(2) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). In Sec. III, and in order to justify the combined use of IBFE þ IBP techniques, we will evaluate a simple diagram, the one-loop triangle diagram with one massive propagator with mass m. To simplify the problem even more, we have assumed that two of the external lines are on mass shell with m ¼ 0. In this way, the diagram will be characterized by only two invariants {m2 ; sg. The above mentioned simplifications, as we will see later, do not imply a loss with respect to the general statements which are the goal of this section. The validation of the combined use of IBFE þ IBP will be realized through a calculation of this diagram, using first only the IBFE technique. Later, the same diagram is evaluated with the combined techniques IBFE þ IBP. We will show that both techniques provide equivalent solutions, both for numerical comparisons for some specific values of the invariants of the problem, as well as for analytical comparisons in certain special limits (m ¼ 0 or s ¼ 0). In Sec. IV, we present a more complex application. The idea is to specify the range of validity of the combined use of the IBFE þ IBP techniques. In particular, we consider the topology associated to the radiative corrections to the propagator with two loops and five internal lines. This diagram can be easily evaluated by means of IBP when all of the propagators are massless with index one. In our case, we will extend the discussion by considering now this diagram with one massive propagator and allowing some arbitrary indices. The analytic solutions for this problem will be given in terms of generalized hypergeometric func2 tions of the form q Fq1 , whose argument can be ( M ) or p2. p2. ( M2 ), according to the relevant kinematical region; p denotes the momentum that flows into the diagram and which fixes the scale, and M is the mass in the massive propagator. The general solutions of Feynman diagrams will correspond always to hypergeometric series, i.e. to natural expansions around a zero value of the argument and with a convergence radius r ¼ 1. The mathematical formalism, the relevant formulas, and applications are presented and discussed in [9–12]). II. IBP AND THE TRIANGLE RULE The well known triangle rule is obtained directly by applying the IBP method to the massless one-loop diagram with three external lines (how this rule is derived can be seen in the book by V. A. Smirnov [13]). The integral that represents the triangle diagram is the following: Z dD q 1 1 1 2a4 2a5 D=2 ðP Þ ðP Þ ½ðP þ qÞ2 a1 i 1 1 ; ½ðP qÞ2 a2 ðq2 Þa3. (1). where P and P identify the independent momenta of the external lines. Then, associating each index {a1 ; a2 ; a3 g with his own propagator, we may introduce a graphical representation for the triangle identity once IBP has been applied to the previous integral.. (2). In this expression, D denotes the number of space dimensions in the integral. The indices ai (i ¼ 1, 2, 3, 4, 5) correspond, as we said, to the powers of the propagators of the corresponding internal lines in the diagram. The indices a4 and a5 will be arbitrary if the corresponding lines are internal lines in a diagram that contains this triangle, and their value will be zero when the lines are external. Obviously, the IBP technique goes beyond this. identity. However, for our purpose this identity will be enough. III. VALIDATION OF THE IBFE þ IBP METHOD In previous articles [7,14], we have shown that the IBFE technique is a useful method for evaluating certain families of Feynman diagrams. In particular, we will use now this technique for handling the following test topology: a one-. 026003-2.

(3) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. loop triangle diagram where one of the propagators has a finite mass and the two external lines are massless. The result for this diagram will be compared with the result that emerges from the combined IBFE and IBP techniques. Explicitly, we will expand the massive propagator according to the IBFE prescription, then extracting out the mass from the diagram and getting finally a series that involves only nonmassive diagrams. Each of them may be reduced topologically, by means of the IBP approach, to a series of bubble diagrams. The IBFE technique will be applied then to each one of these bubble diagrams, obtaining the multiregion expansion (MRE) [7] of the complete diagram which allows us to obtain the solutions in terms of hypergeometric functions of the form q Fq1 . Once we get the solution of this diagram in both ways, we will make an analytical comparison in the limits m ¼ 0 and s ¼ 0.. PHYSICAL REVIEW D 81, 026003 (2010). n1 ;...;n5 ¼ ð1Þn1 þ...þn5. 1 : ðn1 þ 1Þ ðn5 þ 1Þ. (7). The constraints are given by the identities D þ n2 þ n3 þ n4 þ n5 ; 1 ¼ 2 2 ¼ ha1 þ n1 þ n2 þ n3 i;. (8). 3 ¼ ha2 þ n2 þ n4 i; 4 ¼ ha3 þ n5 i: Each constraint, according to the IBFE technique [7], is associated to the replacement of the integral, after expanding the integrand, with the following symbol: Z dxx1 þþ2 1 h1 þ þ 2 i; (9) which also emerged as a result from multinomial expansions that obey the following rule [7]: X X . . . n1 ;...;n An1 1 An ðA1 þ þ A Þ ¼. A. IBFE solution The diagram to be evaluated is:. n1. . (3). n. h þ n1 þ þ n i ; ðÞ. (10). where the terms Ai (i ¼ 1; . . . ; ) and the exponent are quantities that can take arbitrary values. such that the propagator associated to the index one has a mass m; whereas, P21 ¼ P23 ¼ 0 and P22 ¼ s [2]. For this diagram then, the equivalent configuration of invariants is described as follows:. (4). We will start by writing the corresponding Schwinger parametrization for this topology which is given by. 1. Solution for arbitrary indices The properties of the general solutions for Feynman diagrams imply the existence of two interesting kinematical regions: (1) Solution in the region j ms2 j < 1: s G 2 ¼ m ! s a1 þ a2 þ a3 D2 ; a2 ; 2 F1 D m2 2 (11). x1 x2 expð x1 þx sÞ ð1ÞD=2 Z 1 2 þx3 2 dx~ expðx1 m Þ ; (5) G ¼ Q3 ðx1 þ x2 þ x3 ÞD=2 j¼1 ðaj Þ 0. where the factor is defined as ¼ ð1ÞD=2 ðm2 ÞðD=2Þa1 a2 a3. xa1 1 1 xa2 2 1 x3a3 1 dx1 dx2 dx3 .. Some algebra alwhere dx~ ¼ lows us to obtain the MRE [7] form of (5): Q4 ð1ÞD=2 X j¼1 j 2 n n 1 2 ; n1 ;...;n5 ðm Þ ðsÞ G ¼ Q3 D ð 2 þ n2 Þ j¼1 ðaj Þ n1 ;...;n5 (6) where the factor n1 ;...;n5 , defined in [7], is given as follows:. . (12) 2. (2) Solution in the region j ms j < 1:. 2 1 þ a1 þ a2 þ a3 D; a1 þ a2 þ a3 D2 m ¼ 2 F1 G s 1 þ a1 þ a3 D2. 026003-3. ða1 þ a2 þ a3 D2 ÞðD2 a2 a3 Þ : ða1 ÞðD2 Þ. ! 1 þ a2 D2 ; a2 m2 þ 2 F1 s 1 þ D2 a1 a3. ! m2 ; s. (13).

