Why is the Integral of Dimensional Regularization Equal to 0?

[Luca_Mantani]

Hi everyone,
I read this property of dimensional regularization, but i do not understand why it is so.

$$\int d^dp =0$$.

Actually looking for an answer i also saw that a general property of dim. reg. is

$$\int d^dp \, (p^2)^\alpha=0$$

for any value of $\alpha$, so also for $\alpha=0$. Do you have an explanation for this?
Thanks in advance.

 

[vanhees71]

Where is this from? It depends on how you define these integrals. Standing alone these integrals do not make sense since you can define only integrals which exist for some space-time dimensions and then use analytic continuation in $d$ to evaluate formally integrals that are not defined. The usual regularization parameter is $2 \epsilon=d-4$, and Feynman diagrams are evaluated in form of a Laurent expansion in $\epsilon$ the poles in $\epsilon$, which diverge for $\epsilon \rightarrow 0$ define the subtractions ("counter terms") to renormalize (order by order in perturbation theory) the proper vertex functions (aka 1PI truncated diagrams). For an introduction to the technique, see my QFT script:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Luca_Mantani]

I read them in an Eft course, in the video the professor was citing the Collins renormalization book for the proof, but I do not have this book at my disposal now.