Dimensional regularization and the finite part

Let be a dimensional regularized integral

$$\int d^{4-\epsilon}kF(k,m,s)= \frac{2}{\epsilon}+\frac{m^{2}}{3}(\gamma +log(4\pi)-\frac{1}{\epsilon}))$$

then formally if we elmiinate the divergent quantities we may have

$$\int d^{4-\epsilon}kF(k,m,s)_{reg}= \frac{m^{2}}{3}(\gamma +log(4\pi)+log\mu)$$

here $m$ and $s$ are parameters and $\mu$ and energy scale

but is it that enoguh is renormalization simply 'deleting' teh divergent quantities proportional to $\frac{1}{\epsilon ^{k}}$ ?

• I'm not sure how you got your results, but renormalization isn't simply deleting the $\frac{1}{\epsilon}$ terms, we need to choose our counterterms such that they remove these divergences. – Will Jul 3 '13 at 19:44

$$\int \mathrm{d}^{4-\epsilon}k \, \, F(k,m,s) = \frac{2}{\epsilon} + \frac{m^2}{3}\left( \gamma + \log (4\pi) -\frac{1}{\epsilon} \right)$$
Renormalization does not simply 'delete' the $1/\epsilon$ divergences. It is a well-defined procedure which expresses the amplitudes in terms of renormalized, measurable and physical quantities rather than the bare parameters (e.g. $e$ and $m$) which appear in the original Lagrangian. The amplitudes cannot be infinite, and hence if the procedure is executed correctly, it must inevitably remove divergences.