Showing a function is integrable Let $\xi, \zeta\in\mathbb{R}^m$. How might one try to show that $f:\mathbb{R}^m\times\mathbb{R}^m\rightarrow \mathbb{R}$, defined by $\displaystyle\frac{1}{(1+\left|\xi - \zeta\right|)^{k}}$ is or is not absolutely integrable over $\mathbb{R}^{m}\times\mathbb{R}^m$ for sufficiently large $k$? I can see that the problem occurs on the "diagonal", when $\xi = \zeta$, but I'm not sure how to show that this isn't such a big problem that we lose integrability.
Any hints would be welcome.
 A: This isn't integrable for any $k$. Change coordinates to $x=\xi-\zeta$ and $y=\xi+\zeta$. Then the problem becomes
$$\int_{\mathbb R^m}\int_{\mathbb R^m}\frac{1}{(1+|\xi-\zeta|)^k}d\xi d\zeta=\int_{\mathbb R^m}\int_{\mathbb R^m}\frac{1}{(1+|x|)^k}dxdy=\int_{\mathbb R^m}Cdy=\infty.$$
If you're not convinced, ask Wolfram alpha to do it for $m=1$ and $k=10$.
A: Let $A_n$ be the hypervolume of the $(n–1)$-dimensional unit sphere (i.e., the "area" of the surface of the $n$-dimensional unit ball). We have
$$ A_n = \frac{2\pi^{n/2}}{\Gamma(n/2)} \tag{1} $$
and by Cavalieri's principle (i.e. integrating along spherical shells), in $\mathbb{R}^n$ we have:
$$ \int_{|x|\leq R}\frac{d\mu}{1+|x|^k}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{0}^{R}\frac{\rho^{n-1}}{1+\rho^k}d\rho\tag{2}$$
the RHS of $(2)$ is convergent as $R\to +\infty$ iff $\color{red}{k>n}$, and in such a case:
$$ \int_{\mathbb{R}^n}\frac{d\mu}{1+|x|^k} = \color{red}{\frac{2\pi^{1+n/2}}{k\,\Gamma(n/2)\sin\left(\frac{\pi n}{k}\right)}}.\tag{3} $$
In a similar fashion,
$$ \int_{|x|\leq R}\frac{d\mu}{(1+|x|)^k}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{0}^{R}\frac{\rho^{n-1}}{(1+\rho)^k}d\rho \tag{4}$$
is also convergent as $R\to +\infty$ iff $\color{red}{k>n}$, and in such a case:
$$\int_{\mathbb{R}^n}\frac{d\mu}{(1+|x|)^k} = \color{red}{\frac{2\pi^{n/2}\Gamma(k-n)\Gamma(n)}{\Gamma(n/2)\Gamma(k)}}.\tag{5}$$
