Given $f:\mathbb{R}^n \to \mathbb{R}^1$ such that $$f(x_1, x_2, x_3,...,x_n)=\frac{1}{(x_1^2+x_2^2+x_3^2+...+x_n^2)^k},$$ what values of $k$ would make the hypervolume of $f(x_1,x_2,x_3,...x_n)$ within a unit $n$-sphere finite? What values would make it infinite?
My attempt thus far has been to integrate it over all $n$ variables, redefining the bounds in hyperspherical coordinates in order to simplify the demonimator into $\text{radius}^{2k}$.
There is a general concept of "spherical coordinates" in $n$-dimensional Euclidean space. If $r$ denotes the distance from the origin and $dS$ denotes the $(n-1)$-dimensional volume element on the unit sphere, the Euclidean volume element is $dV = r^{n-1}\, dr\, dS$.
The integrand, meanwhile, is $f(x) = r^{-2k}$. The enclosed volume is therefore finite in the sense asked if and only if the improper integral $$ \int_{0}^{1} r^{-2k} r^{n-1}\, dr = \int_{0}^{1} r^{n-2k-1}\, dr $$ converges, if and only if $-1 < n - 2k - 1$, or $k < n/2$. (You may enjoy checking this concretely for polar coordinates in the plane, where $dV = r\, dr\, d\theta$, and ordinary spherical coordinates in three-space, where $dV = r^{2} \sin\phi\, dr\, d\theta\, d\phi$.)
If it matters, the integral of your function over the unit ball is the value of the integral above multiplied by the $(n - 1)$-dimensional volume of the unit sphere ($2\pi$ for $n = 2$, $4\pi$ for $n = 3$, $2\pi^{2}$ for $n = 4$, and so forth).
