# How can one know that a given integral diverges?

In a d-dimensional integral of a rational function,How can we know that an integral diverges or converges depending on d and the highest powers of the numerator and denominators? What I understand is that if a 4-dimensional minkwoski integral is spherically symmetric with respect to the origin ,One can transform $d^{3}k$ to spherical coordinates and so obtain a function of one variable which is easy but I don't know how to generalize this to the general d-dimensional minkwoski spacetime ?

-

## migrated from physics.stackexchange.comJan 18 '13 at 13:09

This question came from our site for active researchers, academics and students of physics.

$$\int^\Lambda d^dk \frac{k^n}{(k^{2}-\mu^2)^m}\approx\int^\Lambda dk k^{d-1} \frac{k^n}{(k^{2}-\mu^2)^m}\approx \Lambda\Lambda^{d-1+n-2m}=\Lambda^{d+n-2m}$$
this is the behaviour of the integral in the ultraviolet regime, namely $k$ that goes to infinity. You have to think $\Lambda$ as a constant (cut-off) and then push it to infinity to understand the kind of divergence. Remind that: $$\int^\Lambda dk \frac{1}{k}=\log\Lambda \quad log\quad divergent$$
$$\int^\Lambda dk =\Lambda \quad linearly\quad divergent$$
$$\int^\Lambda dk k =\Lambda^2 \quad quadratically\quad divergent$$
I used $\approx$ because I am omitting the solid angol in d-dimension. – Gauge Jan 18 '13 at 12:20
Thanks alot , but I'm confused why $d^{d}k$ gives $dkk^{d-1}$ when transforming to spherical coordinates . It seems easy but I can't get it – nabilahmed Jan 18 '13 at 12:24
It's there because the volume of (d-1)-dimensional sphere goes like $k^{d-1}$ .Right? – nabilahmed Jan 18 '13 at 12:29