(4) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). where the different factors are ¼ ð1ÞD=2 ðsÞðD=2Þa1 a2 a3. ða1 þ a2 þ a3 D2 ÞðD2 a1 a3 ÞðD2 a2 a3 Þ ; ða1 Þða2 ÞðD a1 a2 a3 Þ. ¼ ð1ÞD=2 ðsÞa2 ðm2 ÞðD=2Þa1 a3. ða1 þ a3 D2 ÞðD2 a2 a3 Þ : ða1 ÞðD2 a2 Þ. 2. Solution for unitary indices. (15). and the factors. This situation is especially relevant, since when an arbitrary Feynman diagram is computed, the indices associated to the propagators are normally unitary, this case also being useful for numerical computations. Then, with D ¼ 4 2, the Eqs. (11) and (13) are (1) Solution in the region j ms2 j < 1: s G 2 ¼ ð1Þ ðm2 Þ1 m ð1 þ Þ s 1 þ ; 1 F : 2 1 2 m2 ð1 Þ (16) 2. (2) Solution in the region j ms j < 1: 2! 2 m 2 m ¼ 1 F0 þ G s s m2 ! ; 1 ; 2 F1 s 1 . (14). ¼ ð1Þ ðsÞ1. ð1 þ Þð1 Þ2 ; 2 ð1 2Þ. (18). ð1 þ Þ : 2. (19). ¼ ð1Þ ðsÞ1 ðm2 Þ. B. Solution obtained by the application of IBFE þ IBP The first step required for combining these two techniques is to eliminate the mass from the diagram in (4). Once we have done this, we can apply IBP to the massless triangle (2). For this purpose, we will use IBFE on the propagator that contains this mass and expand it according to this technique, as it will be shown next: ðq2. X 1 h þ n1 þ n2 i ; ¼ n1 ;n2 ðm2 Þn1 ðq2 Þn2 2 m Þ ðÞ n1 ;n2 (20). (17) or in a graphically equivalent way:. (21) Therefore, (4) will be given by. (22). The next step is obtained from the application of IBP to the triangle, once the mass has been eliminated from the graph. However, before this, and considering the invariants in (4) we can rewrite the triangle rule (2) as. (23). In order to apply this rule, we will consider special values for the indices: we take a3 ¼ 1 and keep arbitrary values for a1 and a2 (a4 ¼ a5 ¼ 0, since they are external lines). Then, the previous expression can be written as. 026003-4.

(5) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. PHYSICAL REVIEW D 81, 026003 (2010). (24). Applying (24) in (22), we have. (25). For simplicity, let us define. (26). and also. (27). where the identity ¼ ðþ1Þ ðÞ has been used to express everything in terms of Gamma functions. To find the MRE’s associated to the terms A and B, we only need to express the bubble diagrams in these equations through the following identity [14]:. G A ð; ; nj Þ ¼. ð1ÞD=2 X 1 2 3 ; ðÞðÞ nðjþ1Þ ;nðjþ2Þ nj ;nðjþ1Þ ;nðjþ2Þ ðD2 þ nj Þ (29). the constraints being {i g D þ nj þ nðjþ1Þ þ nðjþ2Þ ; 1 ¼ 2 . 2 ¼ h þ nj þ nðjþ1Þ i;. (27) where the one bubble function is given by. 026003-5. 3 ¼ h þ nj þ nðjþ2Þ i:. (30).

(6) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). Through a careful replacement in A and B, we find for A ð1ÞD=2 X ðD þ n2 a2 2Þðn2 þ 1Þ ha1 þ n1 þ n2 i ðm2 Þn1 ðsÞn3 ða1 Þða2 Þ n1 ;...;n5 n1 ;...;n5 ðD þ n2 a2 1Þðn2 Þð1 n2 ÞðD2 þ n3 Þ D þ n3 þ n4 þ n5 h1 n2 þ n3 þ n4 iha2 þ n3 þ n5 i; 2. A¼. (31). and B is given by B¼. ð1ÞD=2 X ðD þ n2 a2 2Þ ðm2 Þn1 ðsÞn3 ha1 þ n1 þ n2 i ða1 Þða2 Þ n1 ;...;n5 n1 ;...;n5 ðD þ n2 a2 1Þðn2 ÞðD2 þ n3 Þ D þ n3 þ n4 þ n5 hn2 þ n3 þ n4 ih1 þ a2 þ n3 þ n5 i: 2. (32). Once we have obtained the MRE’s of the terms A and B, we proceed to find the corresponding solutions according to the 2 quotient j ms j. 1. Solutions from the term A 2. (1) Solution for the region j ms j < 1: From the MRE (31), we get the following solution for this kinematical region: 2 m 2 þ a1 þ a2 D; 2 þ a1 þ a2 D; 1 þ a1 þ a2 D2 ¼ 3 F2 A1 3 þ a1 þ a2 D; 2 þ a1 D2 s. ! 2 m ; s. (33). where the factor is given by ¼ ð1ÞD=2 ðsÞðD=2Þa1 a2 1. ða1 þ a2 þ 1 D2 ÞðD2 a2 ÞðD2 a1 1ÞðD a1 a2 2Þ ða1 Þða2 ÞðD a1 a2 1Þ2. and also the term 2! 2 m m a2 ; 1 þ a2 D2 ; 1 þ a2 D2 ¼ 3 F2 ; A2 D D 2 þ a2 2 ; 2 a1 s s (35). 2. (2) Solution for the region j ms j > 1: In this region we only have one term: 2 m ¼ A3 s 3 F2. where we have defined as ¼ ð1ÞD=2 ðsÞa2 ðm2 ÞðD=2Þa1 1 ða1 þ 1 D2 ÞðD2 1ÞðD2 2ÞðD2 a2 1Þ : ða1 ÞðD2 a2 Þð1 D2 ÞðD2 Þ. (34). a2 ; 1 þ a1 þ a2 D2 ; D D 2; 2. ! s ; m2 . D 2 1 . (37). where. (36). ¼ ð1ÞD=2 ðm2 ÞðD=2Þa1 a2 1. ð1 þ a1 þ a2 D2 ÞðD2 a2 1ÞðD2 1Þð2 þ a2 D2 Þ : ða1 Þð1 þ a2 D2 ÞðD2 Þ2. (38). 2. Solutions from the term B 2. (1) Solution for the region j ms j < 1: For B, we have as solution in this kinematical region 2 m 2 þ a1 þ a2 D; 2 þ a1 þ a2 D; 1 þ a1 þ a2 D2 ¼ 3 F2 B1 3 þ a1 þ a2 D; 1 þ a1 D2 s being. 026003-6. ! 2 m ; s. (39).

