Reduction of one-loop integrals with higher poles by unitarity cut method

Unitarity cut method has been proved to be very useful in the computation of one-loop integrals. In this paper, we generalize the method to the situation where the powers of propagators in the denominator are larger than one in general. We show how to use the trick of differentiation over masses to translate the problem to the integrals where all powers are just one. Then by using the unitarity cut method, we can find the wanted reduction coefficients of all basis except the tadpole. Using this method, we calculate the reduction of scalar bubble, scalar triangle, scalar box and scalar pentagon with general power of propagators.

method to do the reduction. Thus it is naturally to ask if the unitarity cut method can be applied to these more general situations.
In this paper, we consider the reduction of one-loop integrals with general pole structures, i.e., propagators could have general power. The general integration will be given by where N [ℓ] is an arbitrary polynomial function of ℓ. According to the PV-reduction method [17], we can decompose where c t 's are the rational functions and I t are scalar basis of tadpole, bubble, triangle, box and pentagon. Let us consider the c i 's first. The data entering the integral (1.1) are the external momenta, the masses and the polarization vectors (the coupling constants are overall factor, so can be dropped in the discussion).
The c i should be the rational functions of these data with Lorentz invariant contractions. The idea of unitarity cut method is to compare the imaginary part of both sides of (1.2). Since c t 's are rational functions without imaginary contributions, we have Different master integrals have different analytic structures for the imaginary part (we will call them as the "signature"), thus if we can analytically computer the left hand side, we can do the spliting according to the analytic signature of each master integral and find expansion coefficients c t at the right hand side. With this thought, the key of reduction by unitarity cut method is to compute the imaginary part of M [ℓ]. When all a j = 1, the computation of left hand side (1.3) is transformed to the reduced phase space integration with double cuts, which we know how to do it as reviewed in the first paragraph of this section. For general a i , we use following trick to solve the problem. Noticing that 2 the computation of the imaginary part of an one-loop integral is transformed to the reduced phase space integration with double cut for the case with all a i = 1. More explicitly, let us separate both sides of (1.4) to the real and imaginary part, we have Since the η i 's are defined to take real values, we have thus we get For general N [ℓ] in (1.4), we know the expansion (1.8) and the action of d dη will act on both c t and Im(I t [ℓ]). Since the function c t 's have been given in [9,10,11], the unknown piece is the action of d dη on Im(I t [ℓ]) and its expansion. In another words, we just need to consider the reduction of general power with N [ℓ] = 1 in (1.1) for n ≤ 5.
The plan of the paper is following. In the section two, we consider the reduction of bubbles with higher poles. We establish the general recurrence relation and check our results with some examples. Same method has been applied to triangles, boxes and pentagons in the section three, four and five. A brief conclusion is given in the section six.

Bubble
For bubble topology, let us define to be the general scalar bubble integral with higher power of propagators. The master integral of bubble is the case n = m = 1 and for this special case, sometimes we just write it as I 2 [K; M 1 , M 2 ] or just I 2 for simplicity. The reduction of I 2 (a, b)[K; M 1 , M 2 ] will be the following expansion where c 2→1;i means that when reducing the bubble to the tadpole, the i-th propagator has been kept. The tadpole I 1;i [M i ] should be written in the standard form with the proper momentum shifting, i.e., Expansion coefficients in (2.2) can be found by various methods, for example, the IBP method. However, in this paper, we will try to use the unitarity cut method to find expansion coefficients. When we use the unitarity cut method, the tadpole can not be detected, thus tadpole coefficients can not be found by this way. Although we will not consider the tadpole coefficients in this paper, we want to point out that some efforts have been done to fill the gap by using the single cut [19,20,21]. The unitarity cut of I 2 (1, 1) is given by (see references [7,9,8] which is the Landau surface of bubble of the first type of singularities. For the later convenience, let us define Thus by comparing (2.6) with (2.4) we see that C[ . The Bub (n) is well defined for n ≥ 0 and it is easy to derive a recursion relation by integration-by-part Solving it we get We find that when doing the reduction for triangles and boxes, we will meet the form Bub (n) with the negative integer n. For this case, we can use (2.7) to analytically continue from positive n to negative n. For example, using n = 0 in (2.7), we get Having above preparation, now we consider the reduction of I 2 (a, b). According to our general idea in To get a better idea, let us start with Before doing this, we rewrite ∆ as In the reference [9] when we do the general one-loop unitarity cut phase space integration, an overall factor ∆[K,M 1 , has been neglected in later computation (see the equation after eq. (9)). This is fine for the work in [9], but since it depends on the masses, it is crucial for current computation and we must include it back. 4 From the discussion of (1.4) one can see that the role of η is identical with m 2 . For scalar basis, m does not appear in other places, so we can take the derivative of m 2 instead of η without making any mistake.
where each factor is linear in M 2 1 . It is easy to get the n-th derivative of M 2 1 is given by where we have defined (2.14) Using above result, we can find the reduction coefficient of I 2 (n + 1, 1) by (1.7) since by the result (2.13). For the general bubble coefficients of the reduction of I 2 (n + 1, m + 1), using the relation we get reduction coefficient as The general analytic expression of c 2→2 (n + 1, m + 1) will be complicated to write down. Now we check the result (2.18). The first check is by the symmetry. Noticing that when shifting p → p + K, the (2.1) becomes by the IBP method, one can derive the relation and when a 2 = 1 we have To compare with our calculation, we should take M 1 , M 2 → 0, K → q and d → 4 − 2ǫ after all derivatives related to M 2 1 , M 2 2 having been done. We have used the Mathematica to check the massless case of our general result (2.18). The third check is that we have used the LiteRed [23] to explicitly calculate some examples with nonzero M 1 , M 2 and we do find the match.

