# Dimensional Regularization in Three Dimensions

I am looking at the Thirring model in three dimensions, which is non-renormalizeable. I was trying to calculate the one loop self energy of the fermion to see where the infinities crop up that cannot be cancelled. In doing so, I came across the Euclidean space superficially divergent integral $$\int \frac{d^3p}{(2\pi)^3}\frac{1}{p^2+m^2}$$

If you put a UV cutoff $$\Lambda$$ on this integral, it's easy to see this is linearly divergent with the value $$\int \frac{d^3p}{(2\pi)^3}\frac{1}{p^2+m^2}=\frac{1}{2\pi^2}\int_0^{\Lambda}dp\frac{p^2}{p^2+m^2}=\frac{\Lambda}{2\pi^2}-\frac{m}{2\pi^2}\arctan{\Big(\frac{\Lambda}{m}\Big)}\sim \frac{\Lambda}{2\pi^2}$$ There are other ways of regularizing integrals, dimensional regularization being a very useful one. So regularizing it that way, we get $$\int \frac{d^dp}{(2\pi)^d}\frac{1}{p^2+m^2}=\frac{2\pi^{d/2}}{\Gamma(d/2)(2\pi)^d}\int_0^\infty dp\frac{p^{d-1}}{p^2+m^2}=\frac{2\pi^{d/2}}{\Gamma(d/2)(2\pi)^d}m^{d-2}\frac{\Gamma(d/2)\Gamma(1-d/2)}{2\Gamma(1)}\to-\frac{m}{4\pi}$$ Somehow I have avoided all infinities, and the final answer is negative!

Is my answer using dimensional regularization justified? Why are the answers so drastically different?

Edit: I should emphasize that I believe this should be a generic feature of dimensional regularization in three dimensions, since the possible singularity arising from $$\Gamma(m-d/2)$$ with $$m$$ an integer is finite when $$d=3$$.

• If keep the term subleading in $\Lambda$ you find that the finite piece matches what you get from dimensional regularisation. The divergent part is cancelled by counterterms. May 12, 2021 at 5:19

Furthermore, in his book Weinberg points out that renormalization actually has very little to do with divergences -- they are something of a technical annoyance. The content of renormalization is that the presence of interactions means that an interacting theory behaves in a different way than a free theory, so for example the mass of a particle that you can actually measure (eg the pole in a propagator) is not the same thing as the parameter multiplying the $$\phi^2$$ term in a Lagrangian.
Putting this together, while I haven't checked the details of your calculation, it isn't problematic that you find certain integrals are power-divergent with a cutoff scheme, and finite with dimensional regularization. In the end, you should be able to do the renormalization in a way that the physical results are the same (or put differently: if you can't, then that is a genuinely interesting situation). I do know that in four dimensions, there are integrals that are power-law divergent with a cutoff scheme, that actually vanish in dimensional regularization! In general, typically power law divergences with a cutoff scheme are set to zero in dimensional regularization, while log divergences appear in both regularization methods (though in dim reg it may appear as the log of the ratio of an energy scale to the "running scale" $$\mu$$, rather than the log of the ratio of the energy scale to the cutoff $$\Lambda$$ in a cutoff regularization).
Notice that, for small $$m/\Lambda$$, your expression for the hard cutoff is $$\frac{\Lambda}{2 \pi^2} - \frac{m}{4 \pi} + O(m^2/\Lambda),$$ so in the large $$\Lambda$$ limit, you get a divergent piece linear in $$\Lambda$$, but a "universal" piece equal to $$-m/4\pi$$. Indeed, in any regularization scheme, you expect the integral to become $$\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{p^2+m^2} "=" \left(\#\right) \Lambda - \frac{m}{4 \pi} + \mathrm{finite},$$ where some constant $$\left(\#\right)$$ is regularization dependent, the term $$-m/4\pi$$ always come out the same, and the rest vanishes in the limit of small $$m/\Lambda$$.
Here are some other regulators. Pauli-Villars: $$\int \frac{d^3 p}{(2 \pi)^3} \left( \frac{1}{p^2+m^2} - \frac{1}{p^2+\Lambda^2} \right) = \frac{\Lambda}{4 \pi} - \frac{m}{4 \pi}$$ (there are actually no subleading terms in $$m/\Lambda$$ here). Heat kernel regularization: $$\int \frac{d^3 p}{(2 \pi)^3} \frac{e^{-p/\Lambda}}{p^2+m^2} = \frac{\Lambda}{2 \pi^2} - \frac{m}{4 \pi} + O(m^2/\Lambda).$$ As you can see, these are all consistent with the general form above, it's just that the constant I indicated as $$(\#)$$ happens to vanish in dimensional regularization.