Name: Vladyslav Shtabovenko (email_not_shown)
Date: 10/21/17-06:10:25 PM Z


FYI, while Tarcer indeed turns out to be not very useful for processing those 3 integrals, LiteRed can actually reduce them further.

First of all, we can use TarcerToFC to rewrite the J-integrals as FAD's

(TarcerToFC[TJI[d, m1^2, {{3, m2}, {1, m2}, {1, m1}}], {q1, q2}] /.
    d -> D) // FDS[#, q1, q2] & // FCE

-> FAD[{q1, m2}, {q1, m2}, {q1, m2}, {q1 - q2, m2}, {-m1 + q2, m1}]

(TarcerToFC[TJI[d, m1^2, {{2, m2}, {2, m2}, {1, m1}}], {q1, q2}] /.
    d -> D) // FDS[#, q1, q2] & // FCE

-> FAD[{q1, m2}, {q1, m2}, {q1 - q2, m2}, {q1 - q2, m2}, {-m1 + q2, m1}]

(TarcerToFC[TJI[d, m1^2, {{2, m2}, {2, m1}, {1, m2}}], {q1, q2}] /.
    d -> D) // FDS[#, q1, q2] & // FCE

-> FAD[{q1, m2}, {q1, m2}, {q1 - q2, m1}, {q1 - q2, m1}, {-m1 + q2, m2}]

Now with LiteRed:

<< LiteRed`;

SetDim[d];
Declare[{q1, q2, p}, Vector];
Declare[{m1, m2}, Number];
sp[p, p] = m1^2;

NewBasis[j1, {sp[q1] - m2^2, sp[q1 - q2] - m2^2, sp[q2 - p] - m1^2,
  sp[q1, p], sp[q2, p]}, {q1, q2}, AnalyzeSectors -> True,
 FindSymmetries -> True, GenerateIBP -> True]

SolvejSector /@ UniqueSectors[j1];

(*Integral 1*)

IBPReduce[j[j1, 3, 1, 1, 0, 0]]

-((-2 + d)^2*(-m1^2 - 3*m2^2 + d*m2^2)*j[j1, 0, 1, 1, 0, 0])/(32*(-4 + d)*m1^2*(m1 - m2)*m2^4*(m1 + m2)) +
 ((-2 + d)^2*(-3*m1^2 + d*m1^2 - m2^2)*j[j1, 1, 1, 0, 0, 0])/(64*(-4 + d)*m1^2*(m1 - m2)*m2^4*(m1 + m2)) +
 ((-3 + d)*(-8 + 3*d)*(-5*m1^2 + d*m1^2 + m2^2)*j[j1, 1, 1, 1, 0, 0])/(32*(-4 + d)*m1^2*(m1 - m2)*m2^2*(m1 + m2)) +
 ((35*m1^4 - 17*d*m1^4 + 2*d^2*m1^4 - 62*m1^2*m2^2 + 32*d*m1^2*m2^2 - 4*d^2*m1^2*m2^2 + 3*m2^4 - d*m2^4)*j[j1, 2, 1, 1, 0, 0])/
  (8*(-4 + d)*m1^2*(m1 - m2)*m2^2*(m1 + m2))

(*Integrals 2 and 3*)

IBPReduce[j[j1, 2, 2, 1, 0, 0]]

-((-2 + d)^2*(-7*m1^2 + 2*d*m1^2 + 3*m2^2 - d*m2^2)*j[j1, 0, 1, 1, 0, 0])/(16*(-4 + d)*m1^2*(m1 - m2)*m2^4*(m1 + m2)) +
 ((-2 + d)^2*(-5*m1^2 + d*m1^2 + m2^2)*j[j1, 1, 1, 0, 0, 0])/(32*(-4 + d)*m1^2*(m1 - m2)*m2^4*(m1 + m2)) +
 ((-3 + d)*(-8 + 3*d)*(-3*m1^2 + d*m1^2 - m2^2)*j[j1, 1, 1, 1, 0, 0])/(16*(-4 + d)*m1^2*(m1 - m2)*m2^2*(m1 + m2)) +
 ((-3 + d)*(-7*m1^4 + 2*d*m1^4 + 14*m1^2*m2^2 - 4*d*m1^2*m2^2 + m2^4)*j[j1, 2, 1, 1, 0, 0])/(4*(-4 + d)*m1^2*(m1 - m2)*m2^2*(m1 + m2))

The basis still contains j[j1, 2, 1, 1, 0, 0] which seems to be an
irreducible integral in this sector. Probably one could trade it for an integral with loop-momentum dependent numerators...

BTW, I'm seriously considering to add support for LiteRed to the next version of FeynHelpers. It would require some work, but considering the usefulness of this tool, this would be a very useful addition for many FeynCalc users.

Cheers,
Vladyslav

> Dear Tarcer creators,
>
> despite the fact that I never found Tarcer doing anything wrong I frequently came across 'strange' beahviour. Mostly it boiled down to a lack of proper understanding on my side, but one class of cases remained:
>
> (1) TJI[d, m1^2 ,{{3,m2},{1,m2},{1,m1}}]
> (2) TJI[d, m1^2 ,{{2,m2},{2,m2},{1,m1}}]
> (1) TJI[d, m1^2 ,{{2,m2},{2,m1},{1,m2}}]
>
> are not recursed down to basis integrals. The recursion formulas (74) and (81) in hep-ph/9801383 boil down to 0 = 0 for certain kinematical configuarations. In this case one has to go off-shell (m1^2 -> s) and then divide out a zero, such that a nontrivial recursion relation remains. the coefficients of the ladder operators then have single or double zeros of the form (s-m1^2)^(1 or 2).
>
> I could solve the last case in a straightforward fashion and the first two by deriving a simplified relation from (81) which led from (1) to (2) and back. In this way I got a 2x2 system in (1) and (2).
>
> Tarcer.nb is too advanced to allow me to provide a patch (in case one is needed, probably I have just not yet seen how to do the recursion using TarcerRecurse) so i though i'd let you know.
>
> best regards
> robert schoefbeck



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