Recurrence relation
Like the IBP relation, the idea of (1.7) can be used to establish the recurrence relation for reduction coefficients. Let us start from the reduction of I 2 (a, 1). Noticing that Relation (2.27) tell us that staring from c 2→2 (2, 1), we can write down all c 2→2 (a, 1). The same idea can be used to write down the recurrence relation of general c 2→2 (a, b). Noticing that (2.28) we get immediately the recurrence relation where c 2→2 (1, m) can be obtained from c 2→2 (m, 1) by proper replacement as given in (2.20). Result (2.29) shows that knowing the coefficient c 2→2 (2, 1) is enough for general reduction coefficients. The similar idea can be used to the reduction for triangle, box and pentagon as will be shown later.

Triangle
For triangle topology, let us define 6 For the special case, i.e., n 1 = n 2 = n 3 = 1, we get the familiar scalar triangle basis, which for simplicity, we will denote as The unitarity cut of I 3 with the cut momentum K 1 has been given by [9] C( where Z is given by The T ri (n) (Z) has the recurrence relation [8,9] T ri (n) which will be useful later.
For general n i , the I 3 (n 1 , n 2 , n 3 ) will be reduced to the sum of one triangle I 3 , three bubbles I 2;ī , i = 1, 2, 3 (whereī means to set n i = 0 in (3.1) and other n j 's one) and three tadpoles I 1;i , i = 1, 2, 3 (where i means to set n i = 1 in (3.1) and other n j 's zero). In other words, we will have the expansion where for simplicity we have dropped the dependence over K, M of coefficients c's. We want to emphasize that when writing down I 2;ī , we need to do the proper momentum shifting to reach the standard form (2.1). More explicitly, we will have and by using the shifting p → p + K 1 , and finally by using the shifting p → p + K 1 + K 2 . For later convenience, let us give a more general bubbles coming from the triangle reduction (3.1) When we use the unitarity cut method to find the reduction coefficients, a given cut can only detect these coefficients of basis which have the corresponding cut. For each cut, it can detect the triangle coefficient and one bubble coefficient, while all tadpole coefficients can not be found by this way. For triangle, there are three different cuts, i.e, K 1 , K 2 , K 3 . the coefficient of bubble I 2;3 can only be detected by the cut with momentum K 1 and similarly bubble coefficients of I 2;1 and I 2;2 by the cut K 2 and K 3 respectively. Thus using them all, we can get reduction coefficients of triangle and bubbles. Furthermore, there are overlaps of detected reduction coefficients between different cuts. For the triangle case, all three cuts can detect the same triangle coefficient, so we can use the overlap as the cross check.

