λ0ziFxiF+λizixi)+zn+1xn+1]×Fλ0xλi1xλ22xλnnxλn+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 coefficients zi so that n+1i=1ziFxi is a multiple of the function F, that is,

      n+1i=1ziFxi+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 as

      F=12Aijxixj,

      (35)

      where A is a symmetric matrix. Thus, we have

      fiFxi,ˆf=ˆAˆx,ˆf[f1f2fnfn+1],ˆx[x1x2xnxn+1],

      Solving ˆx=ˆA1ˆf, we have

      F=12ˆxTAˆx=12ˆfT(ˆA1)TˆAˆA1ˆf=12ˆfT(ˆA1)TˆfˆfTˆKˆf,K=12A1,

      (36)

      where the coefficient matrix ˆK is a real symmetry matrix. In fact, we can go further. Using

      0=ˆfTˆKAˆf

      (37)

      with any antisymmetric matrix KA, we can add (37) to (36) to obtain a more general form

      F=ˆfTˆKˆf+ˆfTˆKAˆf=ˆfT(ˆK+ˆKA)ˆfˆfTˆRˆf=ˆfTˆRˆAˆxˆfTˆQˆx,ˆQ12ˆI+ˆKAˆA.

      (38)

      Note that because the arbitrary matrix ˆKA is of rank n+1, there are n(n+1)2 free independent parameters, a1,,a(n(n+1))/2, in the matrix ˆQ in (38).

      Now, reinserting (38) into (34), we can solve ˆz as

      ˆfTˆz+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 choose B=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 of ˆQ, we may cancel several unwanted terms. Some examples are shown in later computations to illustrate this trick.

    III.   REDUCTION IN ONE-LOOP INTEGRALS
    • 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.

    • A.   Bubble case

    • 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=[14p2114p21m21+m22+p214p2114p2114p21m21m22+p214p21m21+m22+p214p21m21m22+p214p21Δ24m21m224p21].

      (41)

      Adding the antisymmetric matrix KA, we have

      ˆKA=[0a1a2a10a3a2a30],ˆQ=[1+2a2+2a1m21+2a1m222a1p212a2+2a1m22a1a32a1m211+2a32a1m212a1m22+2a1p212a12a2m21a3(m21+m22p21)2a3m22a2(m21+m22p21)12a22a32].

      (42)

    • 1.   Deriving the recurrence relation
    • Taking B=1/x3 in (34), solution (39) gives zi=(Qijxj)/x3. Expanding (32), we obtain the IBP recurrence relation

      cm,niλ0;m,n+cm+1,niλ0;m+1,n+cm+1,n1iλ0;m+1,n1+cm,n+1iλ0;m,n+1+cm1,n+1iλ0;m1,n+1+cm,n1iλ0;m,n1+cm1,niλ0;m1,n+δ2=0,

      (43)

      where δ2 is the boundary term, which we will compute later. The other coefficients are

      cm,n=Q11(1+m)+Q22(1+n)+Q33(1+λ3)+λ0,cm+1,n=Q31λ3=λ3(a2A11+a3A21),cm+1,n1=Q21n=n(a1A11a3A31),

      cm,n+1=Q32λ3=λ3(a2A12a3A22),cm1,n+1=Q12m=m(a1A22+a2A32),cm,n1=Q23n=n(a1A13a3A33),cm1,n=Q13m=m(a1A32+a2A33).

      (44)

      Because we aim to obtain the reduction in I2(1,2), starting from m=n=0, we want to eliminate terms with the indices (m+1,n) and (m+1,n1) while keeping the term with the index (m,n+1). Thus, we impose cm+1,n=0 and cm+1,n1=0, which can be satisfied by choosing the free parameters

      a2=a1A21A31=a1(m21+m22p21),a3=a1A11A31=2a1m21.

      (45)

      After this choice, the matrix ˆQ becomes

      ˆQ;r=[12a1A31(A22A31A21A32)a1A31(A23A31A21A33)012a1A31(A12A31A11A32)a1A31(A11A33A13A31)0a1A31(A12A21A11A22)12+a1A31(A13A21A11A23)],

      leaving five terms with non-zero coefficients.

      cm,n+1=a1λ3A31(A11A22A12A21)=a1λ3A31|˜A33|=a1λ3,cm1,n+1=ma1A31(A21A32A22A31)=ma1A31|˜A13|=a1m(m21m22p21),cm,n1=na1A31(A11A33A13A31)=na1A31|˜A22|=a1n,cm1,n=ma1A31(A21A33A23A31)=ma1A31|˜A12|=a1m,cm,n=a1A31((1+n)(A11A32A12A31)(λ3+1)(A11A23A13A21))=a1A31(nλ3)|˜A23|=a1A31((nλ3)(m21m22+p21)).

