One-loop pentagon integral with one offshell leg in d dimensions from differential equations in ε-form
OOne-loop pentagon integral with one offshell leg in d dimensionsfrom differential equations in (cid:15) -form Mikhail G. Kozlov
1, 21
Budker Institute of Nuclear Physics, Novosibirsk, 630090 Russia Novosibirsk State University, Novosibirsk, 630090 Russia ∗ Abstract
We apply differential equations technique to the calculation of the one-loop massless diagramwith one offshell legs. Using reduction to (cid:15) -form, we managed to obtain a simple one-fold integralrepresentation exact in space-time dimensionality. Expansion of the obtained result in (cid:15) andanalytical continuation to physical region are discussed. ∗ [email protected]; This work is supported by the RFBR grants No. 16-32-60033 and 16-02-00888 a r X i v : . [ h e p - ph ] D ec . INTRODUCTION In this paper we consider the one-loop integral with massless internal lines and one off-shell leg exactly in the dimension of space-time, which we call below the pentagon integral .This type of integrals arises for example in a system of differential equations on one loopmaster integrals with massless internal lines and with more than five legs, for example onmassless on-shell hexagon integral. Also exact expression for the one-loop pentagon integralis needed to calculate the MHV multiloop amplitudes in the planar limit of perturbationtheory, using Bern–Dixon–Smirnov (BDS) ansatz [1].The expression for pentagon integral has been known in the expansion in (cid:15) up to (cid:15) for a long time [2]. In Ref. [2], it was shown that up to (cid:15) order the pentagon integral in d = 4 − (cid:15) can be expressed via box integrals with one and two off shell legs. Also in [2]it was shown that higher order terms are related to the expansion of the same pentagonintegral in d = 6 − (cid:15) dimensions.In the previous paper [3] we considered the pentagon integral with all on-shell legs exactlyin d dimensions. In this paper we use the same algorithm of calculation as in [3].The result has a one-fold integral representation. We consider the pentagon integral ina subregion of Euclidean region and then perform the analytical continuation to all regionswith real invariants.In the second chapter, we introduce the notation and present the result for the Pentagonintegral. The third chapter is dedicated to bringing the system of differential equations onthe Pentagon to the e-form and obtain solutions of this system. In the fourth chapter, wediscuss the analytic continuation to physical region of invariants. The fifth chapter is theconclusion. In Appendix A, we discuss the computation of the triangle master integral withall off-shell legs. In Appendix B and C we present the calculation of the easy box and thehard box master integrals respectively. II. DEFINITIONS AND RESULT
The pentagon integral is defined as P ( d ) ( s , s , s , s , s ) = (cid:90) d d li π d/ (cid:81) n =0 ( l n + i , (1)2here l n = l − n (cid:88) i =1 p i , (2)and p i are the incoming momenta, p = m = µ , p i = 0 ( i = 2 , . . . , , (cid:88) i =1 p i = 0 , (3)and the invariants s i are defined as s n = ( p n − + p n +2 ) , s ≡ ( s , s , s , s , s , µ ) . (4)Here we adopt cyclic convention for indices, e.g. s n ± = s n . We find it convenient to use thefollowing notation r n = − s µ + (cid:88) i =0 ( − i s n + i s n + i +1 , n = 1 , , ,r = − µ s s s + s µ + (cid:88) i =0 ( − i s i s i , r = − µ s s s + s µ + (cid:88) i =0 ( − i s i s i , ∆ = det (2 p i · p j | i,j =1 ,... ) , S = 4 s s ( s s − s µ ) s ∆ , ∆ = 4 s s − ( µ − s − s ) , S = 4 s s µ ∆ . (5)Using techniques described in detail in the next sections, we obtain the following result (6)exact in d representation for P (6 − (cid:15) ) for real s i in non-equivalent regions R i . These regionsdefined by the list (cid:0) sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( µ ) (cid:1) . Solution for otherregions can be obtained using symmetry s ↔ s , s ↔ s and the identity P (6 − (cid:15) ) ( s ) =3 iπ(cid:15) (cid:104) P (6 − (cid:15) ) ( − s ) (cid:105) ∗ following from Feynman parametrization. P (6 − (cid:15) ) ( s ) = 2 C ( (cid:15) ) (cid:15) (cid:20) − (cid:88) i =1 , , ( − s i ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b i ( t ) (cid:18) arctan r i b i ( t ) − arctan g i ( t ) b i ( t ) (cid:19) ++ ( − s ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:18) arctan r b ( t ) − arctan g ( t ) b ( t ) (cid:19) ++ ( − s ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:18) arctan r b ( t ) − arctan g ( t ) b ( t ) (cid:19) −− ( − µ ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:18) arctan r b ( t ) − arctan g ( t ) b ( t ) (cid:19) ++ Θ( − s )Θ( − s )Θ( − µ ) (cid:16) Θ( s − s − µ ) + Θ( s − s − µ ) + Θ( µ − s − s ) − (cid:17) ×× ( − S ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:18) arctan g − ( t ) b ( t ) − arctan g + ( t ) b ( t ) (cid:19) + ( − S ) − (cid:15) √ ∆ π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) Θ(∆) B ( s ) (cid:21) , (6)where b i ( t ) = (cid:112) ∆( S t/s i − , b ( t ) = (cid:112) ∆( S t/µ − , b ( t ) = (cid:112) ∆( S t/S − , (7) C ( (cid:15) ) = Γ (1 − (cid:15) )Γ(1 + (cid:15) )Γ(1 − (cid:15) ) , (8) g ( t ) = r + 2 s s (1 − t ) ,g ( t ) = r + (1 − t ) 2 s ( s s − s µ ) s − µ , g ( t ) = r + (1 − t ) 2 s ( s s − s µ ) s − µ ,g ( t ) = r + ( s s − s µ ) µ ( s − s − µ ) (cid:20) − (cid:115) s µ ( s − s − µ ) ( t − (cid:21) , (9) g ( t ) = r + ( s s − s µ ) µ ( s − s − µ ) (cid:20) − (cid:115) s µ ( s − s − µ ) ( t − (cid:21) ,g ( t ) = r + s ( µ − s − s ) (cid:20) − (cid:115) s s ( µ − s − s ) ( t − (cid:21) ,g ± ( t ) = − s s + s s − s s ± s (cid:112) ∆ ( t − , (10)and factor B ( s ) is shown in Table I. By Re (cid:0) arctan( √ x/r ) / √ x (cid:1) we understand the function f ( x, r ) = √ x arctan( √ xr ) , x > √− x log (cid:12)(cid:12)(cid:12) r + √− xr −√− x (cid:12)(cid:12)(cid:12) , x < . (11)4 egion B ( s ) R ( − − − − −− ) Θ( s s − s µ ) (cid:104) Θ( µ − s ) + Θ( µ − s ) + Θ( σ ) + Θ( σ ) − (cid:105) R (+ − − − −− ) 0 R ( − + − − −− ) Θ( s µ − s s ) (cid:104) Θ( µ − s ) − Θ( µ − s ) + Θ( S − S )Θ( s s − s s + s s ) − (cid:105) R ( − − + − −− ) 0 R ( − − − − − +) 1 R (+ + − − −− ) Θ( s − µ ) − Θ( s − µ ) + Θ( S − S )Θ (cid:0) s s ( s − s ) /µ + s s − s s (cid:1) R (+ − + − −− ) Θ( s µ − s s ) (cid:104) Θ( s − µ ) − Θ( σ ) − Θ( σ ) + 1 (cid:105) R (+ − − − − +) 0 R ( − + − − − +) 0 R ( − − + − − +) Θ( s s − s µ ) + Θ( s µ − s s ) (cid:0) Θ( s − µ ) − (cid:1) R ( − + + − −− ) − Θ( s µ − s s ) − Θ( s s − s µ ) (cid:0) Θ( s − µ ) − (cid:1) R ( − − + + −− ) Θ( s s − s µ ) (cid:104) Θ( σ ) + Θ( σ ) (cid:105) R ( − + − + −− ) − Θ( s − µ ) + Θ( S − S )Θ (cid:0) s s ( s − s ) /µ + s s − s s (cid:1) R ( − + − − + − ) Θ( s s − s µ ) (cid:104) Θ( s − µ ) + Θ( s − µ ) (cid:105) R (+ + + − −− ) Θ( s s − s µ )Θ( µ − s ) − Θ( s µ − s s ) R (+ + − + −− ) Θ( s s − s µ ) (cid:104) − Θ( µ − s ) + Θ( S − S )Θ( s s − s s + s s ) (cid:105) R (+ + − − + − ) 1 R (+ + − − − +) Θ( s s − s µ ) + Θ( s µ − s s )Θ( s s − s s + s s ) R (+ − + + −− ) 0 R (+ − + − − +) 0TABLE I. Coefficient B ( s ) for non-equivalent regions R i . Each region is marked by the list (cid:0) sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( µ ) (cid:1) . In order to crosscheck our result, we have performed comparison with the results for thepentagon obtained using
Fiesta 4.1 [4] (we compared only two orders in (cid:15) ), and foundperfect agreement. 5
IG. 1. Pentagon, B easy box, B , hard boxes, B , boxes, T triangle and bubble integrals. III. DIFFERENTIAL EQUATIONS IN (cid:15) -FORM
In this section we consider integrals in d = 4 − (cid:15) dimensions in “Euclidean” region s < , s < , s < , s < , s < , µ < . We use integration by part (IBP) reduction, as implemented in
LiteRed package, Ref. [5],to obtain the system of partial differential equations for the pentagon integral P and twelvesimpler master integrals, see Fig. 1. Introducing the column-vector of master integrals J = (cid:0) P, B , B , B , B , B , T, R , R , R , R , R , R (cid:1) T , (12)we can represent the system of differential equations in the matrix form ∂∂s i J = M i ( s , (cid:15) ) J , i = 1 , . . . , ∂∂µ J = M ( s , (cid:15) ) J , (13)where M i ( s , (cid:15) ) are upper-triangular matrices of rational functions of s j , µ and (cid:15) .We know the simpler master integrals, which are the bubbles R i = R ( s i ) = (cid:90) d d liπ d/ l i +1 l i +3 = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) ( − s i ) − (cid:15) , i = 1 , . . . , ,R = R ( µ ) = (cid:90) d d liπ d/ l l = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) ( − µ ) − (cid:15) , (14)the triangle integral with all off-shell legs (see Appendix A) T = T ( s , s , µ ) = (cid:90) d d liπ d/ l l l , (15)6nd the box integrals B i = (cid:90) d d liπ d/ (cid:81) k =3 l i + k , (16)where B is the easy box (see appendix B), B , are hard boxes (see appendix C) and B , are boxes with one off-shell leg. The representation of the easy box and the boxes with oneoff-shell leg in terms of hypergeometric functions were obtained in [2, 6]. The representationof the box integral with one off-shell leg has the form B ( s, t, µ ) = 2 C ( (cid:15) ) (cid:15) s t (cid:26) ( − µ ) − (cid:15) F (cid:16) , − (cid:15) ; 1 − (cid:15) ; µ ( s + t − µ ) s t (cid:17) −− ( − s ) − (cid:15) F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s + t − µt (cid:17) − ( − t ) − (cid:15) F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s + t − µs (cid:17)(cid:27) . (17)In order to reduce the system of differential equations to (cid:15) -form we find the appropriatebasis P = C ( (cid:15) ) (cid:15) s s ( s s − s µ ) s (cid:18) √ ∆ (cid:101) P + 12 (cid:88) i =1 r i (cid:101) B i (cid:19) ,B = C ( (cid:15) ) (cid:15) ( s s − s µ ) (cid:101) B , B i = C ( (cid:15) ) (cid:15) s i +2 s i − (cid:101) B i , i = 1 , . . . , ,T = C ( (cid:15) ) (cid:15) √ ∆ (cid:101) T , R i = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) (cid:101) R i . (18)7he differential equations in the new basis can be written in d log (cid:15) -form d (cid:101) P = − (cid:15) (cid:26) (cid:101) P d (cid:16) log S (cid:17) + (cid:88) i =1 (cid:101) B i d (cid:16) arctanh a i (cid:17) − (cid:101) T d (cid:16) arctan y (cid:17) ++ 2 (cid:88) i =1 , , (cid:101) R i d (cid:16) arctanh a i − arctanh a i +2 − arctanh a i − (cid:17) ++ (cid:101) R d (cid:16) arctanh a − a − arctanh a (cid:17) ++ (cid:101) R d (cid:16) arctanh a − a − arctanh a (cid:17) ++ (cid:101) R d (cid:16) a + arctanh a + arctanh a (cid:17)(cid:27) ; d (cid:101) B = − (cid:15) (cid:26) (cid:101) B d log( − S (1)4 ) − (cid:101) R d log (cid:16) S (1)4 s − (cid:17) − (cid:101) R d log (cid:16) S (1)4 µ − (cid:17) ++ 2 (cid:101) R d log (cid:16) S (1)4 s − (cid:17) + 2 (cid:101) R d log (cid:16) S (1)4 s − (cid:17)(cid:27) ; d (cid:101) B = − (cid:15) (cid:40) (cid:101) B d log( − S (2)4 ) − (cid:101) T d arctan b + 2 (cid:101) R d log (cid:16) S (2)4 s − (cid:17) −− (cid:101) R d log 1 + b ( b − x ) + (cid:101) R d log 1 + b ( b + x ) + (cid:101) R d log 1 + b ( b + x ) (cid:41) ; d (cid:101) B = − (cid:15) (cid:26) (cid:101) B d log( − S (3)4 ) − (cid:101) R d log (cid:16) S (3)4 s − (cid:17) + 2 (cid:101) R d log (cid:16) S (3)4 s − (cid:17) ++ 2 (cid:101) R d log (cid:16) S (3)4 s − (cid:17)(cid:27) ; d (cid:101) T = − (cid:15) (cid:101) T d log (cid:16) − S (cid:17) − (cid:15) (cid:101) R d arctan x − (cid:15) (cid:101) R d arctan x − (cid:15) (cid:101) R d arctan x ; d (cid:101) R i = − (cid:15) (cid:101) R i d (cid:0) log s i (cid:1) , i = 1 , . . . , , d (cid:101) R = − (cid:15) (cid:101) R d (cid:0) log µ (cid:1) . (19)We also need the equations for (cid:101) B and (cid:101) B , which have the same form as the equations for (cid:101) B and (cid:101) B . In Eq. (19) we use the following denotations y = 1 √ ∆∆ (cid:16) − µ s + 2 µ s s − s s − µ s s + s s + µ s s − s s s + 2 µ s s ++ 2 µ s s + 2 s s s − s s s − µ s s − s s s − s s s − s s + s s (cid:17) , a i = r i √ ∆ ,S (1)4 = s s − s µ s + µ − s − s , S (2)4 = − s s ( s − µ )( s − s ) + s s , S (3)4 = s s s − s − s ,x = s − s − µ √ ∆ , x = s − s − µ √ ∆ , x = µ − s − s √ ∆ ,b = 2 s µ s √ ∆ + x , b = 2 s µ s √ ∆ + x . (20)8et us fix the region of invariants R where solutions for triangle and hard boxes are moresimple R = { s | s < , s < , s < , s < , s < , µ < , ∆ > , s s − s µ > , ∆ > , x > , x < , x > } . (21)Here the signs of ∆ and s s − s µ fix the sign of S .Let us now split the above differeintial system. Given a system d (cid:101) J = dM (cid:101) J , we schemat-ically depict the matrix dM by replacing each nonzero element with “ ∗ ”. For system (19)we have: dM = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