Recurrence relation
Before going to explicit calculation, let us establish some recurrence relations like the one (2.27) and (2.29) in the previous section. To prepare for this task, we discuss some symmetric properties of the integral (3.1) first. By shifting the momentum p, we can rewrite (3.1) to different forms, for example, by shifting p → p + K 1 we get (3.14) and similarly for the shifting p → p + K 1 + K 2 . Above relations (3.14) by shifting can be summarized as the cyclic symmetry Z 3 of three ordered lists We can also consider the variable changing p → −p in (3.1) to get 8 i.e., the measure is invariant under the reflection.
which can be summarized as the reflection symmetry Z 2 of three ordered lists When combining g 2 , g 3 together, we generate the permutation group S 3 . Using the symmetric property, we can connect the reduction of one integral to another integral. For example, we know the expansion of and we want to calculate the expansion of Using (3.14), we have Similarly, for bubble part we have (3.20) One simple consequence of above symmetry property is that if we know the expansion of I 3 (1, 1, 2), we know also I 3 (1, 2, 1) and I 3 (2, 1, 1).
Now we show how to use I 3 (1, 1, 2), I 3 (1, 2, 1) and I 3 (2, 1, 1) to get the expansion of general I 3 (n 1 , n 2 , n 3 ) using only differentiation. Let us start from I 3 (1, 1, n 3 ) first. It is easy to see that where in the last line, since the I 2;3 does not depend on m 1 , the action of d dm 2 1 is zero (see (3.10), (3.11) and (3.12)). Now, using the expansion of I 3 (1, 1, 2), I 2 (1, 2) and I 2 (2, 1), we get immediately the recurrence relation Knowing (3.22), it is easy to use (3.15) and (3.17) to get reduction coefficients for I 3 (1, n, 1) and I 3 (n, 1, 1) as shown in (3.18). One point we want to emphasize is that since g 2 , g 3 acts on K, M also, the kinematic dependence of K, M in (3.22) should be carefully identified although for simplicity we have not written them down explicitly.
Knowing the reduction of I 3 (1, 1, n 3 ), now we consider the case I 3 (1, n 2 , n 3 ). Similarly to (3.21), we have thus we can get the recurrence relation for reduction coefficients So if we have known the expansion coefficients of I 3 (1, n 2 − 1, n 3 ) to the basis I 3 and I 2 , we can derive the expansion of I 3 (1, n 2 , n 3 ).
Finally, for I 3 (n 1 , n 2 , n 3 ) with n 1 > 1, similar action lead to the recurrence relation From the recurrence relation in previous subsection, we see that all computations have been reduced to the reduction of I 3 (1, 1, 2). Although there are other methods to reduce I 3 (1, 1, 2) such as the IBP method. In this part, we show how to use unitarity cut method to fulfill the task. Using our idea, we can write To get all reduction coefficients, we need to calculate all three cuts. However, different cuts can be related using the symmetry discussed in previous subsection, thus we will focus on only one cut.

Cut K 1
For this cut, the RHS is given by where we have used the fact that only Z contains the m 1 . Carrying out the derivative, we get Among two terms of ∂ ∂Z T ri (0) (Z), by comparing with (3.3), the second term is For the first term, noting that Putting all together, we get we get the reduction coefficients as where for simplicity we have not expanded Z further (see (3.4)). We want to remark that the coefficients c 3→3;K 1 and c 3→2;3;K 1 are nothing, but the c 3→3 and c 3→2;3 is the reduction I 3 (1, 1, 2) = c 3→3 (1, 1, 2)I 3 + c 3→2;3 (1, 1, 2)I 3;3 + · · · · · · (3.36) The reason that we have added subscript K 1 in (3.34) and (3.35) is to emphasize that these two expressions are calculated using the cut K 1 .

Cut K 2
No we consider the cut K 2 . The use the cut result (3.2), we need to rewrite the form (3.1) into the standard form for the cut K 2 by shifting the integral momentum p → p + K 1 , thus it becomes For the case n 3 = 2, n 1 = n 2 = 1 , there are two terms containing the parameter M 2 in the Cut K 1 (I 3 (1, 1, 2)).

Cut K 3
Now we consider the cut K 3 . Again we need to rewrite the form (3.1) into the standard form for the cut K 3 by shifting the integral momentum p → p + K 1 + K 2 . With similar argument we will have . The calculation is similar to the previous subsection. After simplification, we have with the coefficients After doing the permutation g −1 3 on the coefficients above, we get the c 3→3 and c 3→2;2 in this cut. Again, one can check the c 3→3 is the same with c 3→3;K 1 in (3.34).
For the case I 3 (1, 1, 2) there is another consistent check we can do. Noticing that when we do following changing variable p → −p + K 1 , the (3.1) becomes to Thus for (n 1 , n 2 , n 3 ) = (1, 1, 2), there is the symmetry among the reduction coefficients, i.e., under the the permutation K 2 ←→ K 3 , M 1 ←→ M 2 , we should have c 3→3 invariant and c 3→2;2 ↔ c 3→2;1 . This can be easily checked by MATHEMATICA for our results.

