**Next message:**Vladyslav Shtabovenko: "Re: ToTFI"**Previous message:**Vladyslav Shtabovenko: "Re: ToTFI"**In reply to:**SUN Qingfeng: "About the function TID in FC9.0"**Next in thread:**Sun Qingfeng: "Re: About the function TID in FC9.0"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

Hi,

frankly speaking, I don't quite see what is the issue

with FeynCalc's result. The integral you write is clearly

scaleless, such that in Dimensional Regularization (which is always

assumed in FeynCalc) you can make a shift k->k+q1 and obtain

k^mu / (k-q1)^2 = (k^mu + q1^mu) / k^2 = k^mu /k^2 + q1^mu /k^2

Now the first integral vanishes b/c it is antisymmetric under k-> -k.

The second integral is scaleless and is set to zero in dim.reg,

following the general rule that

\int d^D k (k^2)^a = 0 for any a.

This is how 0 comes out.

I'm not sure that I understand how you arrive to q1^mu / (k-q1)^2

Even if you don't set scaleless integrals to zero, the formal result of

the tensor decomposition is given by

int = FCI[Tdec[{{k, mu}}, {q1}, List -> False] FAD[k - q1]]

or

(int /. k -> k + q1) // MomentumExpand // ExpandScalarProduct

which is of course still scaleless and hence zero in dim. reg.

Cheers,

Vladyslav

Am 29.06.2015 um 00:24 schrieb SUN Qingfeng:

*> Input:
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*> test = FVD[k, \[Mu]] FAD[k - q1] // FCI
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*> TID[test, k]
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*>
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*> Output:
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*> 0
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*>
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*> But I think the answer should be:?
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*> FVD[q1, \[Mu]] FAD[k - q1]
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*>
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*> Which answer should be correct?
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*>
*

*> My version Info:
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*> Mathematica 8.0 FeynCalc 9.0
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*>
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