00 0 0 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
00 0 0 0 0 0 0 0 0 0 0 0 ∗ . (22)Then we interpret this schematic form as adjacency matrix of the directed graph, with “ ∗ ij ”denoting a directed edge i → j . In general, the node i is said to be an ancestor of the node j if there is a directed path from i to j . A leaf is a node which is not an ancestor of anyother node. To each leaf we associate the subgraph consisting of the leaf itself and of all itsancestors. For each such subgraph, we search for a solution of the original system havingthe form of the column vector with zeros put in all entries except the ones correspondingto the nodes of the subgraph [3]. The general solution of differential system is written asthe sum over different leaves. Here we have six leaves, R i ; i = 1 , . . . ,
6. We search for thesolution in the form (cid:101) J = (cid:88) i =1 (cid:101) J ( i ) , (23)9here (cid:101) J (1) = (cid:0) (cid:101) P (1) , (cid:101) B (1)1 , , (cid:101) B (1)3 , (cid:101) B (1)4 , , , (cid:101) R , , , , , (cid:1) T , (cid:101) J (2) = (cid:0) (cid:101) P (2) , , (cid:101) B (2)2 , , (cid:101) B (2)4 , (cid:101) B (2)5 , (cid:101) T (2) , , (cid:101) R , , , , (cid:1) T , (cid:101) J (3) = (cid:0) (cid:101) P (3) , (cid:101) B (3)1 , , (cid:101) B (3)3 , , (cid:101) B (3)5 , , , , (cid:101) R , , , (cid:1) T , (cid:101) J (4) = (cid:0) (cid:101) P (4) , (cid:101) B (4)1 , (cid:101) B (4)2 , , (cid:101) B (4)4 , , , , , , (cid:101) R , , (cid:1) T , (cid:101) J (5) = (cid:0) (cid:101) P (5) , , (cid:101) B (5)2 , (cid:101) B (5)3 , , (cid:101) B (5)5 , (cid:101) T (5) , , , , , (cid:101) R , (cid:1) T , (cid:101) J (6) = (cid:0) (cid:101) P (6) , (cid:101) B (6)1 , (cid:101) B (6)2 , , , (cid:101) B (6)5 , (cid:101) T (6) , , , , , , (cid:101) R (cid:1) T . (24) Equation for (cid:101) P ( k ) , k = 1 , , . Let us consider the differential equation for (cid:101) P ( k ) with k = 1 , ,
4. This equations is thesame as for the massless pentagon case [3] (in that work we had a slightly different basis): d (cid:101) P ( k ) = − (cid:15) (cid:26) (cid:101) P ( k ) d (cid:16) log S (cid:17) + (cid:88) i = k,k ± (cid:101) B ( k ) i d (cid:16) arctanh a i (cid:17) ++ 2 (cid:101) R k d (cid:16) arctanh a k − arctanh a k +2 − arctanh a k − (cid:17)(cid:27) . (25)From Eq. (17) we can identify (cid:101) B ( k ) i : (cid:101) B ( k ) i = 2( − ( k − i ) / ( − s k ) − (cid:15) Re F (cid:18) , − (cid:15) ; 1 − (cid:15) ; s k S (cid:16) − a i (cid:17)(cid:19) == 2( − ( k − i ) / ( − s k ) − (cid:15) (cid:40) − (cid:15) (1 − a i )Re (cid:90) ∞ dt t (cid:15) − t S/s k − a i + i (cid:41) . (26)Using the integral representation for (cid:101) B ( k ) i we arrive at the following differential equation for (cid:101) P ( k ) : d (cid:16) ( − S ) (cid:15) (cid:101) P ( k ) (cid:17) = H ( k ) k da k + H ( k ) k +2 da k +2 + H ( k ) k − da k − ,H ( k ) i = 2 (cid:15) ( − ( k − i ) / (cid:16) Ss k (cid:17) (cid:15) (cid:90) ∞ dt t (cid:15) − t S/s k − a i ,Ss k = 1 + a k − a k +2 − a k ( a k − + a k +2 ) , k = 1 , , . (27)The right-hand side of (27) depends only on free dimensionless variables a n , ( n = i, i ± − S ) (cid:15) (cid:101) P ( k ) = (cid:90) a k − σ k ∞ H ( k ) k ( a, a k +2 , a k − ) da + g ( a k +2 , a k − , (cid:15) ) , (28)10here σ k = sign( a k +2 + a k − ). Next, we can check that g depends only on (cid:15) : ∂g ( a k +2 , a k − , (cid:15) ) ∂a k ± = H ( k ) k ± ( a k → − σ k ∞ ) = 2 (cid:15) (cid:15) − (cid:0) − a k ( a k +2 + a k − ) (cid:1) (cid:15) − (cid:12)(cid:12)(cid:12) a k →− σ k ∞ = 0 . (29)Substituting the explicit form of H ( i ) i , we have( − S ) (cid:15) (cid:101) P ( k ) = 2 (cid:15) Re (cid:90) a k − σ k ∞ da (cid:90) ∞ (cid:0) K ( a ) (cid:1) (cid:15) dt t (cid:15) − K ( a ) t − a + i g ( (cid:15) ) ,K ( a ) = 1 + a k − a k +2 − a ( a k − + a k +2 ) . (30)Note that K ( a ) > t → t/K ( a ), changing integration order we have and integrating over a we have (cid:101) P ( k ) = 2 (cid:15) ( − s k ) − (cid:15) √ ∆Re (cid:90) ∞ dt t (cid:15) − b k ( t ) (cid:26) arctan r k b k ( t ) − arctan g k ( t ) b k ( t ) (cid:27) , (31)where b k ( t ) = (cid:114) ∆ (cid:16) Ss k t − (cid:17) , g k ( t ) = r k + S ∆ s k ( r k +2 + r k − ) (1 − t ) . (32) Equation for (cid:101) P ( j ) , j = 2 , , . Recall that we are working in the region (21). The equation for (cid:101) P (2) (and equationsfor (cid:101) P (5) , (cid:101) P (6) ) is more complicated than the equation for (cid:101) P (1) and has four independentvariables a , a , a and y : d (cid:101) P (2) = − (cid:15) (cid:26) (cid:101) P (2) d (cid:16) log S (cid:17) + (cid:101) B (2)2 d (cid:16) arctanh a (cid:17) + (cid:101) B (2)5 d (cid:16) arctanh a (cid:17) + (cid:101) B (2)4 d (cid:16) arctanh a (cid:17) −− (cid:101) T (2) d (cid:16) arctan y (cid:17) + (cid:101) R d (cid:16) arctanh a − a − arctanh a (cid:17)(cid:27) , (33)where from equations (17), (C18) and (A8) we have (cid:101) T (2) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz , S = s (1 + x ) , (cid:101) B (2)2 = 2 (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) b + z + i , (cid:101) B (2)5 = 2 (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) b − z + i , (cid:101) B (2)4 = 2( − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s