# show that integral converges even if it has a singularity

i am currently reading through a book on generalized functions, and there it is said that:

... $\int_{|x|\le r} |x|^{-t} dx$ converges for $t < n$ (in $n$ dimensions) and diverges for $t \ge n$.

Why does it converge for $t < n$, its area/volume still goes to infinity near zero?

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Use spherical coördinates. The $c_n$ shown below is the surface area measure of $S^{n-1}$.
$$\int_{B_r(0)} |x|^{-t}\, dx = \int_{S^{n-1}}\int_0^r \rho^{-t} \rho^{n-1}\,d\rho\,d\sigma = c_n \int_0^r \rho^{n-1-t}d\rho$$
The last integral converges iff $n-1-t > -1$, or if $n > t$.