Consider a function $f: \mathbb{R}^n \to \mathbb{R}$ that has a singularity at point $x_0$ such that $\lim_{x\to x_0} f(x) = +\infty$. When we can be sure that this singularity is integrable? (you can assume that f is continuous for $x \neq x_0$ and that outside of some neighborhood of $x_0$ it equals to zero).
For the case $n = 1$ it is enough to check that $f(x) \sim |x - x_0|^\alpha$ near $x_0$for some $\alpha > -1$ (are there similar criteria for $n>1$?).
Yes, it suffices if $|f(x)| \lesssim |x - x_0|^{\alpha}$ with $\alpha > -n$.
To see this, use integration in polar coordinates (assume w.l.o.g. that $x_0 = 0$, otherwise first apply a translation):
$$ \int_{\Bbb{R}^n} |f(x)| \,dx = \int_0^\infty r^{n-1} \cdot \int_{S^{n-1}} |f(r\xi)| \,d\xi \, dr = \int_0^R r^{n-1} \cdot \int_{S^{n-1}} |f(r\xi)| \,d\xi \, dr, $$
where $f(x) = 0$ for $|x| > R$.
Here, $d\xi$ denotes the surface measure on $S^{n-1}$. The only thing you have to know here is that $\int_{S^{n-1}} 1 \,d\xi \in(0,\infty)$.
EDIT: For the formula for integration in polar coordinates, see e.g. here Integral formula for polar coordinates
EDIT 2: Using the above calculation, you can also show that $|f|$ is not integrable if $|f(x)| \gtrsim |x-x_0|^\alpha$ with $\alpha < -n$.