S (1 − a ) (cid:17) == 2( − s ) − (cid:15) (cid:40) − (cid:15) Re (cid:90) ∞ (1 − a ) dt t (cid:15) − t S/s − a + i (cid:41) . (34)11he native variables for the hard boxes and the triangle are b , b and x . We have toexpress them through independent variables a , a , a and y . From equalities Ss = (1 + x ) 1 − a b = (1 + x ) 1 − a b , y = b b − a a b a + a b , b + x b − x = a − a a + a (35)we can express b , b , x : b = a y + a (cid:115) (1 + y ) 1 − a − a , b = a y + a (cid:115) (1 + y ) 1 − a − a ,x = − a y + ( a − a − a + a a a ) (cid:115) y (1 − a )(1 − a ) . (36)The relationship between variables b , b , x and a , a , a , y can be obtained using Gr¨obnerbasis. Another way to derive the relationship is to use relations (35).Equations for (cid:101) P (5) and (cid:101) P (6) can be found from the equation for (cid:101) P (2) . The equation for (cid:101) P (5) we obtain by replacing a ↔ a , a → a (or x → x , b ↔ b ) in (33). The equationfor (cid:101) P (6) we derive by replacing (cid:101) P (2) → − (cid:101) P (6) , y → − y , a → − a and a → a (or x → x , b → − b ) in (33). Using expressions (34) we arrive at the following differential equation for (cid:101) P (2) : d (cid:16) ( − S ) (cid:15) (cid:101) P (2) (cid:17) = H dy + H da + H da + H da ,H = − (cid:15) y (cid:18) − a b (cid:19) (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz ,H = − (cid:15) (1 − a ) (cid:15) − (1 + b ) (cid:15) (cid:40) (1 + x ) (cid:15) (cid:15) + Re (cid:90) x + ∞ (1 + z ) (cid:15) dzb + z + i (cid:41) ,H = − (cid:15) (cid:18) (1 + x ) 1 − a b (cid:19) (cid:15) − Re (cid:90) ∞ t (cid:15) − dtt − (1 − a )(1+ b )(1 − a )(1+ x ) + i ,H = − (cid:15) (1 − a ) (cid:15) − (1 + b ) (cid:15) (cid:40) (1 + x ) (cid:15) (cid:15) + Re (cid:90) x + ∞ (1 + z ) (cid:15) dzb − z + i (cid:41) . (37)Let us integrate equation ∂∂a (cid:0) ( − S ) (cid:15) (cid:101) P (cid:1) = H (as simplest equation):( − S ) (cid:15) (cid:101) P (2) = (cid:90) a + ∞ H ( a , z, a , y ) dz + g ( a , a , y, (cid:15) ) (38)where g ( a , a , y, (cid:15) ) is some function to be fixed. It is easy to check that g depends only on (cid:15) : ∂g ( a , a , y, (cid:15) ) ∂a , = H , ( a → + ∞ ) = 0 , ∂g ( a , a , y, (cid:15) ) ∂y = H ( a → + ∞ ) = 0 . (39)12he limit a → + ∞ corresponds to the limit x → + ∞ x (cid:12)(cid:12)(cid:12) a → + ∞ ≈ ( − µ ) √ ∆( s s − µ s ) √ ∆ a → + ∞ . (40)Substituting the explicit form of H , we have( − S ) (cid:15) (cid:101) P (2) = − (cid:15) (cid:90) a + ∞ dz Re (cid:90) ∞ (cid:0) t K ( z ) (cid:1) (cid:15) − dtt − (1 − z ) /K ( z ) + i g ( (cid:15) ) . (41)Here we introduce function K ( z ) (this function is an analog of function K ( a ) in the [3]): K ( z ) = K ( z ) (cid:0) A ( z − z ) (cid:1) , K ( a ) ≡ Ss = 1 − a b (1 + x ) ,K ( z ) = 1 − a b = s ( s s − µ s )∆ µ ∆ , A = b + b a + a = µ √ ∆( s s − µ s ) √ ∆ ,x = − A ( a − z ) , z = a b − a b b + b = s s ( s − s ) + µ ( s s − s s ) µ √ ∆ . (42)Note that K ( z ) > t → t/K ( z ) andchanging the order of integration we have (cid:101) P (2) = 2 (cid:15) ( − S ) − (cid:15) (cid:90) ∞ K ( a ) dt t (cid:15) − Re (cid:90) a K − ( t ) dzt − z + i g ( (cid:15) )( − S ) − (cid:15) ,K − ± ( t ) = z ± (cid:115) A (cid:20) tK ( z ) − (cid:21) . (43)After integration (43) over z and change of variable t → tK ( a ) we obtain (cid:101) P (2) = 2 (cid:15) ( − s ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − (cid:112) K ( a ) t − (cid:18) arctan a (cid:112) K ( a ) t − −− arctan a + x (cid:0) − (cid:112) t + ( t − /x (cid:1) /A (cid:112) K ( a ) t − (cid:19) + g ( (cid:15) )( − S ) − (cid:15) == 2 (cid:15) ( − s ) − (cid:15) √ ∆Re (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:18) arctan r b ( t ) − arctan g ( t ) b ( t ) (cid:19) + g ( (cid:15) )( − S ) − (cid:15) ,b ( t ) = (cid:114) ∆ (cid:16) Ss t − (cid:17) , g ( t ) = r + x r + r b + b (cid:18) − (cid:115) t + t − x (cid:19) , (44)where b ( t ) is defined in (32). It is easy to see that this solution has the same form as (cid:101) P (1) ( the difference is only in the function g ( t ), which is not linear with respect to t ). Let usrepresent the solution in the form (cid:101) P (2) = ( − s ) − (cid:15) F ( a , a , a , y ) . (45)13hen solutions for (cid:101) P (5) and (cid:101) P (6) are (cid:101) P (5) = ( − s ) − (cid:15) F ( a , a , a , y ) , (cid:101) P (6) = − ( − µ ) − (cid:15) F ( − a , a , a , − y ) . (46)Solution for (cid:101) P (5) we can be obtained by replacing s ↔ s and s ↔ s in (44). For (cid:101) P (6) replacement a → a , a → − a and y → − y corresponds to x → x and b → − b . Constant g ( (cid:15) ) Using equations (18), (23) we can write the solution for the pentagon integral in the form P = C ( (cid:15) ) √ ∆ (cid:15) s s ( s s − s µ ) s (cid:18) (cid:88) i =1 (cid:101) P ( i ) + g ( (cid:15) )( − S ) − (cid:15) (cid:19) + r s s s B + s s s s − s µ (cid:88) i =2 r i s i − s i s i +1 B i . (47)Let us fix the constant g ( (cid:15) ). Note that the condition ∆ = 0 implies the existence of a linearrelation between p , . . . , p , therefore we can express the pentagon integral at ∆ = 0 in termsof the box integrals. Moreover, ∆ = 0 is not a branching ponit of P . The only way to satisfythese two conditions is to require that (cid:88) i =1 (cid:101) P ( i ) + g ( (cid:15) )( − S ) − (cid:15) (cid:12)(cid:12)(cid:12) ∆ → → . (48)Consider limit ∆ → s → s s − s s + s s = 0. In the limit s → (cid:101) P (1) ∼ (cid:101) P (5) ∼ (cid:101) P (6) ∼ ∆ − (cid:15) → , (cid:101) P (2) → (cid:15) π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) , (cid:101) P (3) → − (cid:15) π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) Θ( µ − s ) , (cid:101) P (4) → − (cid:15) π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) Θ( µ − s ) . (49)Therefore from Eq. (48) we obtain g ( (cid:15) ) = 2 (cid:15) π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) (cid:16) Θ( µ − s ) + Θ( µ − s ) − (cid:17) . (50) Solution for d = 6 − (cid:15) Let us consider now the dimensional recurrence relation