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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}} $ ?

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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
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1 Answer 1

Taking your integral as an example, i.e.

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

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.

For additional resources regarding renormalization, I recommend:

  1. Peskin and Schroeder's Introduction to Quantum Field Theory which provides a thorough (and sufficiently explicit) introduction to renormalization and the general theory known as the renormalization group which further explores ideas such as scale-dependent quantities, e.g. coupling constant.
  2. Srednicki's Quantum Field Theory which also covers renormalization, and is available for free at http://web.physics.ucsb.edu/~mark/qft.html. Furthermore, the text provides some calculations in much more detail than Peskin and Schroeder, however as a whole I believe that the first resource is superior.
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