A question on dimensional regularization in 4+1D

Question

Let us consider the vacuum energy of a scalar field in $d+1$-dimensional spacetime. We have the integral $$I=\int\frac{d^d k}{(2\pi)^d}\frac{E_k}{2},$$ where $E_k=\sqrt{k^2+m^2}$ and $k$ is a $d$-dimensional vector. Now this integral is apparently divergent. We will employ the dimensional regularzation. We have a useful formula (see arXiv:1701.01554 , page 20) $$\Phi(m,d,A)=\int\frac{d^dk}{(2\pi)^d}\frac{1}{(k^2+m^2)^A}=\frac{1}{(4\pi)^{d/2}}\frac{\Gamma(A-d/2)}{\Gamma(A)}\frac{1}{(m^2)^{A-d/2}}.$$

If we consider a 4-dimensional spacetime, then we take $d=3-2\epsilon$ and the integral has a contribution proportional to $1/\epsilon$ which comes from $\Gamma(-1/2-3/2+\epsilon)=\Gamma(-2+\epsilon)$, indicating the divergence for $d=3$. Now, if we take $d=4-2\epsilon$, we have $$\Gamma(-5/2+\epsilon)=\frac{\Gamma(1/2+\epsilon)}{(-5/2+\epsilon)(-3/2+\epsilon)(-1/2+\epsilon)}=\frac{\Gamma(1/2)+\Gamma(1/2)'\epsilon}{(-5/2+\epsilon)(-3/2+\epsilon)(-1/2+\epsilon)}.$$ The above expression, however, has no pole for $\epsilon\rightarrow 0$. Why isn't the integral divergent? Shouldn't one expect worse behavior for convergence with larger $d$? Is there anything wrong with the above mathematics or a physical reason to explain it?

Answer 1

The fact that the dimensionally-regularized integral does not have a divergent part in 4+1 dimensions is not a problem. In general, one expects that if you use an arbitrary UV regulator $\Lambda$, that the integral will have the form $$ I = \left(\#\right) \Lambda^{d+1} + \mathrm{finite} $$ where $\left(\#\right)$ is some constant which depends on your regulator. It happens that this constant turns out to be zero in dimensional regularization sometimes. If anything, this is useful! You don't need counter-terms in this case.

Answer 2

Dimensional regularization (DimReg) is insensitive to odd divergences. DimReg replaces every even divergence by $1/\epsilon$, but it does give any info about odd divergences. Consider more general integral, $$\int\frac{d^dk}{(2\pi)^d}\frac{k^{2n}}{(k^2+m^2)^A}\rightarrow I=\int_{0}^{\infty}dk\frac{k^{d-1}k^{2n}}{(k^2+m^2)^A}.$$ It is clear that in IR everything is OK and divergence comes from UV region. In UV, we can write $$I\sim\int_{0}^{\infty}dk\,k^{d-1+2n-2A},$$ where $\sim$ means that now we are interested only in divergence. Statement: for even divergences, this integral can be regularized with help of DimReg, whereas in case any of odd divergence one should be more accurate and consider PDS (power divergence substraction) or just use regularization with hard cut-off.