A short summary
In the subsection, we have reduced the I 3 (1,1,2) to triangle I 3 and bubble I 2 . After taking three different cuts, we get the all needed triangle and bubble coefficients of the reduction of I 3 (1, 1, 2) where the · · · represents the tadpoles neglected in the whole paper. The coefficients in (3.46) are The result is confirmed with IBP method using LiteRed [23]. We have also carried out the reduction of I 3 (1, 2, 1), I 3 (2, 1, 1) and I 3 (1, 2, 2) using the method laid out in the paper and found perfect match with the IBP method.

Box
For the box topology, we define for later convenience, where we have labeled the cut momentum as K and the masses of two cut propagators as M 1 and M 2 respectively. For the other two momenta we denote them as P 1 and P 2 respectively. For example, if we choose K 1 as the cut momentum, we will have P 1 = K 1 + K 2 and P 2 = −K 4 , but if the cut momentum is K 1 + K 2 , we will have P 1 = K 1 , P 2 = K 1 + K 2 + K 3 . The special case with n i = 1, i = 1, 2, 3, 4 gives the scalar basis, which we will denote as I 4 . The unitarity cut of I 4 with the cut momentum K has been given by [9] C( with the parameters ∆ in (2.5) and 9 With the definition we can write C(I 4 ) as and there is also a recurrence relation of Box (n) , which is given by [8,9] with the parameter To be calar, we list six possible cuts of a box, with K 1 , K 2 , K 3 , K 4 in clockwise ordering. And there will be two cut triangles related to each given cut momentum K of box: Symmetry analysis: Similar to the case of triangle, there are some symmetries of box by momentum shifting and reflection. Let us define two generatorŝ Furthermore, using the same idea in the subsection 3.1 we can write down similar recurrence relation of reduction coefficients for general I 4 (n 1 , n 2 , n 3 , n 4 ) using the expansion of I 4 (2, 1, 1, 1), I 4 (1, 2, 1, 1), I 4 (1, 1, 2, 1) and I 4 (1, 1, 1, 2). However, by relation (4.12) and (4.13), all three cases I 4 (2, 1, 1, 1), I 4 (1, 2, 1, 1) and I 4 (1, 1, 2, 1) can be reduced to the reduction of I 4 (1, 1, 1, 2), thus we need only to deal with the reduction of I 4 (1, 1, 1, 2). I 4 (1, 1, 1, 2) Now, we use our method to calculate the reduction coefficients of I 4 (1, 1, 1, 2). According our idea, we can write

Box
Comparing with the calculation of triangle, we need to consider six different cuts: four cuts with nearby propagators and two with opposite propagators. By the symmetry, we could just calculate one for each type of cuts, and get the others by proper permutation. To demonstrate our method, let us show the calculation for the cut K 1 only. The reduction coefficients calculated for I 4 (1, 1, 1, 2) has been checked using the LiteRed [23]. For the cut K 1 , we have K = K 1 , P 1 = K 1 + K 2 and P 2 = −K 4 in (4.2). To calculate ∂ ∂m 2 2 C(I 4 ), since the parameter ∆ and b do not contain the m 2 , we could just write it as To simplify ∂ ∂m 2 2 Box (0) , first we rewrite Box (0) as by partial integration and algebraic separation as where the bubble and two triangles are specified by our cut K. Now we consider the action of ∂ ∂m 2 2 . Since only B and D contains the parameter m 2 , we can write directly where the B ′ Box (−1) , while using using (4.8), we can expand Box (−1) as T ri (0) (Z 2 ) • (2) For the second term, using the trick of splitting terms and the recurrence relation, we get where T ri (0) , we could use result (3.32) and (3.28) to rewrite it as Now collecting all results together for (4.17) we get our final result in the cut K 1 : ∂ ∂m 2 2 I 4 = c 4→4;K 1 I 4 + c 4→3;4;K 1 I 3;4 + c 4→3,3;K 1 I 3;3 I + c 4→2;12;K 1 I 2;12;K 1 + · · · (4.21) where I 3;4 = I 4 (1, 1, 1, 0), I 3;3 = I 4 (1, 1, 0, 1), I 2;12;K 1 = I 4 (1, 1, 0, 0) (4. 22) with the coefficients where C 3;Z 1 , C 3;Z 2 and C 2 given in Eq.(4.20), and Having given details for the cut K 1 , the computation of other cuts will be similar, as shown in the section of triangle. For simplicity we will not list them one by one. All coefficients have been checked using LiteRed [23].

