**Next message:**Vladyslav Shtabovenko: "Re: Trouble with C0 Integral"**Previous message:**Vladyslav Shtabovenko: "Re: Additional Bug to #4 in OneLoop or PaVeReduce?"**Maybe in reply to:**Gilberto Tavares Velasco: "Possible bug in OneLoop"**Next in thread:**Jongping Hsu: "F_0(0,0,p^2,....)?"**Reply:**Jongping Hsu: "F_0(0,0,p^2,....)?"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

Dear both,

with the current stable version the issue does not apper anymore.

This

S1 = OneLoop[k,

FVD[k, \[Mu]] SPD[k, p] SPD[

k] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, my}]];

S2 = OneLoop[k,

FVD[k, \[Mu]] SPD[k, p] SPD[

k] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, mu}]];

NewM = S1 - S2;

Simplify[PaVeReduce[NewM /. mu -> my]]

gives zero, as well as

ampy = SP[k] SP[k, p] FourVector[k,

a] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, my}] // FCI //

ChangeDimension[#, D] &

ty = PaVeReduce[OneLoop[k, ampy]];

ampy1 = FDS[ampy];

ty1 = PaVeReduce[OneLoop[k, ampy1]];

ty - ty1

and

T1 = OneLoop[k,

FVD[k, a] SPD[k, p] SPD[

k] FAD[{k, my}, {k + p - q, mw}, {k + p + q, mw}],

DenominatorOrder -> False];

T2 = OneLoop[k,

FVD[k, a] SPD[k, p] SPD[

k] FAD[{k, my}, {k + p - q, mw}, {k + p + q, mw}],

DenominatorOrder -> True];

Simplify[PaVeReduce[T1 - T2]]

I will nevertheless add your examples to our testsuite.

Cheers,

Vladyslav

*> 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:**Vladyslav Shtabovenko: "Re: Trouble with C0 Integral"**Previous message:**Vladyslav Shtabovenko: "Re: Additional Bug to #4 in OneLoop or PaVeReduce?"**Maybe in reply to:**Gilberto Tavares Velasco: "Possible bug in OneLoop"**Next in thread:**Jongping Hsu: "F_0(0,0,p^2,....)?"**Reply:**Jongping Hsu: "F_0(0,0,p^2,....)?"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

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