In general, the best strategy for such complicated integrals is to first
rewrite them in terms of the coefficient functions
amp = Pair[Momentum[k], Momentum[p1]]^3 FAD[{k, lam}, {(k - q),
lam}, {(k - p1), m}, {(k + p2), M}];
res = TID[amp, k, UsePaVeBasis -> True, ToPaVe -> True]
which quickly returns a rather compact result
-3 I \[Pi]^2 PaVe[0, 0,
1, {SPD[p1, p1], SPD[p1, p1] + 2 SPD[p1, p2] + SPD[p2, p2],
SPD[p2, p2] + 2 SPD[p2, q] + SPD[q, q], SPD[q, q], SPD[p2, p2],
SPD[p1, p1] - 2 SPD[p1, q] + SPD[q, q]}, {lam^2, m^2, M^2, lam^2},
PaVeAutoOrder -> True, PaVeAutoReduce -> True] SP[p1, p1]^2 -
...
Then, depending on what one wants to do, one can
1) Evaluate the coefficient functions numerically using LoopTools,
Collier or whatever other package
2) Evaluate the coefficient functions analytically via Package-X by
using PaXEvaluate from the FeynHelpers extension
3) Reduce the coefficient functions to scalar integral, which would of
course generate a huge amount of terms, e.g.
res[[1]] // PaVeReduce
Cheers,
Vladyslav
> I'm trying to do an integral that FeynCalc chokes on. The message returned
> is the usual
>
> FYI: Tensor integrals of rank higher than 3 encountered; Please use the
> option CancelQP -> True or OneLoopSimplify->True or use another program.
>
> However, it appears that CancelQP->True is the default, and OneLoopSimplify
> expresses the results in terms of Contract3, which doesn't seem to exist.
>
> The integrals are box diagrams, and a typical term would look something like
>
> (k.p1)^3 / [k^2-lam^2][(k-q)^2-lam^2][(k-p1)^2-m^2][(k+p2)^2-M^2]
>
> where p1^2=m^2 and p2^2=M^2. This term looks innocent enough, and in fact
> looks to me like it IS of rank 3. By a lot of fudging and manipulating I
> managed to get a result using ScalarProductCancel, but it is hit and miss
> for various terms in the amplitude.
>
> Is there a fix in FeynCalc, or do I have to use another program (and if so,
> which one)?
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