# Cancellation of the quadratic divergence in QCD

I am currently reading about QCD in QFT-Peskin&Schroeder. When calculating 1-loop diagrams for QCD and using dimensional regularization, the 3-vertex boson loop, 4-vertex boson loop and ghost loop yield the following divergent terms which contain $\Gamma(1-\frac{d}{2})$ and $\Gamma(2-\frac{d}{2})$.

The coefficient for $\Gamma(1-d/2)g^{\mu\nu}q^2x(1-x)$ is $(1-\frac{d}{2})(2-d)$. At around $d=2$, this term vanishes clearly. However, for physical dimentions, i.e. $d\rightarrow4$, this is not equal to $0$. The book also says that "the pole at $d=2$ is canceled, $\Gamma(1-\frac{d}{2})$ becomes $\Gamma(2-\frac{d}{2})$". Can anyone help me understand this?

Or is there a better explanation why the cancellation of the divergent quantities here happen to

• They mean nothing more than applying the identity $\Gamma(2-d/2) = (1-d/2)\Gamma(1-d/2).$ – Vibert Dec 22 '12 at 15:30
• @Vibert it's an important identity, so perhaps worth posting in an answer? – twistor59 Dec 22 '12 at 16:17
• I'll wait to see if OP can make some progress. If (s)he has solved the issue and wants to 'close' the question I could post it as an answer. – Vibert Dec 22 '12 at 16:39
• Hi, thanks for the posts. I also guess that they indeed mean the identity $\Gamma(2-\frac{d}{2}) = (1-\frac{d}{2})\Gamma(1-\frac{d}{2})$. – user53997 Dec 24 '12 at 3:48
• Indeed for any positive $\epsilon >0$, $\Gamma(\epsilon)=(-1+\epsilon)\Gamma(-1+\epsilon)$. – user53997 Dec 24 '12 at 3:50