P (6 − (cid:15) ) = s s ( s s − s µ ) s (cid:15) ∆ (cid:32) P (4 − (cid:15) ) − r s s s B (4 − (cid:15) )1 − s s s s − s µ (cid:88) i =2 r i s i − s i s i +1 B (4 − (cid:15) ) i (cid:33) . (51)14his relation can be easily obtained with the LiteRed [5]. Comparing (47) and (51), we get P (6 − (cid:15) ) = C ( (cid:15) ) (cid:15) √ ∆ (cid:20) (cid:88) i =1 (cid:101) P ( i ) + 2 (cid:15) π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) (cid:16) Θ( µ − s ) + Θ( µ − s ) − (cid:17)(cid:21) . (52)Next, we write the solution for the pentagon in d = 6 − (cid:15) in the region R (21) P (6 − (cid:15) ) = 2 C ( (cid:15) ) (cid:15) (cid:20) (cid:88) i =1 (cid:98) P i − H ( s , (cid:15) ) (cid:16) Θ( µ − s ) + Θ( µ − s ) − (cid:17)(cid:21) , (53)where H ( s , (cid:15) ) is the homogenius solution of the differential equation for the pentagon: H ( s , (cid:15) ) = π Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) ( − S ) − (cid:15) √ ∆ , (54)and (cid:98) P i are (cid:98) P i = ( − α i ( − s i ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − b i ( t ) (cid:20) arctan r k ( i ) b i ( t ) − arctan g i ( t ) b i ( t ) (cid:21) ,α i = i = 2 , i = 1 , , , , k ( i ) = i, i = 1 , , k (2) = 4 , k (5) = 3 , k (6) = 1 , (55) g i ( t ) = r i + 4 s s ( s s − s µ ) s s i ( r i +2 + r i − ) (1 − t ) , i = 1 , , ,g ( t ) = r + ( s s − s µ )( s − s − µ ) µ (cid:20) − (cid:115) s µ ( s − s − µ ) ( t − (cid:21) ,g ( t ) = r + ( s s − s µ )( s − s − µ ) µ (cid:20) − (cid:115) s µ ( s − s − µ ) ( t − (cid:21) ,g ( t ) = r + s ( µ − s − s ) (cid:20) − (cid:115) s s ( µ − s − s ) ( t − (cid:21) . (56)Here we assume that s ≡ µ and b i ( t ) = (cid:113) ∆ (cid:0) S t/s i − (cid:1) . IV. ANALYTICAL CONTINUATION
The real part of the integrals (55) is equivalent to half of the sum of the integrals over acontour passing above the singular points and the contour passing under the singular points.Let us introduce a new representation for integrand (55): (cid:98) P i = ( − s i ) − (cid:15) Re (cid:90) ∞ dt t (cid:15) − W i ( s , t )Re W i ( s , t ) = 12 (cid:88) ± (cid:26) r k ( i ) f (cid:18) r k ( i ) ∆ Ss i ( t ± i − ∆ (cid:19) − g i ( t ± i f (cid:18) g i ( t ± i ∆ Ss i ( t ± i − ∆ (cid:19)(cid:27) (57)15here function f ( z ) = √ z arctan( √ z ) is defined on the complex plane with a cut going from −∞ to −
1. The Riemann surface, corresponding to the multivaluated function F ( z ) [3]with the main branch defined by F (0) ( z ) = f ( z ), is glued of set of sheets numbered by n ∈ Z with two cuts. On the n -th sheet the function is defined as F ( n ) ( z ) = − √− z √− z − √− z + iπn √− z , n ∈ Z ,F ( n ) ( x ± i
0) = √ x arctan √ x ± πn √ z = F ( − n ) ( x ∓ i , x > − √− x ln √− x √− x − + iπ ( n ± / √− x = F ( n ± ( x ∓ i , x < − . (58)The integrand of (57) has the following branching points on the real axis of t : • t = 0 is a branching point of the t (cid:15) − , • t ∗ i = s i S , i = 2 , , g i ( t ) , i = 2 , , • t ai = s i S (cid:16) − r k ( i ) ∆ (cid:17) , where the argument of the first function in (57) becomes − • t bi and t ci : g i ( t bi,ci ) = − ∆ (cid:16) Ss i t bi,ci − (cid:17) , where the argument of the second functionin (57) becomes − t bi and t ci for i = 2 , , g i ( t ) , i = 2 , , • t i : g i ( t i ) = 0, where the argument of the second function in (57) becomes 0, • t ∞ i = s i S , where the arguments of both functions in (57) become ∞ .We will carry out an analytic continuation from the region R (21) to the region of interestby the paths lying in the region D = { s | Im s i (cid:62) } as in [3]. Analytical continuation in Euclidean region
Let us now discuss the analytical continuation of the result obtained in the region (21)to the whole Euclidean region.Consider the analytical continuation of (cid:98) P integral from region x > x > x < x >
0. We put s − s − µ = re iφ and change φ form π to 0 taking r sufficiently small16 IG. 2. The figure shows the movement of the singular point t ∗ and the transformation of integra-tion contour. so that the singular point t ∗ was the closest singular point to t = 1. Then, only one singularpoint t ∗ is moving around point t = 1, see Fig 2:( s − s − µ ) + 4 s µ ( t ∗ −
1) = 0 , t ∗ = 1 − r s µ e iφ . (59)From Fig. 2 you can see how the path of integration changes, hence our integral transformsto (cid:98) P → (cid:98) P + ( − s ) − (cid:15) (cid:90) ∞ t ∗ dt t (cid:15) − b ( t ) (cid:20) arctan g ( t ) b ( t ) − arctan ¯ g ( t ) b ( t ) (cid:21) , (60)where ¯ g ( t ) has a different sign before the square root in contrast to g ( t )¯ g ( t ) = r + ( s s − s µ )( s − s − µ ) µ (cid:20) (cid:115) s µ ( s − s − µ ) ( t − (cid:21) . (61)The additional integral in (60) we can rewrite in the following form( − s ) − (cid:15) (cid:90) ∞ t ∗ dt t (cid:15) − b ( t ) (cid:20) arctan g ( t ) b ( t ) − arctan ¯ g ( t ) b ( t ) (cid:21) == ( − S ) − (cid:15) (cid:90) ∞ dt t (cid:15) − b ( t ) (cid:20) arctan g − ( t ) b ( t ) − arctan g + ( t ) b ( t ) (cid:21) − H ( s , (cid:15) ) = G ( s , (cid:15) ) − H ( s , (cid:15) ) (62)where we introduce denotations b ( t ) = (cid:114) ∆ (cid:16) SS t − (cid:17) , g ± = − s s + s s − s s ± s (cid:112) ∆ ( t − . (63)We can generalize the considered case to analytic continuation for any sign of x , x , x inthe Euclidean region. In transition from a region of one sign x i to the region with the othersign of x i , the singular point t ∗ i = s i /S completely bypasses the point t = 1 meshing with17he contour of integration as shown in Figure 2. Hence we can write our solution for anysign of σ i = sign( x i ) for Euclidean region: P = 2 C ( (cid:15) ) (cid:15) (cid:26) (cid:88) i =1 (cid:98) P i + (cid:0) Θ( σ ) − Θ( − σ ) − Θ( − σ ) (cid:1) G ( s , (cid:15) )++ H ( s , (cid:15) ) (cid:16) Θ( − σ ) − Θ( σ ) − (cid:2) Θ( µ − s ) + Θ( µ − s ) − (cid:3)(cid:17)(cid:27) . (64)Let us consider analytical continuation to the region with s s − s µ < > R , µ > s , µ > s and go to s s − s µ < s = s s µ − re iφ ,where φ changes from 0 to − π . Hence