QFT and divergences: what makes the finite part be regularization-independent?

Question

It seems that the "finite part" of divergent loop integrals are the same, irrelevant of the regularization scheme used to regulate the integrals - why is this?

Consider the following momentum-space integrals (with $m>0$ some mass), which presumably could appear in some loop calculation in QFT: $$ I = \int_{m}^\infty dp\ p^2 = \infty $$ and $$ J = \int_{m}^\infty \frac{dp}{p} = \infty $$ Both of these integrals are divergent: regulating with a UV cutoff $\Lambda \gg m$ gives for example $$ I_{\text{UV cutoff}} = \int_{m}^{\Lambda} dp\; p^2 = \frac{\Lambda^3}{3} - \frac{m^3}{3} $$ meanwhile, if you regulate using dimensional regularization with $D \in \mathbb{C}$ and $M > 0$ some arbitrary mass scale (expanding near $D=0$) $$ I_{\text{dim-reg}} = \int_{m}^{\infty} dp\; \left( \frac{p}{M} \right)^D p^2 = - \frac{m^3}{D+3} \left( \frac{m}{M}\right)^D \simeq - \frac{m^3}{3} + \mathcal{O}(D) $$ Similarly we have $$ J_{\text{UV cutoff}} = \int_{m}^{\Lambda} \frac{dp}{p} = \log\left(\frac{\Lambda}{m}\right) $$ and $$ J_{\text{dim-reg}} = \int_{m}^{\infty} \frac{dp}{p} \left( \frac{p}{M} \right)^D = - \frac{1}{D} \left( \frac{m}{M}\right)^D \simeq - \frac{1}{D} - \log\left( \frac{m}{ M} \right) $$

In the above examples, you find the same "finite" part across different regularization schemes ($m^3/3$ for $I$, and $\sim -\log(m)$ for $J$). In the calculations I can remember, this has been the case for diverging integrals.

Question 1: is it true that the finite part is the same across different regularization schemes?

Question 2: if the answer is yes to the above, why is this the case? It seems counter-intuitive, since we can parametrize a divergence any way we want.

Answer 1

It is not independent. To use your own example, consider the following regulator: $$ \int_{m}^{\Lambda+1} dp\; p^2 = \frac{\Lambda^3}{3}+\Lambda^2+\Lambda - \frac{m^3}{3}+\frac13 $$ whose finite part is $-m^3/3+1/3$ instead of just $-m^3/3$.

So, what is the correct claim?

The regulated version of the integral $$ \int dp f(p) $$ can be thought of as a deformation of the form $$ \int dp\, f(p)\chi_\epsilon(p) $$ where $\chi_\epsilon$ is some function that makes the integral absolutely convergent, and such that $\chi_0(p)=1$.

The final answer, after adjusting counter-terms and everything, will be a function of the form $$ \int dp\, F(p)\chi_\epsilon(p) $$ where $$ \int dp\, F(p) $$ is convergent by itself, without the need for the regulator. This last integral is independent of the regularization scheme.

The reason is more or less straightforward: the regulated integrals are all absolutely convervent, and therefore limits commute with integrals. A different regulator is just a different choice for $\chi$; but, clearly, $$ \lim_{\epsilon\to0} \int dp\, F(p)\chi_\epsilon(p)= \lim_{\epsilon\to0} \int dp\, F(p)\chi'_\epsilon(p) $$ for any two functions $\chi,\chi'$, as all integrals are perfectly convergent. This shows that the final answer, the integral of $F(p)$, is independent of the choice of $\chi$.