(7) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. ¼ ð1ÞD=2 ðsÞðD=2Þa1 a2 1. ða1 þ a2 þ 1 D2 ÞðD2 a2 1ÞðD2 a1 ÞðD a1 a2 2Þ ða1 Þða2 ÞðD a1 a2 1Þ2. and also the term 2! 2 m m 1þa2 ; 2þa2 D2 ; 2þa2 D2 ¼ 3 F2 ; B2 D D s 3þa2 2 ; 2 þ1a1 s (41) where ¼ ð1ÞD=2 ðsÞ1a2 ðm2 ÞðD=2Þa1 . PHYSICAL REVIEW D 81, 026003 (2010). ða1 D2 ÞðD2 2 a2 Þða2 þ 1Þ : ða1 Þða2 ÞðD2 1 a2 Þ. (42). 2. (2) Solution for the region j ms j > 1: In this case, we only get one hypergeometric function: s B3 2 ¼ m ! s 1þa2 ; 1þa1 þa2 D2 ; D2 1 ; 3 F2 D D m2 ; 2 2. In the same region of interest, the solution given in (16) obtained by a unique application of IBFE is notoriously different, both from the amount of terms as well as from the structure. Let us discuss this point more in detail: s GIBFE 2 ¼ ð1Þ ðm2 Þ1 m ð1 þ Þ s 1 þ ; 1 F : (47) 2 1 2 m2 ð1 Þ We can do now comparisons to show the equivalence between both techniques (see the Appendix for a general proof). A direct possibility is to analyze the equivalence in certain limit cases, i.e. to compare, for m ¼ 0, the solutions we obtained from the expressions valid in the kinematical 2 region j ms j < 1 and to compare also the solutions for s ¼ 0, this time considering the corresponding expressions for 2 the region j ms j > 1. 1. Analytic equivalence in the limits m ! 0 and s ! 0. (43) where ¼ð1ÞD=2 ðm2 ÞðD=2Þa1 a2 1 ð1þa1 þa2 D2 ÞðD2 a2 1ÞðD2 1Þð1þa2 Þ : ða1 Þða2 ÞðD2 Þ2. (1) Solution for m ¼ 0: In this limit, the Eq. (46) has the form GIBPþIBFE jm¼0 ¼ A1 ð0Þ þ B1 ð0Þ ¼ 2ð1Þ ðsÞ1 ð1 þ Þð1 Þ2 ; (48) 22 ð1 2Þ. (44). C. Comparison of the solutions for the region j ms2 j < 1 (unitary indices) In order to compare the solutions we have obtained with both techniques, we will take the result for the kinematical region j ms2 j < 1. For the case of unitary indices and D ¼ 4 2, the combined technique IBP þ IBFE produces s s s G 2 ¼ B3 2 þ A3 2 ; (45) m m m i.e. GIBPþIBFE. (40). . s ð1 þ Þ ¼ ð1Þ ðm2 Þ1 2 m ð1 Þ2 s 2; 1 þ ; 1 3 F2 2 m 2 ; 2 s 1; 1 þ ; 1 3 F2 2 : m 2 ; 2 (46). 026003-7. whereas (47) GIBFE jm¼0 ¼ ð1Þ ðsÞ1. ð1 þ Þð1 Þ2 : 2 ð1 2Þ (49). Some algebra allows us to show that effectively, GIBPþIBFE jm¼0 GIBFE jm¼0 ¼ 0:. (50). (2) Solution for s ¼ 0: The Eq. (46) becomes now GIBPþIBFE js¼0 ¼ A3 ð0Þ þ B3 ð0Þ ¼ ð1Þ ðm2 Þ1. ð1 þ Þ : ð1 Þ (51). In the same way as (47), in this limit we have GIBFE js¼0 ¼ ð1Þ ðm2 Þ1. ð1 þ Þ : (52) ð1 Þ.

