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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?

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

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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$.

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  • $\begingroup$ You cannot seriously write $d^{d-1}$ in such a situation. $\endgroup$ – Christian Blatter May 25 at 19:25
  • $\begingroup$ 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. $\endgroup$ – Zachary May 25 at 19:52

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