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

Question

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 \begin{equation} 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}|. \end{equation}

It is claimed that

1. The "measure of the integral" is $\xi^{d-4}$, hence $I$ is a constant for $2<d<4$ 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.

Answer 1

Here we try to give a conceptional (rather than a computational) answer to OP's 3 questions:

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 $d>2$.