Expansion of common expression in Dimensional Regularisation

Question

In QFT we often have it with the expression $$\frac{\Gamma(\epsilon)}{16\pi^2}\left(\frac{4\pi}{A^2}\right)^\epsilon.$$ This is expandet as $$\frac{1}{16\pi^2}\left(\frac{1}{\epsilon}+\ln\left(\frac{4\pi}{A^2}\right)-\gamma_E+O(\epsilon)\right).$$ But from where comes the logarithm? And it appers as additional term not as an factor. I don't now how to expand this first expression.

Answer 1

So there are a few things going on here. The first is the expansion of $\Gamma(\epsilon) \simeq 1/\epsilon - \gamma + \mathcal{O}(\epsilon)$. The second is the expansion of the term $(4\pi/A^2)^\epsilon \equiv B^\epsilon$, where the $B$ defined here just makes it easier for me to type out things.

Now $B^\epsilon = e^{\ln(B^\epsilon)}= e^{\epsilon \ln(B)} \simeq 1 + \epsilon \ln(B) + \mathcal{O}(\epsilon)$, so the product gives:

$1/\epsilon - \gamma - \ln(B) + \mathcal{O}(\epsilon)$.

(Up to the prefactor I ignored and plugging in the definition of $B$ this gives the desired result.)