      (46)

      The boundary δ2 term: The δ2 term is given by

      δ2=3i=1δλi,0dΠ(2){ziFλ0xm1xn2xλ3+13}|xi=0,

      (47)

      where λi represents the power of xi. It is worth emphasizing that because zi contains xi, the total power λi of xi is not equal to m,n,λ3 in general. Expanding it, we get

      δ2=δλ1,0dΠ(2)(Q11Fλ0xm+11xn2xλ33+Q12Fλ0xm1xn+12xλ33+Q13Fλ0xm1xn2xλ3+13)|x1=0+δλ2,0dΠ(2)(Q21Fλ0xm+11xn2xλ33+Q22Fλ0xm1xn+12xλ33+Q23Fλ0xm1xn2xλ3+13)|x2=0.

      (48)

      Remembering our extended notation explained under (22), we have

      dΠ(2)F|λ0x1=0xn2iλ0;1,n,dΠ(2)F|λ0x2=0xm1iλ0;m,1,

      (49)

      and the δ2 term can be written as

      δ2;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 and Qij;r indicates that a2 and a3 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 with

      c0,0iλ0;0,0+c0,1iλ0;0,1+δ2;00=0,

      (51)

      with the coefficients

      c0,0=a1(D3)(m21m22+p21),c0,1=a1(D3)(m41+m42p412m21p212m22p212m21m22),δ2;00=Q12;riλ0;1,1+Q13;riλ0;1,0+Q21;riλ0;1,1+Q23;riλ0;0,1,

      (52)

      where

      Q21;r=a1A31(A21A32A22A31)=a1A31|˜A13|,Q23;r=a1A31(A11A33A13A31)=a1A31|˜A22|,Q12;r=a1A31(A21A32A22A31)=a1A31|˜A13|,Q13;r=a1A31(A21A33A23A31)=a1A31|˜A12|.

      (53)

      From this, we can directly write the solution as

      iλ0;0,1=c0,0c0,1iλ0;0,0Q21;rc0,1iλ0;1,1Q23;rc0,1iλ0;0,1Q12;rc0,1iλ0;1,1Q13;rc0,1iλ0;1,0.

      (54)

      Translating back to scalar integrals, it is

      I2(1,2)=c1211I2(1,1)+c1210I2(1,0)+c1220I2(2,0)+c1201I2(0,1)+c1202I2(0,2),

      (55)

      with c1220=0 and

      c1211=(3+D)(m21m22+p21))(m41+(m22p21)22m21(m22+p21),c1210=D22m21(m22+p21)+m41+(m22p21)2c1201=2D2m21(m22+p21)+m41+(m22p21)2,c1202=m21+m22+p212m21(m22+p21)+m41+(m22p21)2.

      (56)

      Using I2(2,0)=D22m21I2(1,0) and I2(0,2)=D22m22I2(0,1), we have our final results for the reduction in I2(1,2),

      I2(1,2)=c22I2(1,1)+c21;ˉ2I2(1,0)+c21;ˉ1I2(0,1),

      (57)

      with the coefficients

      c22=(D3)(m21m22+p21)2m21(m22+p21)+m41+(m22p21)2,c21;ˉ2=D22m21(m22+p21)+m41+(m22p21)2,c21;ˉ1=(D2)(m21+m22p21)2m22(2m21(m22+p21)+m41+(m22p21)2),

      (58)

      which is given in (30).

    • B.   General case for bubbles

    • 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 in D2. Similarly, by choosing different values of a2 and a3, we can obtain another IBP recurrence relation to reduce the integral to those with no higher power in D1. The choice is

      a2=a1A22A32,a3=a1A12A32,

      (59)

      and the corresponding IBP recurrence is

      cm+1,niλ0,m+1,n+cm+1,n1iλ0,m+1,n1+cm,n1iλ0,m,n1+cm1,niλ0,m1,n+cm,niλ0,m,n+δ2;r=0,

      (60)

      with the coefficients

      cm+1,n=(|˜A33|)(D3mn),cm+1,n1=n|˜A23|,cm,n1=n|˜A21|,cm1,n=m|˜A11|,cm,n=|˜A13|(3+2m+nD),

      (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.

    • 1.   Example: I2(1,3)
    • In the example I2(1,3), we simply need to reduce D2 from power 3 to 1. The strategy is to use (46) twice. In the first step, by setting m=0 and n=1 in (46), we get

      I2(1,3)=|˜A23|(D5)2|˜A33|I2(1,2)+|˜A22|(D3)2|˜A33|I2(1,1)+|˜A12|(D3)2|˜A33|I2(0,2)+|˜A13||˜A33|I2(0,3).