we have (cid:98) P → (cid:98) P + iπ − s ) − (cid:15) (cid:90) + ∞ t a dt ( t − i (cid:15) − (cid:114) ∆ (cid:16) − Ss t (cid:17) , (cid:98) P → (cid:98) P − iπ − s ) − (cid:15) (cid:90) + ∞ t b dt ( t − i (cid:15) − (cid:114) ∆ (cid:16) − Ss t (cid:17) , (cid:98) P → (cid:98) P − iπ − s ) − (cid:15) (cid:90) + ∞ t c dt ( t − i (cid:15) − (cid:114) ∆ (cid:16) − Ss t (cid:17) , (cid:98) P → (cid:98) P , (cid:98) P → (cid:98) P , (cid:98) P → (cid:98) P + iπ ( − µ ) − (cid:15) (cid:90) ∞ t ∞ dt ( t − i (cid:15) − (cid:114) ∆ (cid:16) − Sµ t (cid:17) − i π − µ ) − (cid:15) (cid:90) t a + ∞ dt t (cid:15) − (cid:114) ∆ (cid:16) − Sµ t (cid:17) . (65)Using relations t a µ = t a s = t b s = t c s ,iπ ( − s i ) − (cid:15) (cid:90) ∞ t ∞ i ( t + i (cid:15) − dt (cid:114) ∆ (cid:16) − Ss i t (cid:17) = π e iπ(cid:15) S − (cid:15) √ ∆ Γ(1 / − (cid:15) )Γ(1 − (cid:15) ) = H ( s , (cid:15) ) , (66)the sum of (cid:98) P i in (65) is transformed to (cid:88) i =1 (cid:98) P i → (cid:88) i =1 (cid:98) P i + H ( s , (cid:15) ) , (cid:98) P → (cid:88) i =1 (cid:98) P i . (67)Therefore all terms with the homogenius solution vanish. For other subregions of Euclideanregion analytical continuation is performed in the same way. Hence we can write the solutionfor the whole Euclidean region P (6 − (cid:15) ) = 2 C ( (cid:15) ) (cid:15) (cid:18) (cid:88) i =1 (cid:98) P i + (cid:16) Θ( σ ) − Θ( − σ ) − Θ( σ ) (cid:17) G ( s , (cid:15) )++ Θ( s s − s µ ) (cid:16) Θ( − σ ) − Θ( σ ) − Θ( µ − s ) − Θ( µ − s ) + 1 (cid:17) H ( s , (cid:15) ) (cid:19) . (68)18 plus 1 plus 2 plus 3 plus R ( − − − − −− ) R (+ − − − −− ) R ( − + − − −− ) R ( − − + − −− ) R ( − − − − − +) R (+ + − − −− ) R (+ − + − −− ) R (+ − − − − +) R ( − + − − − +) R ( − − + − − +) R ( − + + − −− ) R ( − − + + −− ) R ( − + − + −− ) R ( − + − − + − ) R (+ + + − −− ) R (+ + − + −− ) R (+ + − − + − ) R (+ + − − − +) R (+ − + + −− ) R (+ − + − − +)TABLE II. 20 non-equivalent regions of invariants. Analytical continuation to other regions
Here we consider analytical continuation from Euclidean region to regions with positivesign of invariants. We have 2 regions of invariants with sign + or − , but we have symmetryof the integral and the identity P ( s , s , s , s , s , µ ) = P ( s , s , s , s , s , µ ) , P (6 − (cid:15) ) ( s ) = e iπ(cid:15) (cid:104) P (6 − (cid:15) ) ( − s ) (cid:105) ∗ . (69)Therefore there are 20 non-equivalent regions of invariants (see table II). In the table we usethe denotaion for regions (cid:0) sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( s ) , sign( µ ) (cid:1) .Let us consider the analytic continuation into the most interesting region with s >
0. Westart from Euclidean region with x > , x < , x >
0, ∆ > s s − s µ <
0, wherethere is no homogeneous solution. We put s = | s | e iφ and change φ from π to 0. Whilechanging φ , we track the motion of the singular points t ai , t bi , t ci , t i , t ∞ i , t ∗ i and deform theintegration contours over t in such a way that they do not cross these singular points (and19lso t = 0 ). In final position, when φ = 0 the integrals are written as (cid:98) P → (cid:98) P + π − s ) − (cid:15) √ ∆ (cid:18) i (cid:90) t ∞ t b dt t (cid:15) − (cid:112) − S/s t + (cid:90) ∞ t ∞ dt t (cid:15) − (cid:112) S/s t − (cid:19) , (cid:98) P → (cid:98) P , (cid:98) P → (cid:98) P + π − s ) − (cid:15) √ ∆ (cid:18) − i (cid:90) t ∞ t b dt t (cid:15) − (cid:112) − S/s t − (cid:90) ∞ t ∞ dt t (cid:15) − (cid:112) S/s t − (cid:19) , (cid:98) P → (cid:98) P + π − s ) − (cid:15) √ ∆ (cid:18) − i (cid:90) t ∞ t a dt t (cid:15) − (cid:112) − S/s t + (cid:90) ∞ t ∞ dt t (cid:15) − (cid:112) S/s t − (cid:19) , (cid:98) P → (cid:98) P + iπ − s ) − (cid:15) √ ∆ (cid:90) t ∞ t b dt t (cid:15) − (cid:112) − S/s t − G ( s , (cid:15) ) , (cid:98) P → (cid:98) P − π − µ ) − (cid:15) √ ∆ (cid:90) ∞ t ∞ dt t (cid:15) − (cid:112) S/µ t − G ( s , (cid:15) ) . (70)Using the relations t a s = t c s = t b s = t a µ = s + s − s − µ s s − s µ ; t b s = t a s = t a s = s + s − s s s ; t c s = t a s = t a s = s + s − s s s ; t b s = t b s = t b s = t b µ = s s + ( µ − s )( s − s ) s s ; t c s = t c s = t c s = t c µ = s s + ( µ − s )( s − s ) s s , (71)the sum of the terms in (70) is transformed to (cid:80) i =1 (cid:98) P i without the homogeneous solution.Analytical continuation in other regions can be performed in a similar way. Since there area lot of regions, it is a difficult task. To simplify the analytical continuation it is possibleto use the fact that the pentagon must be finite for d > (cid:15) →
0. Then we look for asolution in an arbitrary domain in the form of (cid:98) P = (cid:88) i =1 (cid:98) P i + aG ( s , (cid:15) ) + b Θ(∆) H ( s , (cid:15) ) , (72)where coefficients a, b are integers. One can pick coefficients a and b with the help ofnumerical calculation of the formula (72) and demanding that the result should be of order (cid:15) . In each non-equivalent region (see table I or II) there are subregions defined by thefollowing thresholds (signs of the following expressions) • sign( s s − s µ ) affects position of t ∞ i singular points, • sign( s − µ ) and sign( s − µ ) affect the location of the singular points in the (cid:98) P and (cid:98) P , 20 IG. 3. Schematic representation of the Triangle master integral • sign( s − s − µ ), sign( s − s − µ ) and sign( µ − s − s ) affect the location of t ∗ i singular points, • S − S > s s − s s + s s ) and sign( s s ( s − s ) /µ + s s − s s )determine the possibility that the integration contour crosses the cut from singularpoint t ∞ i .We have to pick the coefficients a and b for each subregion of region R i . The result foranalytical continuation into the different regions of invariants can be found in the table I. V. CONCLUSION
In this work we have calculated the pentagon integral using a system of differentialequations that was solved by bringing it to the (cid:15) -form. Solution for the integral is a sumof one fold integrals. In dimension d > (cid:15) expansion.
