I'm trying to prove the following: \begin{equation} \int_0^\infty x^a dx = 0, \hspace{2pt} \forall a\in \mathbb{R}. \end{equation} This should work in dimensional regularization. I found a lot of answers in phisycs stack exchange, but they were only for specific cases (i.e. $\frac{1}{x^2}$, $\frac{1}{x^3}$), but I'm not able to adapt them in the general case.
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The result that in dimensional regularization all scale-less integrals vanish is basically a consistency relation. In order to give a precise meaning to dimensional regularization we need to define some functional$^1$ $$\mathcal F_d[f] = \int \text{d}^d\mathbf p f(\mathbf p )$$ that fulfills all needed requirements. A natural choice of requirements (in order to make contact to actual integration) is given by $$ \mathcal F_d[a f + b g]= a\mathcal F_d[ f ] + b\mathcal F_d[g]\quad (\text{linearity}) $$ $$ \mathcal F_d[f(s\times \circ)]= s^{-d}\times\mathcal F_d[f(\circ)]\quad \forall s\in\mathbb R\quad (\text{scaling}) $$ $$ \mathcal F_d[f(\circ + \mathbf q)]= \mathcal F_d[f(\circ)]\quad \forall \mathbf q \in\mathbb R^d\quad (\text{translation}) $$ Now you need to check if such a funcional exists that for integer $d$ is compatible with integration. Let us assume you have done this and found that everything works out in a consistent manner (see Ref.[1]). You find for arbitrairy $s,\alpha$ and $d$ $$ \int \text{d}^d \mathbf p (\mathbf p ^2)^\alpha = \int \text{d}^d \mathbf p ((s\mathbf p) ^2)^\alpha s^{-2\alpha} =s^{-2\alpha}\times \int \text{d}^d \mathbf p ((s\mathbf p )^2)^\alpha \stackrel {\text{(scaling)}}{=}s^{-2\alpha-d}\times\int \text{d}^d \mathbf p (\mathbf p ^2)^\alpha $$ Thus, given the definition of our functional is self-consistant, the scaleless integral vanishes.
$^1$: Note that at this point the notation using the integral sign and all that is just by the (a priori unjustified) requirement to make contact to integration later on.
Resources:
[1] "Renormalization", by J.C. Collins
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