      (63)

      For the first term in (63), setting m=0 and n=0 in (46) again, we have

      I2(1,2)=|˜A23|(D3)|˜A33|I2(1,1)+|˜A22|(D2)|˜A33|I2(1,0)+|˜A13||˜A33|I2(0,2)+|˜A12|(D2)|˜A33|I2(0,1).

      (64)

      Inserting (64) into (63) and using the reduction in the tadpole, we get

      I2(1,3)=c1311I2(1,1)+c1310I2(1,0)+c1301I2(0,1),

      (65)

      with the coefficients

      c1311=(|˜A23||˜A33|+|˜A23|2(D5))(D3)2|˜A33|2,c1310=|˜A22||˜A23|(D5)(D2)2|˜A33|2,c1301=(D2)8|˜A33|2m42A21(2A32|˜A23|(D5)m22+A32At33(D4)4A33|˜A23|(D5)m422A33|˜A33|(D3)m22)A22A31(2|˜A23|(D5)m22+At33(D4))+2A23A31m22(2|˜A23|(D5)m22+|˜A33|(D3)).

      (66)

      The result is confirmed with FIRE6. In this example, we simply need to solve two equations to reduce the bubble topology.

    • 2.   Example: I2(3,5)
    • For this example, we must use (60) to lower the power of D1 and (46) to lower the power of D2. Setting m=1 and n=4 in (60), we can reduce I2(3,5) to I2(2,4), I2(2,5), I2(1,5), and I2(3,4).

      I2(3,5)=|˜A11|(D7)2|˜A33|I2(1,5)+|˜A13|(D9)2|˜A33|I2(2,5)+|˜A21|(D7)2|˜A33|I2(2,4)+|˜A23||˜A33|I2(3,4).

      (67)

      Then, setting m=1 and n=3 in (60), we reduce I2(3,4) to I2(1,4), I2(2,3), I2(2,4), and I2(3,3).

      I2(3,4)=|˜A23||˜A33|I2(3,3)+|˜A13|(D8)2|˜A33|I2(2,4)+|˜A21|(D6)2|˜A33|I2(2,3)+|˜A11|(D6)2|˜A33|I2(1,4).

      (68)

      Using the same idea, we must solve 14 equations to completely reduce I2(3,5). The analytic expressions for these 14 equations have also been confirmed by FIRE6.

    • C.   Triangle case

    • The triangle I3(m+1,n+1,q+1) is given by

      I3(m+1,n+1,q+1)=dDl(l2m21)m+1((lp1)2m22)n+1((l+p3)2m23)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=42λ0mnq=D4mnq.

      (71)

      Using expression (10), we have

      U(x)=x1+x2+x3,V(x)=x1x2p21+x1x3p23+x2x3p22,f(x)=V+Um2ixi=(x1+x2+x3)(x1m21+x2m22+x3m23)x1x2p21x2x3p22x1x3p23,F(x)=U(x)x4+f(x)=(x1+x2+x3)×(m21x1+m22x2+m23x3+x4)x1x2p21x2x3p22x1x3p23.

      (72)

      Thus, we can express the matrices as

      ˆA=[2m21m21+m22p21m21+m23p231m21+m22p212m22m22+m23p221m21+m23p23m22+m23p222m2311110],ˆKA=[0a1a2a3a10a4a5a2a40a6a3a5a60],ˆQ=12ˆI+ˆKAˆA.

      (73)
    • 1.   Deriving the recurrence relation
    • Taking B=1x4 in (39), we get zi=Qijxjx4. Taking this relation into our IBP identities (32), we get

      4i=1dΠ(4){ziFλ0xm1xn2xq3xλ4+14}+δ3=0,

      (74)

      for which we deal with the boundary δ3 term later. After expanding the first term, we get

      cm,n,qiλ0;m,n,q+cm+1,n,qiλ0;m+1,n,q+cm+1,n,q1iλ0;m+1,n,q1+cm+1,n1,qiλ0;m+1,n1,q+cm1,n+1,qiλ0;m1,q+1,q+cm,n+1,q1iλ0;m,n+1,q1+cm,n+1,qiλ0;m,n+1,q+cm,n,q+1iλ0;m,n,q+1+cm,n1,q+1iλ0;m,n1,q+1+cm1,n,q+1iλ0;m1,n,q+1+cm1,n,qiλ0;m1,n,q+cm,n1,qim,n1,q+cm,n,q1iλ0;m,n,q1+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,q1=qQ31,cm+1,n1,q=nQ21,cm,n,q1=qQ34,cm1,n+1,q=mQ12,cm,n+1,q1=qQ32,cm,n+1,q=λ4Q42,cm,n,q+1=λ4Q43,cm,n1,q+1=nQ23,cm1,n,q+1=mQ13,cm1,n,q=mQ14,cm,n1,q=nQ24.