(8) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). In this case, both solutions are also equivalent: GIBPþIBFE js¼0 GIBFE js¼0 ¼ 0:. (53). According to the results we have obtained previously, we can say that IBP þ IBFE is a technique that can be used in an equivalent way to IBFE to evaluate Feynman diagrams. Nevertheless, the former has some technical advantages that extend the usefulness of IBFE beyond the family of diagrams, where it can ideally be applied and which is described in [7]. IV. A TWO-LOOP APPLICATION As an example in a more complex scenario, we will use now the combined method IBP þ IBFE for computing the correction to the propagator given by two loops and five propagators. It is possible to evaluate such a diagram for the massless case and with integer indices in their propagators iterating the use of IBP (if it is necessary), transforming the original topology into a sum of topologies which, for the unitary indices case, generates more simple graphs which can be reduced loop by loop in terms of bubble diagrams. In this work, we will evaluate this diagram taking now one massive propagator, mass M [15], with index a1 , as we can see in the following graphic equation:. (iii) Let us assume the masses m2 ¼ m3 ¼ m4 ¼ m5 ¼ 0 and m1 ¼ M. Let us then consider the loop integral for this graph: Z dD q1 dD q2 1 1 G¼ D=2 D=2 ðq2 M2 Þa1 ½ðq þ pÞ2 a2 i i 1 1 . 1 1 1 : ðq2 þ q1 Þ2 q22 ðp q2 Þ2. (55). The idea is to extract the mass M from the integral in such a way that the resulting integral corresponds to a massless diagram and then, in a second step, to apply the triangle rule (2). This is the place where IBFE can be used. This technique allows us to find the MRE of the propagator that contains the mass, i.e. ðq21. X 1 ha þ n1 þ n2 i : n1 ;n2 ðq21 Þn1 ðM2 Þn2 1 2 a1 ¼ M Þ ða1 Þ n1 ;n2 (56). By replacing then in the integral (55), we get the following expansion (In this expansion, we must keep in mind that the sum indices are not necessarily integers): X ha þ n1 þ n2 i Z dD q1 dD q2 G¼ n1 ;n2 ðM2 Þn2 1 ða1 Þ iD=2 iD=2 n1 ;n2 . 1 1 1 1 1 : ðq21 Þn1 ½ðq1 þ pÞ2 a2 ðq2 þ q1 Þ2 q22 ðp q2 Þ2 (57). We can represent also in a graphic way this result in the following way:. (54). The solutions to this diagram are functions of the variables p2 M2 ) ðM 2 Þ or ð p2 Þ according to the kinematical region of interest. We impose the following conditions to evaluate the diagram: (i) The indices a3 , a4 , and a5 are integer quantities. In particular, we will have a3 ¼ a4 ¼ a5 ¼ 1. This assumption does not imply a loss of generality, since the indices in the loop integrals are unitary. (ii) We will consider arbitrary indices a1 and a2 . This will allow us not only the possibility of taking several values for the indices, but also to implement topological variations of the diagram by inserting corrections to the propagators, having then more loops, or by taking equal or different masses for these propagators. The last point is related to the fact that the extraction of the mass from a certain propagator generates a massless propagator with a variable index, as it is shown in the Eq. (21).. (58). At this point, we can apply the triangle rule to the resulting topology. We have to mention that the multiregion expansion of the massive propagator contains in its structure the expansion for the cases of small and big masses simultaneously, which is precisely what we need to compute this diagram in all of the interesting kinematical zones. A. Application of the triangle rule To apply the formula (2), we will consider the left triangle of the diagram in (58). As a result of this, we get in a simple way the following expression for G:. 026003-8.

(9) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. PHYSICAL REVIEW D 81, 026003 (2010). (59). We will express all the coefficients (only those which contain some sum index) of the previous equation in terms of Gamma functions. This will be useful to form the resulting hypergeometric functions. We use the following identity: ð þ 1Þ ¼ ðÞ ) ¼. ð þ 1Þ ; ðÞ. ðn1 þ 1Þ ; ðn1 Þ. 1 ðD þ n1 a2 2Þ : ¼ ðD þ n1 a2 2Þ ðD þ n1 a2 1Þ. (61). (62). B. Analytical Solution. (60). with this, we may replace in (59) the sum index dependent coefficients ðn1 Þ ¼. and for the denominator,. According to the Eq. (59), we can write the original diagram as a sum of four terms, i.e. G ¼ A1 þ A2 þ A3 þ A4 ;. (63). which are defined as. (64). (65). 026003-9.

(10) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). (66) and finally,. (67). The resulting topologies in each term Ai (i ¼ 1, 2, 3, 4) are appropriate to compute them directly with the IBFE integration technique. With this technique, we will be able to get the analytical solution in the possible kinematical regimes where each term Ai contributes. In what follows, it is necessary to get the equivalent complete MRE for each one of these terms to obtain then the analytical solution. We have two ways to get this expansion: the first one is to expand the parametrical integral that corresponds to the complete diagram present in each term Ai , and the second one, which is the most straightforward procedure, is to use the idea of modular reduction of the diagram [14], that allows us to find very rapidly the MRE of each one of the. diagrams as a product of MRE of their constituent individual loops. These are known as loop functions. We will use this last form to start with the analysis of the remnant diagrams in each term Ai . We will present the complete calculation procedure for the first term of (59), A1 , and then the other terms will be presented in a summarized way. 1. Complete solution of A1 To find the solution from this term, we need first the MRE associated to the diagram. We will use the idea of loop function representations [14] to obtain the expansion which has the form. (68). X ð1ÞD=2 1 2 3 GA ðaj ; ak ; ni Þ ¼ : ni ;nðiþ1Þ ;nðiþ2Þ D ðaj Þðak Þ nðiþ1Þ ;nðiþ2Þ ð 2 þ ni Þ. . D þ ni þ nðiþ1Þ þ nðiþ2Þ ; 2 2 ¼ haj þ ni þ nðiþ1Þ i;. where the loop function GA corresponds to the MRE of each loop, or part of the diagram, and which has the following MRE:. 1 ¼. 3 ¼ hak þ ni þ nðiþ2Þ i:. (70). Clearly, we have to replace the pictorial representation of the propagator by its corresponding algebraic expression:. (69). (71) In this case, the constraints {i g are given by the following expressions:. Some algebra allows us to find the MRE of the diagram in (68):. 026003-10.

