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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 $1/(x-x)^{\omega-2} = 1$

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  • $\begingroup$ Is $z$ a four-vector? I'm confused since it seems like you can just take change integration variables directly to $z \rightarrow z + x$ immediately in your original expression, but then the integral is zero in dim reg. $\endgroup$ – Seth Whitsitt Oct 12 at 23:21
  • $\begingroup$ @SethWhitsitt Yes $x$ and $z$ are 4-vectors... How could the integral be zero? The integral $\int d^4z/z^4$ is logarithmically divergent. $\endgroup$ – Jxx Oct 13 at 7:03
  • $\begingroup$ Does that mean that dim reg "does not see" the logarithmic divergence? I've heard some people say that before. If that is true, how is the correspondence $(2)$ justified? $\endgroup$ – Jxx Oct 13 at 7:21
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    $\begingroup$ I'll just briefly add my interpretation of this "cancellation of IR and UV divergences." I write $\int_0^{\infty} dz \, z^{d-1-\alpha} = \int_0^{1} dz \, z^{d-1-\alpha} +\int_1^{\infty} dz \, z^{d-1-\alpha}$. The first integral is evaluated in the convergent regime $\alpha < d$, obtaining $1/(d-\alpha)$. The second is evaluated in its convergent regime $\alpha > d$ obtaining $1/(\alpha-d)$. So the sum of the two is zero! This should just be interpreted an a mnemonic; the real "proof" is that $\int d^dz \, F(\lambda z) = |\lambda|^{-d}\int d^dz \, F(z)$ is an assumption of dim reg. $\endgroup$ – Seth Whitsitt Oct 13 at 18:39
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    $\begingroup$ I am showing that $\int \frac{d^{d}z}{|z|^{\alpha}} \propto \int_0^{\infty} dz \, z^{d-1-\alpha} = 0$ for all choices of $d$ and $\alpha$. This includes your integral, where $\alpha = 4$ and $d = 2 \omega$, as a special case. $\endgroup$ – Seth Whitsitt Oct 13 at 22:55

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