      (76)

      Now, we can choose particular values for our six parameters a1, a2, a3, a4, a5, and a6 to let the coefficients cm+1,n,q, cm+1,n,q1, cm+1,n1,q, cm1,n+1,q, cm,n+1,q, and cm,n+1,q be zero. The solutions are

      a2=a1A21A42A22A41A31A42A32A41=a1(m21+m22+p21)m21+m22+2(p1p2)+p21,a3=a1(A21A32A22A31)A31A42A32A41=a1(m21m22p21)(m22+m23p22)m21+m22+2(p1p2)+p212a1m22,a4=a1(A11A42A12A41)A31A42A32A41=a1(m21m22+p21)m21+m22+2(p1p2)+p21,a5=a1(A11A32A12A31)A31A42A32A41=a1(m21m22+p21)(m21+m232(p1p2)p21p22)m21+m22+2(p1p2)+p21+2a1m21,a6=a1(A11A22A12A21)A31A42A32A41=a1(m412m21(m22+p21)+(m22p21)2)m21+m22+2(p1p2)+p21.

      (77)

      Then, the matrix ˆQ becomes

      ˆQr=1ΔA[12ΔA0a1|˜A14|a1|˜A13|012ΔAa1|˜A24|a1|˜A23|0012ΔA+a1|˜A34|a1|˜A33|00a1|˜A44|12ΔAa1|˜A43|],ΔA=Det[A31A32A41A42]=A31A42A32A41.

      After this, we obtain the reduced IBP relation, where only the propagator D3=(l+p3)2m23 has one increasing power,

      cm,n,qiλ0;m,n,q+cm,n,q+1iλ0;m,n,q+1+cm,n1,q+1iλ0;m,n1,q+1+cm1,n,q+1iλ0;m1,n,q+1+cm1,n,qiλ0;m1,n,q+cm,n1,qiλ0;m,n1,q+cm,n,q1iλ0;m,n,q1+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=a1λ4A31A42A32A41|˜A44|,cm,n1;q+1=nQ23;r=a1nA31A42A32A41|˜A24|,cm1,n,q+1=mQ13;r=a1mA31A42A32A41|˜A14|,cm1,n,q=mQ14;ra1mA31A42A32A41|˜A13|,cm,n1,q=nQ24;r=a1nA31A42A32A41|˜A23|,cm,n,q1=qQ34;r=a1qA31A42A23A41|˜A33|,

      (79)

      where the subscript r in δ3;r and Qij;r indicates that the parameters a2 to a6 should be replaced by (77).

      The reduction in the boundary δ3 part: Similar to the bubble situation, inserting the value of zi into the δ3 part, we obtain

      δ3;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, and iλ0,1,n,q contribute to the sub-topology of the triangle, that is, the bubble.

    • 2.   Triangle example: I3(1,1,2)
    • Now, we apply the complete recurrence relation to the example I3(1,1,2). Setting m=n=q=0 in (78), we obtain

      c0,0,0iλ0,0,0,0+c0,0,1iλ0,0,0,1+δ3;000=0,

      (81)

      with the coefficients

      c0,0,1=λ4Q43;r=1m21+m22+2(p1p2)+p21×{2a1(D4)(m41p222m21(m22((p1p2)+p22)m23(p1p2)+p22((p1p2)+p21))+m42(2(p1p2)+p21+p22)+m22(2(p1p2)(2(p1p2)+p21+p22)2m23((p1p2)+p21))+p21(m432m23((p1p2)+p22)+p22(2(p1p2)+p21+p22)))},c0,0,0=D2+Q11;r+Q22;r+Q33;r+(D3)Q44;r=2a1(D4)(m21(p1p2)m22((p1p2)+p21)+p21(m23(p1p2)p22))m21+m22+2(p1p2)+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 becomes

      δp;000δp;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)=c3111I3(1,1,1)+c3110I3(1,1,0)+c3101I3(1,0,1)+c3011I3(0,1,1)c3210I3(2,1,0)+c3201I3(2,0,1)+c3120I3(1,2,0)+c3021I3(0,2,1)c3102I3(1,0,2)+c3012I3(0,1,2),

      (84)

      with the coefficients

      c3111=c0,0,0Γ(D3)c0,0,1Γ(D4),c3110=Q34;rΓ(D2)c0,0,1Γ(D4),c3101=Q24;rΓ(D2)c0,0,1Γ(D4),c3011=Q14;rΓ(D2)c0,0,1Γ(D4),c3210=Q31;rΓ(D3)c0,0,1Γ(D4),c3201=Q21;rΓ(D3)c0,0,1Γ(D4),c3021=Q12;rΓ(D3)c0,0,1Γ(D4),c3120=Q32;rΓ(D3)c0,0,1Γ(D4),c3102=Q23;rΓ(D3)c0,0,1Γ(D4),c3012=Q13;rΓ(D3)c0,0,1Γ(D4).

