# Checking if an integral converges (or diverges) using dimensional analysis

Cross posted at Math.SE

I have been watching some online lectures, and the lecturer uses dimensional analysis to make claims such as the following:

Consider the integral $$$$I(\xi, d) = \int_0^\xi \frac{\mathrm{d}^\mathrm{d}q}{(2\pi)^d} \frac{1}{q^2(q^2+1)}\quad\text{where}\quad q=|\boldsymbol{q}|.$$$$

It is claimed that

1. The "measure of the integral" is $$\xi^{d-4}$$, hence $$I$$ is a constant for $$2 in the limit $$\xi\to\infty$$
2. Using the above, $$I$$ diverges with $$\xi$$ for $$d>4$$
3. When $$d=4$$ (then"marginal case"), $$I\sim\log(\xi)$$

I recognise that these claims are imprecise: but that's exactly my question!

1. How is such dimensional analysis used to determine the convergence or lack thereof of integrals in arbitrary dimension $$d$$?

2. What does the "measure of the integral" mean?

3. Furthermore, how do we know that there is no divergence due to the singularity at the origin?

EDIT

I'm accepting Qmechanic ♦'s answer below. For a purely mathematical answer (which I found very clear), see the linked Math.SE cross post.

• Link to online lecture? Which minute? Commented Nov 7, 2022 at 8:03
• Sorry, the lectures are not publicly accessible. The comments are offhand, with no more detail than is given here - hence my confusion! Commented Nov 7, 2022 at 8:39
• What we need isn't so much dimensional analysis as $\mathrm{d}^dq=q^{d-1}\mathrm{d}q\mathrm{d}\Omega$ (though I can see why that's labelled DA), so $I(\infty,\,d)$ diverges due to a $O(q^{d-5})$ integrand for large $q$ provided $d\ge4$, or due to a $O(q^{d-3})$ integrand for small $q$ provided $d\le2$, but converges for $d\in(2,\,4)$. (In fact, if we allow $d$ to be complex in a suitable continuation scheme, the convergence condition is $\Re d\in(2,\,4)$.)
– J.G.
Commented Nov 7, 2022 at 9:24

1. The relevant notion is the superficial degree of (UV) divergence $$D$$, see e.g. my related Phys.SE answer here.
2. The measure of the integral usually refers to $$\mathrm{d}^\mathrm{d}q$$. It seems the lecturer instead means that the measure of the integral is $$\xi^D$$, which is nonstandard terminology.
3. Infrared (IR) singularities from massless fields are often regularized by giving them a small mass. (Also we assume that the integral has been Wick-rotated to Euclidean signature.) In OP's integral the possible IR singularity is the reason for the mentioned lower bound $$d>2$$.
• One of the bounds on $d$ for convergence comes from avoiding UV divergence; the other comes from IR divergence.
• $\uparrow$ Right. Commented Nov 7, 2022 at 9:26