convergence of integral for s>n/2 I want to prove that $\int_{\mathbb{R^n}}\frac{1}{(1+|x|^2)^s}dx$ is convergent for $s>n/2$.
The only thing which comes to my head is using the $\arctan$ function, but then I got stuck.
Any hints are welcome!
Thank you!
 A: This is equivalent to prove $$\int_{|x|\ge1}\frac{1}{|x|^s} dx< \infty   \tag{1} $$ for $s>n$. 
To show this, think about how you show that $$\int_1^{\infty}\frac{1}{x^s}dx<\infty$$ for $s>1$ in dimension 1.
You write $$\int_1^{\infty}\frac{1}{x^s}= \sum_{k=1}^{\infty}\int_k^{k+1}\frac{1}{x^s}dx\le \sum_{k=1}^{\infty}\int_k^{k+1}\frac{1}{(k+1)^s}dx=\sum_{k=1}^{\infty}\frac{1}{(k+1)^s}dx < \infty.$$
So my hint for you is try to write integral $(1)$ as infinite sum and then estimate.
A: Hint: let $B\subset\mathbb{R}^{n}$ denote the unit ball. Then,
$$
\int_{\mathbb{R}^{n}}\frac{1}{\left(1+\left|x\right|^{2}\right)^{s}}\leq\int_{\mathbb{R}^{n}}\frac{1}{\left|x\right|^{2s}}\leq\int_{B}\frac{1}{\left|x\right|^{2s}}+\int_{\mathbb{R}^{n}\setminus B}\frac{1}{\left|x\right|^{2s}}\leq L+\int_{\mathbb{R}^{n}\setminus B}\frac{1}{\left|x\right|^{n}}
$$
where $L$ is some finite number. Therefore, it suffices to show that the integral on $\mathbb{R}^{n}\setminus B$
is finite.
A: This is pretty much the same as the solution from par, but I wrote it out so I'll submit it. I left out positive constants.
$$
\int_{\mathbb{R}^n}\frac{1}{(1+|x|^2)^s}dx\propto \int_{0}^\infty\frac{r^{n-1}}{(1+r^2)^s}dr 
$$
Substitute $u=1+r^2$
$$
\int_{0}^\infty\frac{r^{n-1}}{(1+r^2)^s}dr \propto\int_{1}^\infty\frac{(u-1)^{n/2-1}}{u^s}du<\int_1^\infty u^{n/2-1-s}du
$$
This integral will converge only when
$$
n/2-1-s < -1$$
or
$$s>n/2$$