      (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)=c33I3(1,1,1)+c32;ˉ3I3(1,1,0)+c32;ˉ2I3(1,0,1)+c32;ˉ1I3(0,1,1)+c31;ˉ2ˉ3I3(1,0,0)+c31;ˉ1ˉ3I3(0,1,0)+c31;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.

    • 3.   General case in triangles
    • 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 propagator Di. Now, we can use eqi with i=1,2,3 to calculate the general case for triangles. Let us denote

      eq1:(a1+1++a1+31+3+a1+21+2+a33+a22+a11+a0)iλ0,m,n,q+δ3;r,eq1=0,eq2:(b2+2++b2+32+3+b12+12++b33+b22+b11+b0)iλ0,m,n,q+δ3;r,eq2=0,eq3:(c3+3++c23+23++c13+13++c33+c22+c11+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 setting m=0, n=1, and q=2 in eq1, we can reduce I3(2,2,3) to I3(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, setting m=0, n=0, and q=2 in eq1, we can reduce I3(2,1,3) to I3(1,1,2), I3(1,1,3), and I3(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 from I3(2,2,3) to all scalar bases.

    • D.   Box case

    • 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=l2m21,D2=(lp1)2m22,D3=(lp1p2)2m23,D4=(l+p4)2m24.

      (89)

      The parametric form of I4(n1+1,n2+1,n3+1,n4+1) can be written as

      I4(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(xj1),λ0=D2λ5=5n1n2n3n42λ0=D5n1n2n3n4,

      (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+m22p21)x1x2+[m21+m23(p1+p2)2]x1x3+[m21+m24(p1+p2+p3)2]x1x4,+(m22+m23p22)x2x3+[m22+m24(p2+p3)2]x2x4+(m23+m24p23)x3x4,F(x)=U(x)x5+f(x)=m21x21+m22x22+m23x23+m24x24+(m21+m22p21)x1x2+[m21+m23(p1+p2)2]x1x3+[m21+m24(p1+p2+p3)2]x1x4+[m22+m23p22]x2x3+[m22+m24(p2+p3)2]x2x4+[m23+m24p23]x3x4+x1x5+x2x5+x3x5+x4x5=(x1+x2+x3+x4)(m21x1+m22x2+m23x3+m24x4+x5)x1x2p21x1x3(p1+p2)2x1x4(p1+p2+p3)2x2x3p22x2x4(p2+p3)2x3x4p23.

      (92)

      Now, the matrices are given by

      ˆA=[2m21m21+m22p21m21+m23p212m21+m24p2131m21+m22p212m22m22+m23p22m22+m24p2231m21+m23p212m22+m23p222m23m23+m24p231m21+m24p213m22+m24p223m23+m24p232m24111110],KA=[0a1a2a3a4a10a5a6a7a2a50a8a9a3a6a80a10a4a7a9a100],

      (93)

      where pijpi+pi+1pj.

    • 1.   Deriving the recurrence relation
    • Taking B=1x5 in (39), we get

      {cn1+1,n2,n2,n41++cn1+1,n2,n3,n41+4+cn1+1,n2,n31,n41+3+cn1+1,n21,n3,n41+2+cn1,n2+1,n3,n42++cn1,n2+1,n3,n412+4+cn1,n2+1,n31,n42+3+cn11,n2+1,n3,n42+1+cn1,n2,n3+1,n43++cn1,n2,n3+1,n413+4+cn1,n21,n3+1,n43+2+cn11,n2,n3+1,n43+1+cn1,n2,n3,n4+14++cn1,n2,n31,n4+14+3+cn1,n21,n3,n4+14+2+cn11,n2,n3,n4+14+1+cn1,n2,n3,n414+cn1,n2,n31,n43+cn1,n21,n3,n42+cn11,n2,n3,n41+cn1,n2,n3,n4}in1,n2,n3,n4+δ4=0,

      (94)

      where

      j+in1njnk=in1nj+1nk,jin1njnk=in1nj1nk.

      (95)

      Similarly, we can choose particular values of the parameters a2 to a10 with a free a1 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 of ˆA.

