QFT and divergences: what makes the finite part be regularization-independent? 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.
 A: 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 $\chi$.
