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

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?

• Do you know about spherical coordinates in arbitrary dimension? This would be of great help to this problem. – Redundant Aunt May 25 at 14:00
• @RedundantAunt I haven't seen it – probably123 May 25 at 15:15

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 $$C_k$$
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 $$r\to\infty$$.
• You cannot seriously write $d^{d-1}$ in such a situation. – Christian Blatter May 25 at 19:25
• Hello. Sorry, I am from a physics background where it is not uncommon to write the $d$-dimensional measure as $d^d x$, though I do know that some references us $d^D x$ in order to better distinguish the dimension and the differential. I do understand, however, that this notation can be confusing, and I apologize if that is the case here (if this is indeed what your comment is pointing to). I myself like this notation, and prefer it to alternatives I've seen. – Zachary May 25 at 19:52