      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 the i,jth rows and k,lth columns. Then, the matrix ˆQ becomes

      ˆQr=1ΔBox[12ΔBox00a1|˜A15|a1|˜A14|012ΔBox0a1|˜A25|a1|˜A24|0012ΔBoxa1|˜A35|a1|˜A34|00012ΔBox+a1|˜A45|a1|˜A44|000a1|˜A55|12ΔBoxa1|˜A54|].

      We then obtain the simplified recurrence relation

      cn1,n2,n3,n4+1in1,n2,n3,n4+1+cn1,n2,n31,n4+1in1,n2,n31,n4+1+cn1,n21,n3,n4+1in1,n21,n3,n4+1+cn11,n2,n3,n4in11,n2,n3,n4+cn1,n2,n3,n41in1,n2,n3,n41+cn1,n2,n31,n4in1,n2,n31,n4+cn1,n21,n3,n4in1,n21,n3,n4+cn11,n2,n3,n4in11,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 the δ4 term and take the values of the parameters a2 to a10 into the δ4 part. Subsequently, we get

      δ4;r=δn1+1,0Q11;ri1,n2,n3,n4+δn1,0Q12;ri1,n2+1,n3,n4+δn1,0Q13;ri1,n2,n3+1,n4+δn1,0Q14;ri1,n2,n3,n4+1+δn1,0Q15;ri1,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 to a10.

    • 2.   Example: I4(1,1,1,2)
    • Now, we can use recurrence relation (96) to calculate the example I4(1,1,1,2). Letting n1=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;0000δ4;r|n1=n2=n3=n4=0. Translating to I, we obtain the following result:

      I4(1,1,1,2)=c41111I4(1,1,1,1)+c41110I4(1,1,1,0)+c41101I4(1,1,0,1)+c41011I4(1,0,1,1)+c40111I4(0,1,1,1)+c42110I4(2,1,1,0)+c42101I4(2,1,0,1)+c42011I4(2,0,1,1)+c41210I4(1,2,1,0)+c41201I4(1,2,0,1)+c40211I4(0,2,1,1)+c41120I4(1,1,2,0)+c41021I4(1,0,2,1)+c40121I4(0,1,2,1)+c41102I4(1,1,0,2)+c41012I4(1,0,1,2)+c40112I4(0,1,1,2),

      (99)

      with the coefficients

      c41111=c0,0,0,0c0,0,0,1(D5)=TrˆQij;r+(D5)Q55;rD2Q54;r,c40111=Q15;rΓ(D3)c0,0,0,1Γ(D5),c41011=Q25;rΓ(D3)c0,0,0,1Γ(D5),c41101=Q35;rΓ(D3)c0,0,0,1Γ(D5),c41110=Q45;rΓ(D3)c0,0,0,1Γ(D5),c40211=Q12;rΓ(D4)c0,0,0,1Γ(D5),c40121=Q13;rΓ(D4)c0,0,0,1Γ(D5),c40112=Q14;rΓ(D4)c0,0,0,1Γ(D5),c42011=Q21;rΓ(D4)c0,0,0,1Γ(D5),c41021=Q23;rΓ(D4)c0,0,0,1Γ(D5),c41012=Q24;rΓ(D4)c0,0,0,1Γ(D5),c42101=Q31;rΓ(D4)c0,0,0,1Γ(D5),c41201=Q32;rΓ(D4)c0,0,0,1Γ(D5),c41102=Q34;rΓ(D4)c0,0,0,1Γ(D5),c42110=Q41;rΓ(D4)c0,0,0,1Γ(D5),c41210=Q42;rΓ(D4)c0,0,0,1Γ(D5),c41120=Q43;rΓ(D4)c0,0,0,1Γ(D5).

      (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)=c44I4(1,1,1,1)+c43;ˉ1I4(0,1,1,1)+c43;ˉ2I4(1,0,1,1)+c43;ˉ3I4(1,1,0,1)+c43;ˉ4I4(1,1,1,0)+c42;ˉ1ˉ2I4(0,0,1,1)+c42;ˉ1ˉ3I4(0,1,0,1)+c42;ˉ1ˉ4I4(0,1,1,0)+c42;ˉ2ˉ3I4(1,0,0,1)+c42;ˉ2ˉ4I4(1,0,1,0)+c42;ˉ3ˉ4I4(1,1,0,0)+c41;D1I4(1,0,0,0)+c41;D2I4(0,1,0,0)+c41;D3I4(0,0,1,0)+c41;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.

    • E.   Pentagon case

    • 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=l2m21,D2=(lp1)2m22,D3=(lp1p2)2m23,D4=(lp1p2p3)2m24,D5=(l+p5)2m25.

