$f(x)=1/(|x|+1)^k$ defined on $\Bbb R^k$ is not integrable

Question

Consider the function $f(x)=1/(|x|+1)^k$ defined on $\Bbb R^k$. I am trying to show that $f$ is "not" integrable on $\Bbb R^k$, i.e., $\int_{\Bbb R^k}f~dx=\infty$. I tried to show that $\int_{B(0,r)}f~dx \to \infty$ as $r\to \infty$, but it does not work so welk. Any hints?

Answer 1

As suggested by Redundant Aunt above, the whole things is a trivial if one uses polar coordinates (to its generalization to $n$ dimensions). Still, one can show that the integral diverges through simple inequalities.

Without loss of generality, assume $|x|=\max_j|x_j|$, the $\ell_\infty$ norm on $\mathbb{R}^k$. With this norm, the ball $B(0;r)=[-r,r]^k$. Then \begin{aligned} \int_{R^k}\frac{1}{(1+|x|)^k}dx&=\sum^\infty_{n=0}\int_{n<|x|\leq n+1}\frac{1}{(1+|x|)^k}dx\geq \sum^\infty_{n=0}\frac{(n+1)^k-n^k}{(1+(n+1))^k}\\ &\geq C_k\sum^\infty_{n=0}\frac{n^{k-1}}{(1+(n+1))^k)} \end{aligned} for some constant $C_k$

Answer 2

We work in $d=k$ dimensions. Expanding on Redundant Aunt's comment, we can assign an expression for the integral in terms of spherical coordinates as follows, $$\int_{S^{d-1}} d^{d-1}\Omega \int_0^\infty \frac{r^{d-1}}{(1+r)^d}\,dr,$$ where $r^{d-1}\,d^{d-1}\Omega\,dr$ is the spherical measure in $d$ dimensions. The radial integrand is $1/r + \mathcal{O}\left( 1/r^2\right)$ as $r\to\infty$ so that the integral is divergent. In fact, I think you can show that $\int_{B(0,r)}f \, d^d x =c(d)\log r + \mathcal{O}\left(1\right)$ as $ r\to\infty$.