# Cutoff-Scheme Renormalization and Order of Integration in QFT

The following is the result of Fubini's Theorem, describing when you can replace a double integral with an iterated integral safely:

For a set $X \times Y \subset \mathbb{R}^2$, if $\iint |f(x,y)| d(x,y)$ is a finite number the we have the equality: $$\iint_{X\times Y} f(x,y) d(x,y) = \int_X \left[ \int_Y f(x,y) dy \right] dx = \int_Y \left[ \int_X f(x,y) dx \right] dy$$

So this tells us: We can safely switch the order of integration if integrating the absolute value of the integrand gives a finite number.

If the condition is not met, then the iterated integrals may converge to different values. For example (as given in the wiki) we have the following: $$\int_{0}^{1} \left[ \int_{0}^{1} \frac{x^2-y^2}{(x^2+y^2)^2} dx \right] dy = - \frac{\pi}{4}\\ \int_{0}^{1} \left[ \int_{0}^{1} \frac{x^2-y^2}{(x^2+y^2)^2} dy \right] dx = \frac{\pi}{4}$$

Renormalization: When we encounter a divergent Feynman integral, a renormalization method is to introduce a cutoff so as to characterize 'how' the integral diverges (ie, like a logarithm, or power-law etc.). For example; $\int_{\mathbb{R}^4} \frac{d^4p}{p^2 - m^2}$ is UV divergent, and diverges like $\propto \Lambda^2 - m^2 \log\left( \frac{\Lambda^2}{m^2} \right)$ (for $\Lambda \gg m$ a UV-cutoff).

My Question: What if we have a Feynman integral that diverges differently depending on the order in which you integrate it?

As an example, I have the following contrived integral: $$\mathcal{I} := \int_0^\infty dt \int_0^\infty dx\ \frac{\ln(x)^2}{x(1-x^2)(\ln(x)^2+t^2)}$$

This integral diverges at a few places, but let's just consider how it diverges near $(t,x) = (0,0)$.

CASE I: Integrate $t$ away first, and we are left with the integral $\int_0^\infty \frac{dx}{x(1-x^2)|\ln(x)|}$. This integral diverges at $x=0$ in the following way $\propto \ln\left(\ln(\tfrac{1}{x})\right)$ (for $x \approx 0$) (you can check the asymptotics on this).

CASE II If you integrate $x$ away first (define a new variable $q = \ln(x)$ and manipulate it to get $\int_0^\infty \frac{q^2}{(q^2+t^2)^2} = \frac{\pi}{4|t|}$), we are left with the integral $\int_0^\infty \frac{\pi}{4t}$, which diverges like $\propto \log(t)$ for $t \approx 0$.

So we see that the integral diverges differently depending on the way in which you integrate this.

What if one were to encounter such a situation physically? Is it not allowed for some reason? How do we renormalize things if the way our integral diverges can vary? Furthermore, what if the variable upon which we need to place our cutoff is different depending on the order of integration? This seems very concerning!

Is this situation too contrived, and would not happen in an actual physically relevant QFT problem?

• Minor comment: $\Lambda\gg0$ is a meaningless statement. You probably meant $\Lambda\gg m$. – AccidentalFourierTransform Oct 31 '17 at 18:11
• Note that the form of the divergences in a healthy QFT is very constrained (cf. Weinberg's theorem). In any case, the proper way to proceed is to consider regulated theories, where all the integrals are always absolutely convergent (so that Fubini is safe). – AccidentalFourierTransform Oct 31 '17 at 18:15
• Forgive my ignorance, but which weinberg theorem are you referring to? Is it in his QFT textbooks? – Greg.Paul Oct 31 '17 at 19:37
• Yes: chapter 12 is in essence a detailed discussion of the possible forms of divergent terms in an arbitrary Feynman diagram. – AccidentalFourierTransform Oct 31 '17 at 19:44

Usually, when calculating Feynman's integrals, we perform a Wick's rotation so that: $$\int f(p^2)d^4p\rightarrow\int f(p_E^2)d^4p_E$$ where: $p^2 = p_0^2-\vec{p}^2$, $p_E^2 = p_0^2+\vec{p}^2$, then do the integration in the 4-dimensional Euclidean space by using the 4-dimensional solid angle: $d^4p_E = p_E^3d\Omega_4$, where $p_E$ is the norm in 4-dimensional space having the meaning of energy.
The function in the Feynman integral with $p_E$ is symmetric in term of $p_0$ and $\vec{p}$, so, i think it is not necessary to concern about the Fubini's theorem here.