Dimensional regularization of a divergent integral 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 $\epsilon\to0$ ? 
 A: 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:


*

*Translations: 


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


*Dilatations:


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


*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.
A: 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 
