I have been watching some online lectures in Physics, 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
- The "measure of the integral" is $\xi^{d-4}$, hence $I$ is a constant for $2<d<4$ in the limit $\xi\to\infty$
- Using the above, $I$ diverges with $\xi$ for $d>4$
- When $d=4$ (then"marginal case"), $I\sim\log(\xi)$
I recognise that these claims are imprecise: but that's exactly my question! How is such dimensional analysis used to determine the convergence or lack thereof of integrals in arbitrary dimension $d$? What does the "measure of the integral" mean? Furthermore, how do we know that there is no divergence due to the singularity at the origin?
The way to analyze these kind of integrals is to consider the $d^d q$ integration as radial (since there is no angular component to $q^2$ (actually $|\vec q|^2$). So that up to $\pi$'s etc $d^d q \sim q^{d-1} dq$. Now consider $$ \int_0^\xi dq {q^{d-1} \over q^2 (1+q^2)}.$$ The lower bound is controlled as long as $d>2$ as expected. For the upper bound, the term $(1+q^2)$ can be approximated as $q^2$ so the tail of the integral behaves like ${q^{d-1}\over q^4} =q^{d-5}$ so the integral diverges when $d>4$.
At $d=4$ the integral still diverges but as $\int^\xi q^{-1}\sim \log(\xi)$ This is called "marginal" because it is not a power divergence, so that reasonable "controlling" of the integral towards infinity could turn this into a convergent integral.