# 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.

• Nov 24, 2021 at 3:17
• I think this is a common misconception: what is the same are the terms thare are non-analytic in the external momenta or in the masses. Those are the predictions of the theory that no change in renoremalization scheme or changing the counter-terms (as long as local) could possibly alter. In certain class of theories (i.e. those renormalizable) also some analytic behavior (say of the amplitude) in the external momenta higher than some value cannot be changed without spoiling the renormalization, that is without interpreting the theory as an EFT. Nov 24, 2021 at 7:13
• You might also want to try sticking $e^{-\alpha p}$ in the integrand and expanding the result around $\alpha = 0$. Nov 24, 2021 at 12:58

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 $$-m^3/3$$.
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 $$\chi$$.