(11) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. PHYSICAL REVIEW D 81, 026003 (2010). (72). whose constraints are determined as follows: D ð1Þ þ n3 þ n4 þ n5 ; ð1Þ 1 ¼ 2 ¼ hn1 þ 1 þ n3 þ n4 i; 2 D ð1Þ ð1Þ þ n6 þ n7 þ n8 ; 3 ¼ ha2 þ n3 þ n5 i; 4 ¼ 2 ð1Þ 5 ¼ h1 þ n6 þ n7 i;. ð1Þ 6 ¼ h1 þ n6 þ n8 i:. Starting from the expression (74), it is possible to find the p2 solutions in the two possible kinematical regions j M 2 j < 1 2 and j M j < 1, where the solutions for each kinematical p2 region can be divided into groups, as a sum of generalized hypergeometric functions2 and where the sets will be dep M2 fined by the terms A1 ( M 2 ) and A1 ( p2 ), respectively. We will show now the analytical solutions we found for the term A1 according to the corresponding kinematical regions: p2 (1) Solution in the region j M 2 j < 1: 2! 2 p p 1þa1 þa2 D2 ; a2 ; D2 1 A1 2 ¼ 3 F2 D D ; M M2 2 2. (73). Finally, by introducing the necessarily replacements, we have obtained the MRE for the first term of the solution of the diagram G. We have added a superindex in the constraints. This only refers to the fact that these constraints are associated to the first term of G, A1 . So, we get then the MRE for this term:. (76). ð1ÞD X A1 ¼ ðM2 Þn2 ðp2 Þn3 þn6 ða1 Þða2 Þ n1 ;...;n8 n1 ;...;n8 . ðD þ n1 a2 2Þ ðD þ n1 a2 1Þðn1 Þ. . ð1Þ ð1Þ 1 6 ; ðD2 þ n3 ÞðD2 þ n6 Þ. where the factor is defined as ¼ ð1ÞD ðp2 ÞðD=2Þ2 ðM2 ÞðD=2Þ1a1 a2 (74). ð1þa1 þa2 D2 ÞðD2 a2 Þð2 D2 ÞðD2 1Þ3 : ða1 ÞðD2ÞðD2 Þ2 (77). where we have also defined an extra constraint associated to the MRE of the massive propagator: ¼ ha1 þ n1 þ n2 i:. 2. (2) Solution in the region j M j < 1: p2. (75). 2 M 2 þ a1 þ a2 D; 2 þ a1 þ a2 D; 1 þ a1 þ a2 D2 A1 2 ¼ 3 F2 2 þ a1 D2 ; 3 þ a1 þ a2 D p 2! M 1 þ a2 D2 ; 1 þ a2 D2 ; a2 ; 3 F2 D D p2 a ; 2þa 2. 1. 2. ! M2 2 þ p (78). 2. being the factors ¼ ð1ÞD ðp2 ÞD3a1 a2. ð1 þ a1 þ a2 D2 ÞðD2 a2 Þð2 D2 ÞðD2 a1 1Þ ðD a1 a2 2ÞðD2 1Þ2 ; (79) ða1 Þða2 ÞðD 2Þ ðD a1 a2 1Þ2. and ¼ ð1ÞD ðp2 ÞðD=2Þ2a2 ðM2 ÞðD=2Þ1a1. ð1 þ a1 D2 ÞðD2 a2 1Þð2 D2 ÞðD2 1Þ2 : ða1 ÞðD 2ÞðD2 a2 Þ. (80). Nevertheless, the solution of the diagram G is by no means complete, since we have to consider also the terms that can be generated from the MRE associated to {A2 ; A3 ; A4 g. Then, we have to add them algebraically, according to the kinematical. 026003-11.

(12) IVÁN GONZÁLEZ AND M. LOEWE. region of interest, i.e. G¼. PHYSICAL REVIEW D 81, 026003 (2010). . 2. 2. 2. 2. p p p p A1 ðM for M2 > jp2 j 2 Þ þ A2 ðM2 Þ þ A3 ðM2 Þ þ A4 ðM2 Þ; M2 M2 M2 M2 A1 ð p2 Þ þ A2 ð p2 Þ þ A3 ð p2 Þ þ A4 ð p2 Þ; for jp2 j > M2 :. (81). We will proceed in an analogous way, as we did for the calculation of A1 , for the other terms Ai : 2. Contributions to the solution generated by A2 For the diagram that appears in A2 , the MRE is obtained as l:. (82). and we use then the expression (69) to replace the one-loop functions GA in the expansion of this graph, getting in an explicit way the MRE of the term A2 :. A2 ¼. . D þ n3 þ n4 þ n5 ; 2. ð2Þ 3 ¼ h1 þ n3 þ n5 i;. ð2Þ 2 ¼ h1 þ n3 þ n4 i; D ð2Þ þ n6 þ n7 þ n8 ; 4 ¼ 2. ð2Þ 5 ¼ hn1 þ 1 þ n6 þ n7 i;. ð1ÞD X ðM2 Þn2 ðp2 Þn6 ða1 Þ n1 ;...;n8 n1 ;...;n8 . ðD þ n1 a2 2Þðn1 þ 1Þ ðD þ n1 a2 1Þðn1 Þ. . ð2Þ ð2Þ 1 6 ; (83) ðD2 þ n3 ÞðD2 þ n6 Þðn1 þ 1Þða2 n3 Þ. where the constraints correspond to the following expressions:. ¼ ð1ÞD ðM2 ÞD3a1 a2. ð2Þ 1 ¼. ð2Þ 6 ¼ ha2 n3 þ n6 þ n8 i:. (84) With this information, we can find in this case the solutions for both relevant kinematical regions: p2 (1) Solution in the region j M 2 j < 1: 2 p A2 2 ¼ M 2! p 3þa1 þa2 D;2þa2 D2 ;1 ; 3 F2 D 2 ; 2 M 2 (85) where the factor is given by the following expression:. ðD 3 a2 Þð3 þ a1 þ a2 DÞð4 þ a2 DÞð2 D2 ÞðD2 1Þ2 : ða1 ÞðD 2ÞðD2 Þð3 þ a2 DÞ. (86). 2. (2) Solution in the region j M j < 1: p2 ! 2! 2 D M2 M M 3 þ a D; 1 þ a 4 þ a1 þ a2 3D ; 2 þ a þ a D 2 2 1 2 2 2 þ 2 F1 ; (87) A2 2 ¼ 2 F1 D D 2 p p2 2 þ a1 2 p 2 a1 . where the factors are defined as ¼ ð1ÞD ðp2 ÞD3a1 a2 . ð3 þ a1 þ a2 DÞðD a2 2Þð2 D2 ÞðD2 a1 1Þ ða1 þ 1Þða1 ÞðD 2Þ. ðD a1 a2 2ÞðD2 1Þ2 ð1 a1 Þ ; D ðD a1 a2 1Þð3D 2 a1 a2 3Þð2 þ a2 2 Þ. 026003-12. (88).

