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Suppose there is an integral in four dimension Euclidean space \begin{equation} I_{d=4}=\int_0^\infty d^4x\frac{1}{|x|^2},~ \end{equation} which is divergent. $|x|$ is the length of the vector. Can one use dimensional regularization to compute this integral by using $d^4x \to d^dx$,with $d=4-\epsilon$ ?

Or more abstractly my question is that If I want to compute an integral $I_{d=4}$, but it divergent for example at range $2<d<5$, can we use dimensional regularization by writing $d=4+\epsilon$. Then at the end of calculation let $\epsilon\to0$ ?

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  • $\begingroup$ Did you mean the integrals to be over $(-\infty, \infty)$ rather than $(0, \infty)$? $\endgroup$
    – Prof. Legolasov
    Commented Oct 25, 2019 at 23:30
  • $\begingroup$ Yes, actually I meet the following integral in two dimension euclidean plane \begin{equation}\int_c d^2z\frac{1}{(z-z_i)^2(\bar{z}-\bar{z}_i)^2}\end{equation}where $z=x+iy=\rho e^{i \theta}$ and $z_i$ is a constant. By translation this integral is independent of $z_i$. In polar coordinate, this integral is just \begin{equation} 2\pi\int_0^\infty \frac{d\rho}{\rho^3} \end{equation} My question is can we use dimensional regularization to evaluate this integral. $\endgroup$
    – phys_student
    Commented Oct 26, 2019 at 1:40
  • $\begingroup$ This integral doesn't make sense in any number of (integer) dimensions, as $\int d\rho \rho^{-n}$ diverges for any $n$. I therefore can't think of a way to analytically continue to a complex-valued dimension. Why not simply use a momentum-space infrared cutoff $\rho \ge \varepsilon$? $\endgroup$
    – Prof. Legolasov
    Commented Oct 26, 2019 at 3:37
  • $\begingroup$ Since there are some other integral of the form for example \begin{equation}\int d^2z\frac{1}{(z-z_i)(\bar{z}-\bar{z}_j)}\end{equation} which can be computed by dim reg. , so I want to using only one regularization schem in the whole calculation. And I don't know how to compute the latter integral by introducing a cutoff. (The latter integral can be found in 1902.01434) $\endgroup$
    – phys_student
    Commented Oct 26, 2019 at 3:49
  • $\begingroup$ Voting -1 for the XY Problem (xyproblem.info) $\endgroup$
    – electronpusher
    Commented Mar 14, 2020 at 11:52

2 Answers 2

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The method of dimensional regularization in QFT comes with a few definitions which are crucial to evaluating integrals of this type. Following Zinn-Justin, they are the properties of these integrals under the following:

  1. Translations:

$$ \int d^d p \, F(p + q) = \int d^d p \, F(p) $$

  1. Dilatations:

$$ \int d^d p \, F(\lambda p) = |\lambda|^{-d} \int d^d p \, F(p) $$

  1. Factorizations:

$$ \int d^d p \, d^{d'}q \, F(p) G(q) = \left( \int d^d p \, F(p) \right) \left( \int d^{d'}q \, G(q) \right) $$

From these properties, you can already address some of the integrals you have mentioned. In particular, the first two properties immediately imply the "identity" $$ \int \frac{d^d p}{(2 \pi)^d} \frac{1}{(p + q)^{2\alpha}} = 0, $$ for all $d$ and $\alpha$.

In the comments, you have also mentioned the integral $$ \int_{\mathbb{C}} \frac{d^2 z}{(z - z_i)(\bar{z} - \bar{z}_j)}. $$ You can consider applying dimensional regularization to this integral, either by introducing multiple copies of $\mathbb{C}$ or writing it as an integral over $\mathbb{R}^2$ and then generalizing to an integration over $\mathbb{R}^d$. You'll find that if $z_j = z_i$, the integral is zero in dimensional regularization, but if $z_j \neq z_i$, I see no reason why it should vanish.

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  • $\begingroup$ "the first two properties immediately imply the "identity" $\int \frac{d^d p}{(2 \pi)^d} \frac{1}{(p + q)^{2\alpha}} = 0$ for all $d$ and $\alpha$" : not true for $d=2\alpha$, i.e. logarithmically divergent case. $\endgroup$
    – MadMax
    Commented Oct 28, 2019 at 15:09
  • $\begingroup$ That's true - one should also ask that the integrals are holomorphic functions of $d$ with $\alpha$ fixed (and vice-versa), so that divergences take form of poles which may be subtracted. Given that the integral is zero for all $d \neq 2 \alpha$, the only value you can assign to it at $d = 2 \alpha$ consistent with this requirement is also zero. $\endgroup$
    – Seth Whitsitt
    Commented Oct 28, 2019 at 18:13
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In dimensional regularisation this integral would normally be set to zero - the reason is that the integrand contains no dimensionful parameter upon which the result can depend. This is curious in qft because it removes ir and uv divergences at the same time

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  • $\begingroup$ Actually I want to using dimension regularization to compute the following 2D integral on complex plane\begin{equation}\int_c d^2z\frac{1}{(z-z_i)^2(\bar{z}-\bar{z}_i)^2}\end{equation} (please see the comments I wrote in above) Is this integral can be set to zero in dimension regularization? $\endgroup$
    – phys_student
    Commented Oct 26, 2019 at 1:44

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