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Hi Xing,

The short answer: the problem with your code is the D->4 limit, that you

are actually not

allowed to take. The B0 function is UV divergent, so by putting D to 4 you

are loosing a contribution to the finite part which leads to the

discrepancy.

The long answer:

First let me observe, that if you evaluate PaVeReduce

with the global option $LimitTo4 set to False, tid1 and tid2 are apparently

the same:

FCClearScalarProducts[];

SPD[p1, p1] = 0;

SPD[p2, p2] = 0;

SPD[p1, p2] = s/2;

tid1 = (ToPaVe[

TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}], q,

UsePaVeBasis -> False], q]);

tid2 = (ToPaVe[

TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}], q,

UsePaVeBasis -> True], q]);

res = Simplify[tid1 - tid2]

$LimitTo4 = False;

res // PaVeReduce

-> 0

Otherwise, you obtain

$LimitTo4 = True;

res // PaVeReduce // Simplify //FCE

((-I/4)*Pi^2*(-2 + D + (-4 + D)*B0[-s, m^2, m^2] +

2*(-4 + D)*m^2*C0[0, 0, -s, m^2, m^2, m^2])*

(-2*(FVD[p1, nu]*FVD[p2, mu] + FVD[p1, mu]*FVD[p2, nu]) + s*MTD[mu, nu]))/

((-2 + D)*s)

As you have observed, this result doesn't vanish when you put D to 4.

This is expected,

since the B0 function is of course UV divergent and produces a

1/Epsilon pole. When you

set D=4, you are missing all the terms of order D/Epsilon which are not

vanishing for

D= 4-2Epsilon (or 4-Epsilon, whatever you prefer). To see that the

output is indeed zero,

as it should be, one has to compute the integrals explicitly.

At this point I would like to explain the meaning of the $LimitTo4

option, since many people

seem to be confused by it. A general Passarino-Veltman coefficient

function PaVe might be

UV-divergent. Hence, we can parametrize it as

PaVe = a/(D-4) + b + O(Epsilon), where

b is the finite part (0th order term in the Epsilon expansion) and a is

the prefactor of the 1/Epsilon pole.

At 1-loop we don't care about terms that are higher order in Epsilon, so

the Epsilon-expansion is truncated here.

Now, in the reducton of PaVe-functions to scalar functions

(A0,B0,C0,D0), which is done

using Equations from the Section 4.2 of arXiv:0709.1075, these

PaVe-functions can be multiplied

by coefficients that are rational functions in D, e.g. D, 1/(D-2), 1/D

etc. Let us denote such a coefficient

as f(D). Then, again we can parametrize it as

f(D) = f(4) + f'(4) (D-4) + O(Epsilon^2),

where f'(4) is the derivative of f(D) w.r.t to D-4 evaluated at D=4.

Notice that here I expanded f(D) up to second order in Epsilon (actually

in D-4), since when I multiply it with a PaVe, the linear term f'(4)

(D-4) multiplied by the a/(D-4) term will produce a term finite in the

D->4 limit.

So I can write

f(D) PaVe = (f(4) + f'(4) (D-4)+ O(Epsilon)^2 )(a/(D-4) + b +

O(Epsilon)) =

f(4)(a/(D-4) + b) + a f'(4) + O(Epsilon) =

f(4) PaVe + a f'(4) + O(Epsilon).

Obviously, if I know the prefactor a of the 1/Epsilon pole of a PaVe

function, then I

can safely rewrite the product f(D)*PaVe as f(4)*PaVe + some finite

part. This is a completely

general statement in dimensional regularization that doesn't depend on a

scheme etc.

Of course, these 1/Epsilon pole coefficients of PaVe functions are well

known and can

be found in the literature, e.g. in the Appendix A of

arXiv:hep-ph/0509141. FeynCalc knows

some of them and can therefore simplify certain f(D)*PaVe products, when

the options $LimitTo4 is set

to True. It is also clear that when we write f(4)*PaVe, the PaVe

function itself still has to

be computed using dimensional regularization and will exhibit a

1/Epsilon pole. On the other hand,

when $LimitTo4 is set to False, no simplifications of f(D)*PaVe can occur.

All this discussion applies for Passarino-Veltman functions that are IR

finite. This is because

in dim reg, IR divergences are usually regulated in the same way as UV

divergences, so that one

cannot distinguish between the two. An IR divergene would modify the

value of 1/Epsilon pole

coefficient which would invalidate the whole approach. So,

$LimitTo4=True should not be used

for IR divegent loop integrals, unless the IR divergence is regulated in

a different way (i.e. with

a fictitious mass). This doesn't apply to your case, as your integral is

IR finite.

Coming back to you problem, there is in fact also an automatic way to

see that the differnce between

tid1 and tid2 vanishes also in the case when we use PaVeReduce with

$LimitTo4=True. Hiren Patel's

Package-X (https://packagex.hepforge.org/) contains a library of

analytic results of PaVe functions

with up to three legs (i.e. A, B and C-functions).

My add-on FeynHelpers (https://github.com/FeynCalc/feynhelpers) can call

Package-X directly from FeynCalc in a transparent way. To install it, use

Import["https://raw.githubusercontent.com/FeynCalc/feynhelpers/master/install.m"]

InstallFeynHelpers[]

Then restart Mathematica and execute

$LoadAddOns = {"FeynHelpers"};

<< FeynCalc`

FCClearScalarProducts[];

SPD[p1, p1] = 0;

SPD[p2, p2] = 0;

SPD[p1, p2] = s/2;

tid1 = (ToPaVe[

TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}], q,

UsePaVeBasis -> False], q]);

tid2 = (ToPaVe[

TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}], q,

UsePaVeBasis -> True], q]) // PaVeReduce;

res = Simplify[tid1 - tid2]

PaXEvaluate[res]

which returns 0.

Of course, you can also evaluate tid1 and tid2 separatly, if you want to

see explicitly

that they are the same

r1 = PaXEvaluate[tid1]

r2 = PaXEvaluate[tid2]

Simplify[r1 - r2]

which gives you zero as expected.

Cheers,

Vladyslav.

Am 19.02.2016 um 20:58 schrieb Xing:

*> Hi Vladyslav,
*

*> When I tried to do TID to a rank-2 3-point loop tensor, TID with UsePaVeBasis set to be true or false give me different results (tid2-tid1 doesn't vanish). Do you know why this happens.
*

*>
*

*> In[24]:=
*

*> ClearScalarProducts;
*

*> SPD[p1, p1] = 0;
*

*> SPD[p2, p2] = 0;
*

*> SPD[p1, p2] = s/2;
*

*> tid1 = (ToPaVe[
*

*> TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}],
*

*> q, UsePaVeBasis -> False], q] /. D -> 4);
*

*> tid2 = (ToPaVe[
*

*> TID[FV[q, mu] FV[q, nu] FAD[{q, m}, {q + p1, m}, {q + p2, m}],
*

*> q, UsePaVeBasis -> True], q] /. D -> 4 // PaVeReduce);
*

*> FullSimplify[tid1 - tid2]
*

*>
*

*> Out[30]= -((I \[Pi]^2 (s Overscript[g, _]^(munu)-2 (Overscript[p2, _]^mu Overscript[p1, _]^nu+Overscript[p1, _]^mu Overscript[p2, _]^nu)))/(4 s))
*

*>
*

*>
*

*> Cheers,
*

*> Xing
*

*>
*

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