**Next message:**Gilberto Tavares: "More on the OneLoop puzzle"**Previous message:**Frederik Orellana: "Re: Problem using OneLoop"**In reply to:**Gilberto Tavares Velasco: "Possible bug in OneLoop"**Next in thread:**Vladyslav Shtabovenko: "Re: Possible bug in OneLoop"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

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).

**Next message:**Gilberto Tavares: "More on the OneLoop puzzle"**Previous message:**Frederik Orellana: "Re: Problem using OneLoop"**In reply to:**Gilberto Tavares Velasco: "Possible bug in OneLoop"**Next in thread:**Vladyslav Shtabovenko: "Re: Possible bug in OneLoop"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

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