How to prove $\int_{\mathbb R^d}{(1+|x|^{2})^{-m}dx}$ is integrable?

Question

$\forall x \in \mathbb R^d$ and $m > d/2$.

I know that $\int_{\mathbb R^d}{(1+|x|^{2})^{-m}dx}$ is integrable, but how can I prove?

Thanks!

Answer 1

Note that $\dfrac{1}{1+x^{2}} \leq \dfrac{1}{x^{2}}.$ Let $B$ be a ball centered at the origin and of radius $\epsilon.$ Let me split up the integral into two parts. The first, without the ball $B,$ and the second one within that ball. Changing to polar coordinates, you see: $\displaystyle \int _{\mathbb{R}^{d}-B}\dfrac{1}{(1+x^{2})^{m}} \leq \int _{\mathbb{R}^{d}-B}\dfrac{1}{x^{2m}}=\int_{\epsilon}^{\infty}\int_{S^{d-1}}\dfrac{1}{r^{2m}}r^{d-1} dr d\sigma_{d-1}=\omega(n)\int_{\epsilon}^{\infty} r^{d-1-2m}. $ If $m$ is greater that $d/2,$ we have $2m=d+\delta$ for some positive $\delta$ and then our integral becomes $\int_{\epsilon}^{\infty}\dfrac{1}{x^{1+\delta}}dx$ which is clearly integrable.

For the integral within $B$, just note that you are integrating a bounded function in finite-measure set! so no problem there.

I hope this helps

Answer 2

Use spherical coordinates, so that the integral becomes $$ I = \int{\mathrm{d}\Omega_{d-1}}\int_0^{+\infty}{\frac{r^{d-1}}{(1+r^2)^m}\mathrm{d}r}, $$ which is obviously summable at $ r = 0 $, while to have integrability at infinity you need the integrand to decay more rapidly than $ r^{-1} $. Since the integrand's rate of decay at infinity is $ r^{d-1-2m} $ this is achieved for $$ d - 1 - 2m < -1 \implies m > \frac{d}{2}. $$