Is it possible to expand the measure in dimensional regularization?

Question

In the dimensional regularization scheme, four-dimensional integrals are analytically continued from their $d$-dimensional counterparts, i.e.,

$$\int d^4 x\, f(x) \longrightarrow d^d x\, f(x)\,, \tag{1}$$

with, e.g., $d := 4 - \varepsilon$.

My question is simple: can the measure be expanded in $\varepsilon$? I doubt it, but if yes I guess it should take the form

$$d^d x \sim d^4 x - \varepsilon A\, d^4 x + \ldots\,, \tag{2}$$

where $A$ is to be determined.

Answer 1

1. In practice dimensional regularization (DR) is not done by expansion of the measure factor $$d^dx$$ (whatever that is mathematically supposed to mean).

2. Instead one typically considers the exact analytic expression for the full integral $$I(d)~=~\int\! d^dx~f(x,d)$$ [of the specific class of integrands $f(x,d)$ that one is interested in] as a function of the integer dimension $d\in\mathbb{N}_0$.

   In favorable cases, there exists a natural analytic continuation of $I(d)$ to (a region of) the complex $d$-plane, which then serves as a definition for the DR scheme.