(13) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. PHYSICAL REVIEW D 81, 026003 (2010). ¼ ð1ÞD ðp2 ÞðD=2Þ2a2 ðM2 ÞðD=2Þ1a1 . 3. Contributions to the solution generated by A3. ð1 þ a1 D2 ÞðD2 a2 1Þð2 D2 Þ2 ðD2 1Þ3 : ða1 ÞðD 2ÞðD2 a2 ÞðD2 Þð1 D2 Þ. We will follow the same procedure for getting this term, i.e. to find the MRE of A3 and their corresponding analytical solutions:. (89). (90). ð1ÞD X A3 ¼ ðM2 Þn2 ðp2 Þn3 þn6 ða1 Þða2 Þ n1 ;...;n8 n1 ;...;n8 . ð3Þ 1 ¼. ð3Þ 3 ¼ ha2 þ 1 þ n3 þ n5 i;. ðD þ n1 a2 2Þ ðD þ n1 a2 1Þ ðD2 þ. D þ n3 þ n4 þ n5 ; 2. ð3Þ 5 ¼ h1 þ n6 þ n7 i;. ð3Þ 1. ð3Þ 6 ; n3 ÞðD2 þ n6 Þðn1 Þ. (91). ð3Þ 2 ¼ hn1 þ n3 þ n4 i; D þ n ð3Þ ¼ þ n þ n 6 7 8 ; 4 2. ð3Þ 6 ¼ h1 þ n6 þ n8 i:. (92). 2. p (1) Solution in the region j M 2 j < 1: 2! 2 p p 1þa1 þa2 D2 ;D2 1;1þa2 A3 2 ¼3 F2 ; DD M2 M 2;2. with the constraints. (93) ¼ ð1ÞD ðM2 ÞðD=2Þ1a1 a2 ðp2 ÞðD=2Þ2. ð1 þ a1 þ a2 . D D 2 Þð 2. a2 1Þð1 þ a2 Þð2 ða1 Þða2 ÞðD 2ÞðD2 Þ2 :. D D 2 Þð 2. . 1Þ3. :. (94). 2. j < 1: (2) Solution in the region j M p2 2 M 2 þ a1 þ a2 D; 2 þ a1 þ a2 D; 1 þ a1 þ a2 D2 A3 2 ¼ 3 F2 1 þ a1 D2 ; 3 þ a1 þ a2 D p M2 ! 2 þ a2 D2 ; 2 þ a2 D2 ; 1 þ a2 3 F2 2 ; p 1 a1 þ D2 ; 3 þ a2 D2. ! M2 þ p2 (95). where the different factors are given by ðD a1 a2 2ÞðD2 a2 1Þð2 D2 ÞðD2 1Þ2 ðD2 a1 Þð1 þ a1 þ a2 D2 Þ ¼ ð1ÞD ðp2 ÞD3a1 a2 ; (96) ða1 Þða2 ÞðD 2Þ ðD a1 a2 1Þ2 ¼ ð1ÞD ðp2 ÞðD=2Þ3a2 ðM2 ÞðD=2Þa1. ða1 D2 Þð1 þ a2 Þð2 D2 ÞðD2 a2 2ÞðD2 1Þ2 : ða1 Þða1 ÞðD 2ÞðD2 a2 1Þ. (97). 4. Contributions to the solution generated by A4. (98). 026003-13.

(14) IVÁN GONZÁLEZ AND M. LOEWE. A4 ¼ . PHYSICAL REVIEW D 81, 026003 (2010). D þ n3 þ n4 þ n5 ; 1ð4Þ ¼ 2. ð1ÞD X ðM2 Þn2 ðp2 Þn6 ða1 Þða2 Þ n1 ;...;n8 n1 ;...;n8. ð4Þ 2 ¼ h1 þ n3 þ n4 i; D þ n ð4Þ ¼ þ n þ n 6 7 8 ; 4 2. ð4Þ 3 ¼ h1 þ n3 þ n5 i;. ðD þ n1 a2 2Þ ðD þ n1 a2 1Þ. ð4Þ 5 ¼ hn1 n3 þ n6 þ n7 i;. ð4Þ ð4Þ 1 6 ; D ð 2 þ n3 ÞðD2 þ n6 Þðn1 n3 Þ. ð4Þ 6 ¼ ha2 þ 1 þ n6 þ n8 i:. (99). (100) 2. p (1) Solution in the region j M 2 j < 1: ! 2 2 D p p D 2; a1 ; 2 1 þ A4 2 ¼ 3 F2 3D D a2 1; 2 a2 3 M2 M p2 ! 2 þ a2 D2 ; 3 þ a2 þ a1 D; a2 þ 1; 1 ; 4 F3 D M2 ; 2; 4 D þ a 2 2. (101). being ¼ ð1ÞD ðp2 ÞD3a2 ðM2 Þa1 ¼ ð1ÞD ðM2 ÞD3a1 a2. ð3 þ a2 DÞðD2 a2 1ÞðD a2 2ÞðD2 1Þ2 ; ða2 Þð3D 2 a2 3ÞðD a2 1Þ. ðD a2 3Þð3 þ a2 þ a1 DÞð2 D2 ÞðD2 1Þ2 ð1 þ a2 Þ : ða1 Þða2 ÞðD 2ÞðD2 Þ. (102). (103). 2. (2) Solution in the region j M j < 1: p2 2 M 4 þ a1 þ a2 3D 2 ; 2 þ a1 þ a2 D; A4 2 ¼ 3 F2 D 2 þ a p 1 2 ; 3 þ a1 D 2! M 2 þ a2 D2 ; D 2; a2 ; 3 F2 D p2 D a 1; 1. 2! M a1 þ p2 (104). 2. with ¼ ð1ÞD ðp2 ÞD3a1 a2 . ðD2 a2 1ÞðD a1 2ÞðD a1 a2 2Þ ða2 ÞðD 2Þ. ð2 D2 Þð3 þ a1 þ a2 DÞðD2 1Þ2 ; D ðD a1 a2 1Þð3D 2 a1 a2 3Þð2 þ a1 2 Þ. (105). . ¼ ð1ÞD ðp2 Þ1a2 ðM2 ÞDa1 2. GðM¼0Þ. ð2 þ a1 DÞð2 D2 ÞðD2 1Þ2 ð1 þ a2 Þða2 Þ : ða1 Þða2 ÞðD2 Þð1 a2 Þ (106). C. A particular case: Solution for M ¼ 0 with indices a1 ¼ a2 ¼ 1. 2 2 2 M M M ¼ A1 2 þ A2 2 þ A3 2 p p p 2 M þ A4 2 : p M¼0. (107). It is easy to verify that in this limit only four terms will survive. Using some algebra, we can get the following result: GIBFEðM¼0Þ ¼ 2ð1ÞD ðp2 ÞD5. In what follows, we will test the solution we got for the diagram (54). For this, we will compute the massless case, which has a trivial solution when integration by parts is used. In this case, we have to consider the solution obtained 2 by IBP þ IBFE in the region j M j < 1 and to take its limit p2 M ¼ 0, i.e.. 026003-14. ðD2 2ÞðD2 1Þ2 ð2 D2 ÞðD 4Þ ðD 2Þ D ð 2 1Þð3 D2 Þ ð5 DÞ : D ðD 3Þ2 ð3D 2 5Þð3 2 Þ . (108).