ACKNOWLEDGMENTS
This work is supported by the RFBR grants No. 16-32-60033 and 16-02-00888.21 ppendix A: Triangle integral
We use IBP reduction to obtain the system of partial differential equations for the triangleintegral T and three simpler master integrals R , R , R , where R , = R ( s , ) , R = R ( µ ) . (A1)For the triangle integral we introduce the column-vector J T = ( T, R , R , R ) T . Next, we find an appropriate basis to reduce the system to (cid:15) -form T = C ( (cid:15) ) (cid:15) √ ∆ (cid:101) T , R i = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) (cid:101) R i . (A2)The differential equations in the new basis can be written in d log-form d (cid:101) T = − (cid:15) (cid:101) T d log (cid:16) − S (cid:17) − (cid:15) (cid:101) R d arctan x − (cid:15) (cid:101) R d arctan x − (cid:15) (cid:101) R d arctan x ,d (cid:101) R i = − (cid:15) (cid:101) R i d log s i , i = 2 , , d (cid:101) R = − (cid:15) (cid:101) R d log µ . (A3)where x = s − s − µ √ ∆ , x = s − s − µ √ ∆ , x = µ − s − s √ ∆ . (A4)We search solution in the form (cid:101) J T = (cid:101) J (1) T + (cid:101) J (2) T + (cid:101) J (3) T , where (cid:101) J (1) T = ( (cid:101) T (2) , (cid:101) R , , T , (cid:101) J (2) T = ( (cid:101) T (5) , , (cid:101) R , T , (cid:101) J (3) T = ( (cid:101) T (6) , , (cid:101) R ) , . (A5)Therefore we arrive at the following differential equation for (cid:101) T (the other equations are thesame): d (cid:16) ( − S ) (cid:15) (cid:101) T (2) (cid:17) = − (cid:15) (cid:0) S /s (cid:1) (cid:15) d arctan x . (A6)The right-hand side of Eq. (A6) depends only on x . In particular S = s (1 + x ) = s (1 + x ) = µ (1 + x ) . (A7)Let us consider the region of invariants x > , x < , x > , ∆ >
0. Then from thedifferential equations we have (cid:101) T (2) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz , (cid:101) T (5) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x −∞ (1 + z ) (cid:15) − dz (cid:101) T (6) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz (A8)22 IG. 4. Schematic representation of the easy box master integral
Our solution has the following form T = C ( (cid:15) ) (cid:15) √ ∆ (cid:18) (cid:101) T (2) + (cid:101) T (5) + (cid:101) T (6) + g ( (cid:15) )( − S ) − (cid:15) (cid:19) , (A9)where g ( (cid:15) ) is a constant of integration. In the limit ∆ → T (cid:12)(cid:12)(cid:12) ∆ → ≈ − C ( (cid:15) ) (cid:15) (1 − (cid:15) ) (cid:18) ( − s ) − (cid:15) s − s − µ − ( − s ) − (cid:15) s − s − µ + ( − µ ) − (cid:15) µ − s − s (cid:19) + C ( (cid:15) ) g ( (cid:15) ) (cid:15) √ ∆ (cid:16) − s s µ ∆ (cid:17) − (cid:15) . (A10)Therefore g ( (cid:15) ) = 0 and for x > , x < , x > T = C ( (cid:15) ) (cid:15) √ ∆ (cid:18) (cid:101) T (2) + (cid:101) T (5) + (cid:101) T (6) (cid:19) . (A11) Appendix B: Easy box
In this section we consider Easy box master integral in d = 4 − (cid:15) dimensions in “Eu-clidean” region s < , µ < , s < , s < J E = (cid:0) B , R ( s ) , R ( µ ) , R ( s ) , R ( s ) (cid:1) T . (B1)Next, we find an appropriate basis to reduce the system to (cid:15) -form B = C ( (cid:15) ) (cid:15) ( s s − s µ ) (cid:101) B , R ( s i ) = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) (cid:101) R ( s i ) , (cid:101) R ( s i ) = ( − s i ) − (cid:15) . (B2)23he differential equations in the new basis have (cid:15) -form and can be written in d log-form d (cid:101) B = − (cid:15) (cid:26) (cid:101) B d log( − S e ) − (cid:101) R ( s ) d log( y − − (cid:101) R ( µ ) d log( y − (cid:101) R ( s ) d log( y −
1) + 2 (cid:101) R ( s ) d log( y − (cid:27) , (B3)where ( − S e ) = s s − s µ s + s − s − µ = ( − s ) y = ( − µ ) y = ( − s ) y = ( − s ) y . (B4)We search solution in the form (cid:101) J e = (cid:80) i =1 (cid:101) J ( i ) where (cid:101) J (1) = ( (cid:101) B (1)1 , (cid:101) R ( s ) , , , , (cid:101) J (2) = ( (cid:101) B (2)1 , , (cid:101) R ( µ ) , , , (cid:101) J (3) = ( (cid:101) B (3)1 , , , (cid:101) R ( s ) , , (cid:101) J (4) = ( (cid:101) B (4)1 , , , , (cid:101) R ( s )) , (B5)hence for (cid:101) B (1)1 we have the following equation d (cid:16) ( − S e ) (cid:15) (cid:101) B (1)1 (cid:17) = 2 (cid:15) (cid:16) S e s (cid:17) (cid:15) d log( y −
1) = 2 (cid:15)y (cid:15) d log( y − . (B6)Then from the differential equation we have (cid:101) B (1)1 = 2 (cid:15) ( − S e ) − (cid:15) Re (cid:90) y ∞ y (cid:15) dyy − i − (cid:15) ( − s ) − (cid:15) Re (cid:90) ∞ z (cid:15) dzz − s /S e + i − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s S e (cid:17) , (B7) (cid:101) B (1)1 = 2( − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s S e (cid:17) , (cid:101) B (2)1 = 2( − µ ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; µ S e (cid:17) , (cid:101) B (3 , = − − s , ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s , S e (cid:17) . (B8)We can write the solution for the easy box integral in the form B = C ( (cid:15) ) (cid:15) ( s s − s µ ) (cid:18) (cid:88) i =1 (cid:101) B ( i )1 + g ( (cid:15) )( − S e ) − (cid:15) (cid:19) . (B9)Let us consider Gramm determinant for the easy box∆ e = det (cid:16) p i , p j ) (cid:17) = − s + s − s − µ )( s s − s µ ) . (B10)Therefore in the limit ∆ e → S e →
0) we have (cid:101) B (1)1 → − S e ) − (cid:15) Γ(1 − (cid:15) )Γ(1 + (cid:15) ) . (B11)24ince ∆ e = 0 is not a branching point of B hence g ( (cid:15) ) = 0.Solution for the easy box has the form B = 2 C ( (cid:15) ) (cid:15) ( s s − s µ ) (cid:40) ( − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s ( s + µ − s − s ) s s − s µ (cid:17) ++ ( − µ ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; µ ( s + µ − s − s ) s s − s µ (cid:17) −− ( − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s ( s + µ − s − s ) s s − s µ (cid:17) −− ( − s ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; s ( s + µ − s − s ) s s − s µ (cid:17)(cid:41) . (B12) Appendix C: Hard box
Our hard box integral depends on four variables s, t, m = µ , m = µ (see Fig. 5). Inthis section we consider integrals in d = 4 − (cid:15) dimensions in “Euclidean” region s < , t < , µ < , µ <