      (103)

      The parametric form of I5(n1+1,n2+1,n3+1,n4+1,n5+1) can be written as

      I5(n1+1,n2+1,n3+1,n4+1,n5+1)=i(1)5+n1+n2+n3+n4+n5Γ(λ0)5i=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(xj1),λ0=D2,λ6=(D6)n1n2n3n4n5,

      (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)x1x2p21x1x3p212x1x4p213x1x5p214x2x3p22x2x4p223x2x5p224x3x4p23x3x5p234x4x5p24,F(x)=(x1+x2+x3+x4+x5)(m21x1+m22x2+m23x3+m24x4+m25x5+x6)x1x2p21x1x3p212x1x4p213x1x5p214x2x3p22x2x4p223x2x5p224x3x4p23x3x5p234x4x5p24,

      (106)

      where pijpi+pi+1+pj1+pj. Now the matrix are given by

      ˆA=[2m21m21+m22p21m21+m23p212m21+m24p213m21+m25p211m21+m22p212m22m22+m23p22m22+m24p223m22+m25p2241m21+m23p212m22+m23p2232m23m23+m24p23m23+m25p2341m21+m24p213m22+m24p223m23+m24p232m24m24+m25p241m21+m25p21m22+m25p224m23+m25p234m24+m25p242m251111110],

      (107)

      ˆKA=[0a1a2a3a4a5a10a6a7a8a9a2a60a10a11a12a3a7a100a13a14a4a8a11a130a15a5a9a12a14a150].

      (107)

      Taking B=1/x6, and inserting zi into the IBP identities,

      6i=1xi{ziFλ0xn11xn22xn33xn44xn55xλ6+16}+δ5=0,

      (108)

      where δ5 is given by

      δ5=5i=1δλi,0dΠ(5){ziFλ0xn11xn22xn33xn44xn55xλ6+16}|xi=0.

      (109)
    • 1.   Deriving the recurrence relation
    • Similar to the previous subsections, by expanding the IBP relation, we get

      {cn1+1,n2,n3,n4,n51++cn1+1,n21,n3,n4,n51+2+cn1+1,n2,n31,n4,n51+3+cn1+1,n2,n3,n41,n51+4+cn1+1,n2,n3,n4,n511+5+cn1,n2+1,n3,n4,n52++cn11,n2+1,n3,n4,n512++cn1,n2+1,n31,n4,n52+3+cn1,n2+1,n3,n41,n52+4+cn1,n2+1,n3,n4,n512+5+cn1,n2,n3+1,n4,n53++cn11,n2,n3+1,n4,n513++cn1,n21,n3+1,n4,n523++cn1,n2,n3+1,n41,n53+4+cn1,n2,n3+1,n4,n513+5+cn1,n2,n3,n4+1,n54++cn11,n2,n3,n4+1,n514++cn1,n21,n3,n4+1,n524++cn1,n2,n31,n4+1,n534++cn1,n2,n3,n4+1,n514+5+cn1,n2,n3,n4,n5+15++cn11,n2,n3,n4,n5+115++cn1,n21,n3,n4,n5+125++cn1,n2,n31,n4,n5+135++cn1,n2,n3,n41,n5+145++cn11,n2,n3,n4,n51+cn1,n21,n3,n4,n52+cn1,n2,n31,n4,n53+cn1,n2,n3,n41,n54+cn1,n2,n3,n4,n515+cn1,n2,n3,n4,n5}in1,n2,n3,n4,n5+δ5=0.

      (110)

      We can choose particular values for parameters a2 to a15 to ensure the coefficients of the first three line of (110) equal zero. The solution is

      a2=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++cn11,n2,n3,n4,n5+1;r15++cn1,n21,n3,n4,n5+1;r25++cn1,n2,n31,n4,n5+1;r35++cn1,n2,n3,n41,n5+1;r45+cn11,n2,n3,n4,n5;r1+cn1,n21,n3,n4,n5;r2+cn1,n2,n31,n4,n5;r3+cn1,n2,n3,n41,n5;r4+cn1,n2,n3,n4,n51;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,n5iλ0;n1,ni+1,n5,iiλ0;n1,n2,n3,n4,n5iλ0;n1,ni1,n5,

      (113)

      with the coefficients

      c0,0,0,0,1=Q65;rλ6,c1,0,0,0,1=n1Q15;r,c0,1,0,0,1=n2Q25;r,c0,0,1,0,1=n3Q35;r,c0,0,0,1,1=n4Q45;r,c1,0,0,0,0=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+((D6))Q66;rrD2+n1Q11;r+n2Q22;r+n3Q33;r+n4Q44;r+n5Q55;r,

      (114)

      while the matrix ˆQ becomes

      ˆQr=1Δpen[12Δpen000a1|˜A1,6|a1|˜A1,5|012Δpen00a1|˜A2,6|a1|˜A2,5|0012Δpen0a1|˜A3,6|a1|˜A3,5|00012Δpena1|˜A4,6|a1|˜A4,5|000012Δpen+a1|˜A5,6|a1|˜A5,5|0000a1|˜A6,6|12Δpena1|˜A6,5|].

