I am not quite sure that my question below is appropriate for this site, probably it should be addressed to the physical commutity. But I hope that some (mathematical) physicists do attend MO. I have two questions on dimensional regularization used in the renormalization theory (they should be very basic, I am a mathematician, even not a mathematical physicist).

1) Is there a mathematically rigorous exposition of the dimensional regularization?

2) Let $d$ denote the dimension of the space time. My impression is that the method of dimensional regularization works better for even $d$ rather than for odd. Namely for some integrals which are obviously divergent in odd dimensions, the method of dimensional regularization gives convergent expressions. Below I give a simple example of such a situation for $d=3$. Thus my second question is how to resolve this apparent contradiction, and whether the method can be modified to work in odd dimensions as well, even at the physical level of rigour.

Here is the example. Consider the theory $\phi^4$ in Euclidean space-time. Consider the Feynmann diagram with just one vertex, one self-loop, and two exterior lines (though, I guess, one can construct many other examples). The corresponding integral, up to some factors containing the interaction constant, is equal to $\int \frac{1}{p^2+1}d^dp$ (we take the mass $m=1$ for simplicity; it is physically impossible for dimensional considerations, but does not influence the analysis of convergence issues). We have \begin{eqnarray*} \int \frac{1}{p^2+1}d^dp=\frac{2\pi^{d/2}}{\Gamma(d/2)}\int_0^\infty \frac{r^{d-1}}{r^2+1}dr=:A. \end{eqnarray*} The integral $A$ diverges for $d\geq 2$. After some change of variables and standard computations it becomes \begin{eqnarray*} 2\pi^{d/2} \Gamma(-\frac{d}{2}+1). \end{eqnarray*}

The last expression has no poles at odd $d$, in particular at $d=3$. This apparently contradicts the above mentioned divergence of the integral $A$. On the other hand, at even $d$, $\Gamma(-\frac{d}{2}+1)$ does have a pole as expected, and the method works well.


2 Answers 2


There a number of papers by Alain Connes on Dimensional Regularization (Dim Reg) in the context of noncommutative field theory. Some of his papers cite

P. Breitenlohner and D. Maison, "Dimensional renormalization and the action principle," Comm. Math. Phys. Vol. 52, Number 1 (1977).

so I presume this might be a useful reference for you.

Regarding your observation, this is a well known fact among particle physicists and is regarded as a feature rather than a bug. Part of the standard lore is that Dim Reg replaces log divergences by poles but that linear and higher divergences often just give zero. Physicists are most interested in log divergences since these govern violations of scale invariance, determine beta functions and so on. In your example the integral has a pole at d=2 where the integral has a log divergence but vanishes at d=3 where the divergence is linear.

  • $\begingroup$ Thanks for the reference, I will have a look. I did not understand your remark that "Physicists are most interested in log divergences". The physical problem I had in mind is perturbative computation of Green functions and S-matrix. Here log divergences and other ones have equal importance. The example I have considered in the question appears, say, in computation of the exact propagator. Absense of a pole in the gamma function for d=3 implies that the counter term vanishes (if one uses the minimal subtraction prescription). That mean that the divergence of the integral $A$ is not cancelled. $\endgroup$
    – asv
    Jul 29, 2013 at 3:04
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    $\begingroup$ Once one defines the integral via DimReg then I would have said there is no divergence to cancel and hence no counterterm needed. Counterterms are needed to cancel divergences which in DimReg appear as poles, so if there is no pole why do you say there is a divergence that is not cancelled? $\endgroup$ Jul 29, 2013 at 3:09
  • $\begingroup$ Possibly this your comment almost answers my question. May be my mistake was that I automatically ascribed to the integral A the infinite value because the integral obviously diverges. Instead, one had to ascribe the value using meromorphic continuation. However what I still do not understand is that the overall degree of divergence of the considered diagram is d−2=1 for d=3, namely the integral is expected to diverge linearly, and hence there should be counter terms! For larger odd $d=5,7,9...$ the problem is the same. $\endgroup$
    – asv
    Jul 29, 2013 at 9:25
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    $\begingroup$ @semyonalesker If you were to regularize the integral with a momentum space cutoff then you would find the expected divergences and you would need the counterterms you expect. This is a perfectly good way of regularizing the theory and it seems in accord with what you expect intuitively. DimReg is another way of making sense of the divergent integrals and has the special properties you noted. The fact that the continuation in d leads to results without some of the expected divergences doesn't lead to any physical inconsistencies, or mathematical ones as far as I know. $\endgroup$ Jul 29, 2013 at 16:47
  • $\begingroup$ This seems to be the answer to my question. Apparently your last sentence is based on your experience rather than on a precise theorem (it would be great if I am wrong). Thus I will believe in it until I get enough experience of computing enough examples by different regularization methods. I must admit however that now DimReg seems to me to be even more mysterious than previously. Thank you so much for your answers! $\endgroup$
    – asv
    Jul 29, 2013 at 18:04

Albeit rather late I would like to add the following reference as another answer to your question (1): dimensional regularization is also treated rigorously in

P. Etingof, Note on Dimensional Regularization in Vol 1, pp 597–607, of Quantum Fields and Strings: A Course for Mathematicians (see also here).


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