Date: 01/26/17-01:38:01 PM Z

https://github.com/FeynCalc/feyncalc/wiki/FAQ#limitto4

why you should better not.

Cheers,

Am 26.01.2017 um 13:34 schrieb Vladyslav Shtabovenko:
> Hi,
>
> at the moment I'm not able to reproduce this bug.
>
> With FeynCalc 9.2.0 on Mathematica 11 I get
>
> << FeynCalc`
> int = PaVe[1, 2, {0, s, 0}, {mf^2, mf^2, mf^2}];
> res = PaVeReduce[int]
>
> -(((4 - D) PaVe[0, {s}, {mf^2, mf^2}])/(2 (2 - D) s)) - (
> 2 mf^2 PaVe[0, {0, 0, s}, {mf^2, mf^2, mf^2}])/((2 - D) s)
>
>
> which can be checked e.g. with FeynHelpers
>
> << FeynCalc`
>
> int = PaVe[1, 2, {0, s, 0}, {mf^2, mf^2, mf^2}];
> res = PaVeReduce[int]
>
> PaXEvaluate[int] === PaXEvaluate[res]
>
> True
>
> to be the correct decomposition.
>
> Cheers,
>
> Am 26.01.2017 um 12:45 schrieb Davide Racco:
>> Hi,
>> first of all I thank you for making available this extremely useful
>> package.
>> I would like to report what could be a possible bug in the function
>> PaVeReduce in FeynCalc 9.2.0.
>>
>> If I try to PaVeReduce the function
>> PaVe[1, 2, {0, s, 0}, {mf^2, mf^2, mf^2}]
>> with FeynCalc 8.2.0 (I had it installed on my previous laptop) the
>> output is
>> (mf^2*PaVe[0, {0, 0, s}, {mf^2, mf^2, mf^2}])/s + 1/(2 s)
>> while now with FeynCalc 9.2.0 I obtain just the first term
>> (mf^2*PaVe[0, {0, 0, s}, {mf^2, mf^2, mf^2}])/s
>>
>> I checked what should be the right answer by evaluating the three
>> expressions with LoopTools 2.13 (for mf->173, s->400), and I obtain
>> respectively
>> -1.39467*10^-6
>> -1.39467*10^-6
>> -0.00125139
>> therefore the output of FC 8 seems to be the right one.
>>