    • F.   Reducing the δ5 term

    • Similar to the former situation, the δ6;r term is given by

      δ5;r=Q11;rδn1,1i1,n2,n3,n4,n5+Q12;rδn1,0i1,n2+1,n3,n4,n5+Q13;rδn1,0i1,n2,n3+1,n4,n5+Q14;rδn1,0i1,n2,n3,n4+1,n5+Q15;rδn1,0i1,n2,n3,n4,n5+1+Q16;rδn1,0i1,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)
    • G.   Example: I5(1,1,1,1,2)

    • 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;00000δ5;r|n1=n2=n3=n4=n5=0.

      Comparing them with our scalar basis, we have the result

      I5(1,1,1,1,2)=c55I5(1,1,1,1,1)+c501111I4(0,1,1,1,1)+c510111I5(1,0,1,1,1)+c511011I5(1,1,0,1,1)+c511101I5(1,1,1,0,1)+c511110I5(1,1,1,1,0)+c520111I5(2,0,1,1,1)+c521011I5(2,1,0,1,1)+c521101I5(2,1,1,0,1)

      +c521110I5(2,1,1,1,0)+c502111I5(0,2,1,1,1)+c512011I5(1,2,0,1,1)+c512101I5(1,2,1,0,1)+c512110I5(1,2,1,1,0)+c501211I5(0,1,2,1,1)+c510211I5(1,0,2,1,1)+c511201I5(1,1,2,0,1)+c511210I5(1,1,2,1,0)+c501121I5(0,1,1,2,1)+c510121I5(1,0,1,2,1)+c511021I5(1,1,0,2,1)+c511120I5(1,1,1,2,0)+c501112I5(0,1,1,1,2)+c510112I5(1,0,1,1,2)+c511012I5(1,1,0,1,2)+c511102I5(1,1,1,0,2),

      (117)

      with the coefficients

      c55=(D6)c0,0,0,0,0c0,0,0,0,1,c501111=(D6)(5D)Q16;rc0,0,0,0,1,c54;10111=(D6)(5D)Q26;rc0,0,0,0,1,c54;11011=(D6)(5D)Q36;rc0,0,0,0,1,c54;11101=(D6)(5D)Q46;rc0,0,0,0,1,c54;11110=(D6)(5D)Q56;rc0,0,0,0,1,c520111=(D6)Q21;rc0,0,0,0,1,c521011=(D6)Q31;rc0,0,0,0,1,c521101=(D6)Q41;rc0,0,0,0,1,c521110=(D6)Q51;rc0,0,0,0,1,c502111=(D6)Q12;rc0,0,0,0,1,c512011=(D6)Q32;rc0,0,0,0,1,c512101=(D6)Q42;rc0,0,0,0,1,c512110=(D6)Q52;rc0,0,0,0,1,c501211=(D6)Q13;rc0,0,0,0,1,c510211=(D6)Q23;rc0,0,0,0,1,c511201=(D6)Q43;rc0,0,0,0,1,c511210=(D6)Q53;rc0,0,0,0,1,c501121=(D6)Q14;rc0,0,0,0,1,c510121=(D6)Q24;rc0,0,0,0,1,c511021=(D6)Q34;rc0,0,0,0,1,c511120=(D6)Q54;rc0,0,0,0,1,c501112=(D6)Q15;rc0,0,0,0,1,c510112=(D6)Q25;rc0,0,0,0,1,c511012=(D6)Q35;rc0,0,0,0,1,c511102=(D6)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.

    IV.   ANALYTIC RESULTS OF THE COEFFICIENTS
    • The analytic results are provided in the Mathematica notebooks, which are publicly available at https://github.com/Wanghongbin123/oneloop_parametric.

    V.   SUMMARY AND FURTHER DISCUSSION
    • 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 solve x_i using {{{\partial }} F}/{{{\partial }} x_i} with several free parameters. Then, combining all the IBP identities with a particular factor z_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 form x_{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 function F(x) is of degree L+1 , where L is the number of loops. For the high loop case, we should consider how to construct the coefficients z_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.

    ACKNOWLEDGMENTS
    • I would like to thank Bo Feng for the inspiring discussion and guidance.

Reference (32)

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