Hadronic production of a Higgs boson in association with two jets at ...

2.3. Forde’s Laurent expansion method 43powers of t vanish,∫ 2∏∫d 4 l δ(l 2 i)A 1 A 2 A 3 = f 0i=0dtJ t + ∑boxes{l}d l D cut0 . (2.36)If this parameterisation is possible then one has successfully isolated the coefficientof the scalar triangle (which is represented by ∫ dtJ t ) from the previously knownbox coefficients (although if unknown these can be extracted also). In general thecoefficient is given by,c 3;P = −[Inf t A 1 A 2 A 3 ](t)∣ . (2.37)t=0All that remains to be done is to define the correct momentum parameterisationsuch that integrals over non-zero powers of t vanish. The basis is inspired by oneproposed in [125] and relies on the following massless projections of the (potentially)massive vectors K 1 and K 2 (the two momenta which appear as legs of the triangle),K ♭,µ1 = Kµ 1 − (S 1 /γ)K µ 21 − (S 1 S 2 /γ 2 ) , K♭,µ 2 = Kµ 2 − (S 2 /γ)K µ 11 − (S 1 S 2 /γ 2 ) . (2.38)Here γ ± = (K 1 · K 2 ) ± √ ∆ and ∆ = (K 1 · K 2 ) 2 . When determining the trianglecoefficient we must average over the γ solutions. In terms of these basis vectors theloop momenta has the following form,withl µ = α 02 K ♭,µ1 + α 01 K ♭,µ2 + t 2 〈K♭,− 1 |γ µ |K ♭,−2 〉 + α 01α 022t〈K ♭,−2 |γ µ |K ♭,−1 〉, (2.39)α 01 = S 1(γ − S 2 )(γ 2 − S 1 S 2 ) , α 02 = S 2(γ − S 1 )(γ 2 − S 1 S 2 ) . (2.40)The other two on-shell loop momenta l i = l − K i have a similar basis expansionand the coefficients α i1 and α i2 are given explicitly in [122]. Importantly it has beenshown [122,125] that using this basis∫ ∫1dtJ tt = 0, dtJ n t t n = 0, for n ≥ 1, (2.41)which ensures the validity of eq.(2.37). Thus to extract a triangle coefficient onemerely has to parameterise the loop momenta in the above basis, and extract thecoefficient of t 0 in an expansion around t = ∞.