0. Let us introduce useful equalities∆ h = − s (cid:0) ( t − µ )( t − µ ) + s t (cid:1) , S h = 2 s t ∆ h , ∆ = 4 µ µ − ( s − µ − µ ) , S = 4 sµ µ ∆ . (C1)We use IBP reduction to obtain the system of partial differential equations for the hard boxintegral H and five simpler master integrals. Introducing the column-vector J H = (cid:0) H, T, R ( s ) , R ( t ) , R ( µ ) , R ( µ ) (cid:1) T , (C2)we may represent the system in the matrix form ∂ J H ∂s i = M i ( s , (cid:15) ) J H , s = ( s, t, µ , µ ) . (C3)We find an appropriate basis in order to reduce the system to (cid:15) -form: H = C ( (cid:15) ) (cid:15) s t (cid:101) H , T = C ( (cid:15) ) (cid:15) √ ∆ (cid:101) T , R ( s i ) = C ( (cid:15) ) (cid:15) (1 − (cid:15) ) (cid:101) R ( s i ) . (C4)The differential equation in the new basis can be written in d log-form d (cid:101) H = − (cid:15) (cid:16) (cid:101) Hd log( − S h ) − (cid:101) T d arctan b + 2( − t ) − (cid:15) d log( a − −− ( − s ) − (cid:15) d log 1 + b ( b − x ) + ( − µ ) − (cid:15) d log 1 + b ( b + x ) + ( − µ ) − (cid:15) d log 1 + b ( b + x ) (cid:17) , (C5)25 IG. 5. Schematic representation of the hard box master integral where we use denotations x = µ − µ − s √ ∆ , x = µ − µ − s √ ∆ , x = s − µ − µ √ ∆ ,b = t ( s − µ − µ ) + 2 µ µ t √ ∆ , a = s t ( t − µ )( t − µ ) + s t ,S h = ( − t ) a = ( − s ) 1 + x b = ( − µ ) 1 + x b = ( − µ ) 1 + x b = ( − S ) 11 + b . (C6)We search for the solution in the form (cid:101) J H = (cid:88) i =1 (cid:101) J ( i ) , (C7) (cid:101) J (3) = (cid:16) (cid:101) H (3) , (cid:101) T (3) , (cid:101) R ( s ) , , , (cid:17) , (cid:101) J (1) = (cid:16) (cid:101) H (1) , (cid:101) T (1) , , , (cid:101) R ( µ ) , (cid:17)(cid:101) J (2) = (cid:16) (cid:101) H (2) , (cid:101) T (2) , , , , (cid:101) R ( µ ) (cid:17) , (cid:101) J (4) = (cid:16) (cid:101) H (4) , , , (cid:101) R ( t ) , , (cid:17) . (C8)Let us consider the following region of invariants: ∆ > , x > , x < , x >
0. Thereforethe triangle integral has form (cid:101) T = (cid:88) i =1 (cid:101) T ( i ) , (cid:101) T (3) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz , (cid:101) T (1) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) − dz, (cid:101) T (2) = − (cid:15) ( − S ) − (cid:15) Re (cid:90) x −∞ (1 + z ) (cid:15) − dz , (C9)where S = 4 µ µ s µ µ − ( s − µ − µ ) = ( − s )(1 + x ) = ( − µ )(1 + x ) = ( − µ )(1 + x ) . (C10)26sing the expression for the triangle integral we arrive at the following differential equa-tions for (cid:101) H ( i ) d (cid:16) ( − S h ) (cid:15) (cid:101) H ( i ) (cid:17) = G ( i ) ( x i , b ) db + F ( i ) ( x i , b ) dx i , i = 1 , , , (C11) d (cid:16) ( − S h ) (cid:15) (cid:101) H (4) (cid:17) = − (cid:15)a (cid:15) d log( a − , (C12)where F (1 , ( x , , b ) = 2 (cid:15) (cid:18) x , b (cid:19) (cid:15) b − x , , F (3) ( x , b ) = − (cid:15) (cid:18) x b (cid:19) (cid:15) b + x , (C13) G (1 , ( x , , b ) = 2 (cid:15) (1 + b ) − − (cid:15) (1 + x , ) (cid:15) (cid:18) − (cid:15) (cid:90) x , ±∞ (1 + z ) (cid:15) − dz (1 + x , ) (cid:15) + b − b b − x , (cid:19) ,G (3) ( x , b ) = 2 (cid:15) (1 + b ) − − (cid:15) (1 + x ) (cid:15) (cid:18) − (cid:15) (cid:90) x + ∞ (1 + z ) (cid:15) − dz (1 + x ) (cid:15) + b − b b + x (cid:19) . (C14)It is easy to check that (C11) is a total differential ∂F ( i ) ( x i , b ) ∂b = ∂G ( i ) ( x i , b ) ∂x i (C15)as it should be. Then from differential equation ∂ (cid:101) H ( i ) ∂x i = F ( i ) , i = 1 , , − S h ) (cid:15) (cid:101) H ( i ) = (cid:90) x i sign( x i ) ∞ F ( i ) ( z, b ) dz + g ( b, (cid:15) ) . (C16)It is easy to check that g ( b, (cid:15) ) depends only on (cid:15) . Indeed, ∂∂b (cid:16) ( − S h ) (cid:15) (cid:101) H ( i ) (cid:17) = (cid:90) x i sign( x i ) ∞ ∂F ( i ) ( z, b ) ∂b dz + ∂g ( b, (cid:15) ) ∂b (cid:90) x i sign( x i ) ∞ ∂F ( i ) ( z, b ) ∂b dz = (cid:90) x i sign( x i ) ∞ ∂G ( i ) ( z, b ) ∂z dz = G ( i ) ( x i , b ) − G ( i ) ( x i → sign( x i ) ∞ , b ) = G ( i ) ( x i , b ) , (C17)where we use asymptotic G ( i ) ( x i → sign( x i ) ∞ , b ) = 2 (cid:15)b (1 + b ) − − (cid:15) | x i | (cid:15) →
0. Therefore g ( b, (cid:15) ) = g ( (cid:15) ). Substituting the explicit form of F ( i ) we have (cid:101) H (1) = 2 (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) dzb + z + i , (cid:101) H (2) = 2 (cid:15) ( − S ) − (cid:15) Re (cid:90) x −∞ (1 + z ) (cid:15) dzb + z + i (cid:101) H (3) = 2 (cid:15) ( − S ) − (cid:15) Re (cid:90) x + ∞ (1 + z ) (cid:15) dzb − z + i . (C18)27et us integrate the differential equation for (cid:101) H (4) (C12). We have (cid:101) H (4) = − (cid:15) ( − S h ) − (cid:15) Re (cid:90) a + ∞ z (cid:15) dzz − i − − t ) − (cid:15) (cid:18) − (cid:15)a Re (cid:90) ∞ z (cid:15) − dzz − /a + i (cid:19) == − − t ) − (cid:15) Re F (cid:16) , − (cid:15) ; 1 − (cid:15) ; 1 a + i (cid:17) . (C19)It is easy to see that (cid:101) H is expressed by the same function as in the easy box or the boxwith one off-shell leg. Using equations (C4), (C18), (C19) we can write the solution for thehard box integral in the form H = C ( (cid:15) ) (cid:15) s t (cid:18) (cid:88) i =1 (cid:101) H ( i ) + g ( (cid:15) )( − S h ) − (cid:15) (cid:19) . (C20)Let us calculate the constant g ( (cid:15) ) in the limit ∆ h →
0. We assume that ( x > , x < , x > t = −√ µ µ , s = − ( √− µ − √− µ ) (1 + δ ) , δ → . (C21)In the limit δ → (cid:101) H (1) ≈ ( − µ ) − (cid:15) , (cid:101) H (2) ≈ ( − µ ) − (cid:15) , (cid:101) H (3) ≈ ( − s ) − (cid:15) , (cid:101) H (4) ≈ − − t ) − (cid:15) . (C22)Therefore from (C20) we obtain H ≈ C ( (cid:15) ) (cid:15) s t (cid:18) ( − µ ) − (cid:15) + ( − µ ) − (cid:15) + ( − s ) − (cid:15) − − t ) − (cid:15) + g ( (cid:15) ) (cid:16) √ µ µ δ (cid:17) − (cid:15) (cid:19) . (C23)Since ∆ h = 0 is not a branch point, then g ( (cid:15) ) = 0 . Our solution for the hard box integralin the Euclidean region and for x > , x < , x > , ∆ > H = C ( (cid:15) ) (cid:15) s t (cid:88) i =1 (cid:101) H ( i ) . (C24)28
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