(15) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. PHYSICAL REVIEW D 81, 026003 (2010). On the other side, the conditions under which we now have to evaluate the diagram are such that the solution can be found in a conventional way, i.e. applying IBP through the triangle rule. So, we find. (109). X 1 ha þ n1 þ n2 i ; n1 ;n2 ðk2 Þn1 ðM2 Þn2 1 2 a1 ¼ ða1 Þ ðk M Þ n1 ;n2. and we can rewrite this as GIBPðM¼0Þ ¼. 2 D Gð1; 1Þ Gð2; 1Þ G 2; 3 ðD 4Þ 2 . ðp2 ÞD5 ;. 2. (115). (110). where the factors Gð; Þ are given by Gð; Þ ¼ ð1ÞD=2. ð þ D2 ÞðD2 ÞðD2 Þ : ðÞðÞðD Þ (111). Finally, we obtain the following result for this two-loop diagram: GIBPðM¼0Þ ¼ ð1ÞD ðp2 ÞD5. 2 ðD 4Þ. ð2 D2 ÞðD2 1Þ2 ðD2 2Þ ðD 2Þ D ð 2 1Þð3 D2 Þ ð5 DÞðD 3Þ 3D : ðD 3Þ ð 2 5Þð3 D2 Þ. . finding, when replacing this result in (114), X ha þ n1 þ n2 i Z dD k G¼ n1 ;n2 ðM2 Þn2 1 ða1 Þ iD=2 n1 ;n2 . GIBPðM¼0Þ GIBFEðM¼0Þ ¼ 0:. (113). iD=2 ðk2 Þn1 ½ðp kÞ2 a2 ðn1 þ a2 D2 ÞðD2 þ n1 ÞðD2 a2 Þ ¼ ð1ÞD=2 ðn1 Þða2 ÞðD þ n1 a2 Þ 1 (117) 2 n þa ðD=2Þ : ðp Þ 1 2 We replace this result in (116) getting the following MRE for G: G ¼ ð1ÞD=2. This result gives support to the most general solution for this diagram in the case of one massive propagator. V. CONCLUSIONS AND COMMENTS It is not always possible to combine the IBFE technique with another conventional integration method. For example, let us evaluate the one-loop diagram where one of the propagators has a mass M: G¼. Z dD k 1 : 2 M2 Þa1 ½ðp kÞ2 a2 D=2 ðk i. (114). As a first step, we will take out the mass M of the massive propagator in such a way that we may associate the resulting integral to a massless diagram. We need to express this propagator in terms of a MRE, i.e.. (116). If the idea is to use only IBFE, the next step would be to find the corresponding MRE for the resulting loop integral. However, instead of doing this we will evaluate directly, and in a conventional way, the corresponding integral. This integral is elementary and represents a massless bubble diagram whose solution is given by Z dD k 1. (112) The solutions (108) and (112) seem to be apparently different, but by using the properties of the Gamma function and some algebra, it is possible to show that both terms are equal, i.e.. 1 : ðk2 Þn1 ½ðp kÞ2 a2. ðD2 a2 Þ 2 ðD=2Þa X 2 ðp Þ n1 ;n2 ðM2 Þn2 ða1 Þða2 Þ n1 ;n2. ðp2 Þn1 ha1 þ n1 þ n2 i. ðn1 þ a2 D2 ÞðD2 þ n1 Þ : ðn1 ÞðD þ n1 a2 Þ (118). According to the prescription of IBFE, we notice immediately that the maximum number of possible terms we can extract to form the solution is 2 ¼ 2: 1 In fact, one of these terms corresponds to an hypergeometric series which is solution for the kinematical region 2 jM j < 1, and the other one would give us the solution p2 2. p for the region j M 2 j < 1. However, this solution is not complete, since some terms are missed [2,3]. In fact, it remains to add one term to the corresponding solution in. 026003-15.