Pentagon
For the pentagon, let us define I 5 (n 1 , n 2 , n 3 , n 4 , n 5 )[K 1 , K 2 , K 3 , K 4 , where the second form is suitable for the discussion of unitarity cut with cut momentum K in various cuts. The master basis of pentagon is given by I 5 ≡ I 5 (1, 1, 1, 1, 1) and the cut part with K is given by [9] C( K (see also (4.5)) and S[Q 3 , Q 2 , Q 1 , K] is a rational function defined as follows The form of C(I 5 ) is like an addition of three dirrefent C(I 4 ), with the common factor √ 1 − u. To simplify further, noticing that there is a common factor (1 − u) 2 between T jik and T 2 , we can define (please notice that T jik = T ijk ) where T ij is independent of u and T 2,re is a liner function of u. Furthermore, with parameters thus A, B, C, D defined in (4.4) for box cut and (5.3) can be simplified as and With above new notations, we can rewrite the expression of C(I 5 ) as We can see that in (5.8) only parameters α i contains m 2 i , so For later convenience, we define three functions thus the C(I 5 ) could be written as Furthermore, using the same idea in the subsection 3.1 we can write down similar recurrence relation for general I 5 (n 1 , n 2 , n 3 , n 4 , n 5 ) using the expansion of I 5 (2, 1, 1, 1, 1), I 5 (1, 2, 1, 1, 1), I 5 (1, 1, 2, 1, 1), I 5 (1, 1, 1, 2, 1) and I 5 (1, 1, 1, 1, 2). However, by relation (5.13) and (5.14), all other four cases can be reduced to the reduction of I 5 (1, 1, 1, 1, 2). To reduce I 5 (1, 1, 1, 1, 2), according to our ideas, we should calculate ∂ ∂m 2 3 I 5 (1, 1, 1, 1, 1) in 5 2 = 10 different cuts. Again, in the main part, we present only the computation of the cut K 1 . For this case, since the analytic checking using LiteRed [23] is too hard, we have checked only numerically.

Cut
For this cut, we will choose the parameters of the second form in (5.1) as K = K 1 , P 1 = K 12 , P 2 = K 123 and P 3 = −K 5 . Since the m 3 is not contained in ( ∆ 4K 2 ) −ǫ , we could just drop the factor ( ∆ 4K 2 ) −ǫ . Furthermore the parameter m 3 is only contained in B, D, H 0 and T 12 , T 13 and T 23 , thus we have we need to calculate these three terms respectively.
To prepare the reduction of three integral L 1 , L 2 , and L 3 , we rewrite pen (n) , which is defined in (5.10), as following To prepare the reduction of three integral L 1 , L 2 , and L 3 , we want to rewrite pen (n) defined in (5.10) in the following by doing the partial integration du n−ǫ = (n − ǫ)u n−1−ǫ du The first line in (5.18) has the form of L 1 in (5.17) since H 0 does not depend on u as defined in (5.3). Now we consider the third line in (5.18). Using algebraic separation it becomes Among these two terms in (5.20), the second term will be canceled by the second line in (5.18). For the first term, using the factorization form of $ ij in (5.19) we get with the coefficients Among three terms in (5.21), the first term is spurious and will be canceled when summing contributions from three pen ij terms. The second and third terms are essentially the triangle part. Putting all coefficients back, we find the third line in (5.18) is given by There are three terms. The first term could be split into two triangles, as we have done before. The second term is a spurious term, and is canceled with the same term in the integral L 2 . And the last term could also be split into three pieces by using (5.23) with n = −1. Putting all together we have Collecting above results for L 1 , L 2 , L 3 we have In (5.38), the last term will be canceled when summing over three ∂ and T ri (0) (Z 2 ) ij + q ij;Z i T ri (0) (Z i ) + q ij;Z j T ri (0) (Z j ) + q ij;2 Bub (0) (5.42) with the coefficients q ij;5 = l ij;5 = −