I would like to extract the divergence of this integral in 4d Euclidean space:
$$\int d^4z \frac{1}{(x-z)^4}\tag{1}$$
This divergence is expected to cancel with other divergences, which I got using dimensional regularization. I know how to extract the log divergence of $(1)$ using a cutoff, but then I guess I have to use a correspondence between divergences
$$\log \epsilon \longleftrightarrow \frac{1}{\omega-2}\tag{2}$$
as done here (at the bottom of p.11). I am not so fond of this correspondence, and would love to extract the divergence of $(1)$ with dimensional regularization if that is possible, so to be consistent with my other results.
When I try to do regular dimensional regularization, here is what happens:
$$\begin{align}\int d^{2\omega}z \frac{1}{\left[(x-z)^2\right]^{2\omega-2}} &= \int d^{2\omega}z \frac{1}{\left[(x-z)^2(x-z)^2 \right]^{\omega-1}} \\ &= \frac{\Gamma(\omega-2)\Gamma^2(1)}{\Gamma^2(\omega-1)\Gamma(2)} \frac{\pi^\omega}{(x-x)^{\omega-2}} \tag{3}\end{align}$$
The integral was performed using Feynman parametrization and an integral in the appendix of the QFT book by Ramond. This expression is ambiguous about the divergence, since $\Gamma(\omega-2)$ diverges at $\omega-2$, while it is not clear what the behavior of $1/(x-x)^{\omega-2}$ is. Indeed, if I expand the result with $\epsilon=\omega-2$, Mathematica returns:
$$\frac{\Gamma(\epsilon)}{0^\epsilon} \approx \frac{1}{\epsilon} + \infty + \mathcal{O}(\epsilon)\tag{4}$$
I think the problem may be resolved by adding a $+i\epsilon$ somewhere, but I am not sure how to proceed. In any case, the result that I am expecting to find comes out when I set $1/(x-x)^{\omega-2} = 1$