(16) IVÁN GONZÁLEZ AND M. LOEWE. PHYSICAL REVIEW D 81, 026003 (2010). 2 jM p2. the region j < 1. The terms which can be extracted (118) are correct, but they represent only a part of the solution, and this is the problem. We will find now the MRE for this diagram using IBFE as the only integration method. Then, we will compare with the previous result shown in (118). Schwinger’s parametric expansion of (114) is G¼. ð1ÞD=2 Z 1 dx1 dx2 xa1 1 1 xa2 2 1 ða1 Þða2 Þ 0 . x2 expðx1 M2 Þ expð xx1 1þx p2 Þ 2. ðx1 þ x2 ÞD=2. :. (119). After some algebra, the following MRE for the diagram G is obtained: G¼. D=2. X. 2 n1. 2 n2. ð1Þ ðp Þ ðM Þ ða1 Þða2 Þ n1 ;...;n4 n1;...; n4 ðD2 þ n1 Þ. 3 Y. j ;. j¼1. (120) where. tributed to the solution, with terms associated to the corresponding interesting kinematical regions. The conclusion is that IBP is a technique compatible with IBFE. In this work, we have emphasized the role of the triangle identity and the fact that, under certain particular mass distributions, it is possible to evaluate a series of diagrams with the procedure described above. For example,. Although these examples do not necessarily generate a hypergeometric series of one variable, the procedure can be directly applied to these topologies. Another option for generalizing is the insertion of bubbles in the propagators, with arbitrary indices, which can be massless or massive. As an example, let us consider the following case:. D þ n1 þ n3 þ n4 ; 1 ¼ 2 2 ¼ ha1 þ n1 þ n2 þ n3 i;. (121). 3 ¼ ha2 þ n1 þ n4 i: We observe that this time, the possible number of series representations that can be extracted from (120) is given by the combinatory 4 ¼ 4: 3 This shows that when conventional techniques, combined with IBFE, are used, the MRE for a certain diagram is more simple than the MRE obtained only by means of IBFE. This means that the conventional techniques do not generate sums or Kroneker deltas, and therefore the number of different ways for summing, using the constraints of the MRE, becomes also smaller. In general, the solutions that are obtained from these MRE’s do not correspond necessarily to the correct solution. They can be incomplete. In summary, once we have chosen to use IBFE as an integration technique, the whole resulting process must be done with this technique. However, to simplify the problem, it is possible to apply other techniques before IBFE, but not viceversa. The exception to this rule is IBP, where the integration in terms of the topological cancellation of one of the graph lines do not produce the disappearance of terms in the solution. The explanation for this relies in the fact that the IBP expresses the original diagram as a sum of diagrams. In the example given at the beginning of this work, we used IBP in the intermediate step of getting the MRE. The result was a sum of MRE’s in which each one con-. We have shown that IBP þ IBFE is a technique that may extend the classes of diagrams that can be evaluated through this simple procedure, beyond the scenario given in [7] and, although here we have not considered the IBP in his most general version, we assume that the combination IBP þ IBFE can be also useful beyond the triangle identity. ACKNOWLEDGMENTS We acknowledge support from Fondecyt under Grant No. 3080029 and Grant No. 1095217 and also support from the Centro de Estudios Subatomicos, Valparaı́so, Chile. APPENDIX In what follows, we will prove that the solutions that we obtained for the one-loop example given in Eq. (4); the first one using only IBFE (47) and the second one through the combined technique IBP-IBFE (46) are equivalent. Basically, we have to show that the functions defined below, G and G , are equal: ð1 Þ 2; 1 þ ; 1 G ¼ F x 3 2 2 ; 2 ð2 Þ 1; 1 þ ; 1 (A1) 3 F2 x ; 2 ; 2 and. 026003-16.

(17) FEYNMAN DIAGRAMS AND A COMBINATION OF THE . . .. G ¼ 2 F1. . 1 þ ; 1 x : 2 . (A2). The variable x represents the quotient ( ms2 ). In order to prove the equivalence of both solutions, we need to remind the series expression for an hypergeometric function: 1 X ða1 Þk . . . ðaq Þk xk fag ; (A3) F x ¼ q q1 fbg ðb Þ . . . ðbq1 Þk k! k¼0 1 k then we can rewrite (A1) as a series, as follows: 1 ð1 Þ X ð2Þk ð1 þ Þk ð1 Þk xk G ¼ ð2 Þ k¼0 ð2 Þk ð2 Þk k! 1 X ð1Þk ð1 þ Þk ð1 Þk xk ð2 Þk ð2 Þk k! k¼0 X ð1 Þ 1 ð1 þ Þk ð1 Þk xk : ½ð2Þk ð1Þk ¼ ð2 Þ k¼0 ð2 Þk ð2 Þk k! (A4). PHYSICAL REVIEW D 81, 026003 (2010). which implies that (A4) reduces to the following equation: 1 ð1 Þ X ð1Þk ð1 þ Þk ð1 Þk xk : ½1 þ k G ¼ ð2 Þ k¼0 ð2 Þk ð2 Þk k! (A6) Then, we can transform the expression (1 þ k ) as follows: ð1 þ k Þ ¼. ð2 þ k Þ ð2 Þð2 Þk ¼ : ð1 þ k Þ ð1 Þð1 Þk. By replacing this result in (A6), some algebra shows finally that X 1 ð1Þk ð1 þ Þk xk G ¼ ð2 Þk k! k¼0 ! 1 þ ; 1 ¼ 2 F1 (A7) x ¼ G : 2 . ð2 þ kÞ ¼ ð1 þ kÞð1 þ kÞ ¼ ð1 þ kÞð1Þk ; (A5) ð2Þ. Although the equivalence has been shown for the kinematical region j ms2 j < 1, it is also valid for the region j ms2 j > 1, since both regions are related through an analytical continuation.. [1] N. Usyukina, Theor. Mat. Fiz. 22, 210 (1975). [2] E. Boos and A. Davydychev, Theor. Math. Phys. 89, 1052 (1991). [3] C. Anastasiou, E. W. N. Glover, and C. Oleari, Nucl. Phys. B572, 307 (2000); B565, 445 (2000). [4] A. T. Suzuki, E. S. Santos, and A. G. M. Schmidt, Eur. Phys. J. C 26, 125 (2002); J. Phys. A 36, 4465 (2003). [5] A. T. Suzuki and A. G. M. Schmidt, J. Phys. A 31, 8023 (1998); Can. J. Phys. 78, 769 (2000); J. High Energy Phys. 09 (1997) 002; Eur. Phys. J. C 5, 175 (1998); Phys. Rev. D 58, 047701 (1998); J. Phys. A 35, 151 (2002). [6] I. G. Halliday and R. M. Ricotta, Phys. Lett. B 193, 241 (1987). [7] I. Gonzalez and I. Schmidt, Nucl. Phys. B769, 124 (2007). [8] F. V. Tkachov, Phys. Lett. 100B, 65 (1981); K. G. Chetyrkin and F. V. Tkachov, Nucl. Phys. B192, 159 (1981).. [9] W. N. Bailey, Generalized Hypergeometric Functions (Stechert-Hafner Service Agency, New York, 1964). [10] L. J. Slater, Generalized Hypergeometric Functions (Cambridge University Press, Cambridge, United Kingdom, 1966). [11] L. S. Gradshteyn and L. M. Ryzhik, Table of Integrals, Series, and Products (Academic Press, New York, 2000), 6th ed.. [12] H. Exton, Multiple Hypergeometric Functions and Applications (Ellis Horwood, Westergate, United Kingdom, 1976). [13] V. A. Smirnov, Evaluating Feynman Integrals (SpringerVerlag, Berlin, 2004). [14] I. Gonzalez and I. Schmidt, Phys. Rev. D 78, 086003 (2008). [15] D. J. Broadhurst, J. Fleischer, and O. V. Tarasov, Z. Phys. C 60, 287 (1993).. On the other side, we have that ð2Þk ¼. 026003-17.

(18)