Name: blunden@jlab.org (email_not_shown)
Date: 04/24/03-01:03:15 AM Z


I don't have an answer to your question. Howver, I agree
that there is a bug, or at least a terrible inconsistency, in how
OneLoop handles this kind of UV divergent integral. This concerns me
because I've been using FeynCalc to manipulate some integrals whose
sum is finite, but with individual terms that are UV divergent.

In fact, one does not have to rename the mass variable to reveal a problem.
The answer depends on the order of the propagators.

==========================

SetOptions[OneLoop,Prefactor->1/(I Pi^2)];

First we define, as you did:

  ampy = SP[k] SP[k,p] FourVector[k,a] FAD[{k+p-q,mw},{k+p+q,mw},{k,my}]//FCI;

                            k^2 k.p k[a]
ampy = ----------------------------------------------------------
         ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2) (k^2 - my^2)

The one loop integral is:

  ty = PaVeReduce[OneLoop[k,ampy]];

Now instead put it in standard order using FeynAmpDenominatorSimplify:

  ampy1 = FDS[ampy];

                            k^2 k.p k[a]
ampy1 = ----------------------------------------------------------
         (k^2 - my^2) ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2)

  ty1 = PaVeReduce[OneLoop[k,ampy1]];

ty is not equal to ty1. The difference, after simplifying B0[0,m0,m1] is

          -(2 mw^2 + 4 my^2 - p^2 - q^2 - 2 p.q) (p[a] + q[a])
ty-ty1 = ----------------------------------------------------
                                  48

If we now change the mass my -> mu, and so define ampu = ampy/.my->mu,
and ampu1 = ampy1/.my->mu, then the analogous tu and tu1 ARE in fact
equal. However, as you discovered, tu does not equal ty
(after replacing mu->my), with the difference being

         (3 mw^2 - 2 q^2) q[a]
tu-ty = ---------------------
                  6

Going back to the ampy amplitude, if we apply ScalarProductCancel we find
(equivalent to regrouping k^2 -> k^2-my^2 + my^2 and cancelling propagators):

  SPC[ampy]

                         k.p k[a]
    = --------------------------------------------- +
          ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2)
 
                    my^2 k.p k[a]
  ----------------------------------------------------------
  ((k + p + q)^2 - mw^2) (k^2 - my^2) ((k + p - q)^2 - mw^2)

Here the one loop integral gives the same result as the standard ordered
result ty1. Interestingly, evaluating the one loop integral of SPC[ampu]
also gives the SAME result as ty1 (after replacing mu->my). That this
should happen is clear from the form of the amplitude above, since the
first term is independent of my^2, and the second is explicitly
multiplied by my^2. So I suspect that this form gives the "correct"
answer (or perhaps the "preferred" answer).



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