Does the integral $\int_{\mathbb{R}^n\setminus B(0,1)}\frac{1}{1+|y|^n}dy$ converge? Does the integral $\int_{\mathbb{R}^n\setminus B(0,1)}\frac{1}{1+|y|^n}dy$ converge?
I'm trying ultimately to show that $\int_{\mathbb{R}^n}\frac{1}{1+|y|^{2s}}dy$ converges if $2s>n$ by breaking this up into two integrals: 
$\int_{\mathbb{R}^n\setminus B(0,1)}\frac{1}{1+|y|^n}dy$ + $\int_{B(0,1)}\frac{1}{1+|y|^n}dy$
where the second integral is bounded above by 1, and clearly converges.
Any help appreciated!
 A: The surface area of $\|y\|=R$ in $\mathbb{R}^n$ is given by $\frac{2\pi^{n/2}}{\Gamma(n/2)}R^{n-1}$, hence you can do much better than estimating such integral, you can compute it.
$$ \int_{\|y\|\geq 1}\frac{d\mu_n}{1+\|y\|^m} =\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{1}^{+\infty}\frac{R^{n-1}}{1+R^m}\,dR=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{0}^{1}\frac{R^{m-n}}{R(R^m+1)}\,dR$$
is convergent as soon as $m>n$, and in such a case it equals
$$ \frac{2\pi^{n/2}}{m\,\Gamma(n/2)}\int_{0}^{1}R^{1-n/m}\left(\frac{1}{R}-\frac{1}{R+1}\right)\,dR=\frac{2\pi^{n/2}}{m\,\Gamma(n/2)}\sum_{k\geq 0}\frac{(-1)^k}{(k+1)-\frac{n}{m}}. $$
In a similar fashion, for any $m>n$ we have:
$$ \int_{\mathbb{R}^n}\frac{d\mu_n}{1+\|y\|^n} = \frac{2\pi^{n/2}}{\Gamma(n/2)}\int_{0}^{+\infty}\frac{R^{n-1}}{1+R^m}\,dR=\frac{2\pi^{1+n/2}}{\Gamma(n/2)\,m \sin\frac{\pi n}{m}}.$$
A: $$ \int_{\mathbb{R}^\setminus B(0,1)}\frac{1}{1+|y|^m}dy \le \int_{\mathbb{R}^n\setminus B(0,1)}\frac{1}{|y|^m}dy=\sum_{k=1}^{\infty}\int_{2^{k-1}<|y|\le2^k}\frac{1}{|y|^m}dy \le \sum_{k=1}^{\infty}\int_{2^{k-1}<|y|\le2^k}\frac{1}{|2|^{m(k-1)}}dy =\sum_{k=1}^{\infty}\frac{v_n(2^k)^n-v_n(2^{k-1})^n}{|2|^{m(k-1)}} =v_n\sum_{k=1}^{\infty}\frac{2^{kn}-2^{nk-n}}{2^{m(k-1)}}   =v_n\frac{1-2^{-n}}{2^{-m}}\sum_{k=1}^{\infty}\frac{2^{kn}}{2^{mk}}= v_n\frac{1-2^{-n}}{2^{-m}}\sum_{k=1}^{\infty}2^{(n-m)k}$$
The latter sum converges if $m > n$, Using the dominated convergence theorem, you can now prove the first integral exists.
The second inequality follows since the volume of the region where $2^{k-1}<|y|\le2^k$ is equal to $v_n(2^k)^n-v_n(2^{k-1})^n$ where $v_n$ is the volume of the unit sphere in $R^n$
