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When introducing the renormalization group many textbooks start by stating that scattering amplitudes often contain logarithms of the form $\log\frac{\mu^2}{p^2}$, where $p^2$ is the characteristic energy scale of the process and $\mu^2$ is a parameter related to renormalization. $\mu$ either comes from dimensional regularization where it is introduced to keep the right dimensions or it could be a momentum cut-off (then usually called $\Lambda$). It appears to me (for example from reading Schwartz QFT chapter 23) that these large logarithms always arise, independently of the regularization scheme used. I can see how they would enter in dimensional regularization, since loop integrals will always be proportional to $\mu^{4-\epsilon}$ in $4-\epsilon$ dimensions which gives a term containing $\log \mu$ if expanded in $\epsilon$. However, if I used a momentum cutoff $\Lambda$ on a quadratically divergent integral the result would of course be proportional to $\Lambda^2$ and no $\log \Lambda$ enters. So my question is: Is the "problem of large logarithms" tied to dimensional regularization or am I missing something?

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Large logarithms actually occur in different ways depending on the regularization scheme. With a hard cutoff they do appear as $\log \Lambda$, but actually $\log \mu$ is not a divergent logarithm in dimreg. That is because, when working with dimreg, $\mu$ is not a cutoff, but rather could be thought of as an auxiliary scale. The divergence in dimensional regularization is actually the term $\frac{1}{\epsilon}$. However, dimensional regularization will not be able to express power divergences that would occur on a hard cutoff regularization, which are often not required for practical calculations anyway.

As you pointed out, though, not every divergence is logarithmic. You can also have power divergences which, as I mentioned, are actually not captured by dimensional regularization. Dimreg can only pick up the large logarithms, but for many applications that is enough. I expect that the integral you computed will be convergent in dimreg as you take the limit $\epsilon \to 0$. $\mu$ is just an auxiliary scale, and hence $\log\mu$ is not really a large logarithm.

In short, large logarithms are not an artifact of dimensional regularization.

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  • $\begingroup$ I know that $\log \mu$ itself is not divergent, but it leads to a logarithm in a scattering amplitude of the form $\log \frac{\mu^2}{p^2}$, if $p^2$ is the energy scale of the process and $\mu^2$ is the energy at which the renormalization is fixed. This can become large if one of the scales is much larger than the other. What I'm saying is that this dependence on the energy scale will always be logarithmic in dimrep, but with a cut-off we may get a different dependence which confuses me, since the "problem of large logarithms" seems to be a very general problem in most textbooks $\endgroup$
    – F.Burton
    Jun 17, 2023 at 14:44

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