Fourier transform has finite norm.

Question

Let $f:\mathbb R^d\rightarrow \mathbb R,x\mapsto \frac{1}{1+\|x\|_2^d}$. Show that $\hat{f}\in L^2(\mathbb R^d)$
My idea is to show that $\|\hat{f}\|_2<\infty$. With the theorem of Plancherel I get $\|\hat{f}\|_2^2=\|f\|_2^2=\int_{\mathbb R^d}\frac{1}{(1+\|x\|_2^d)^2}dx$. Why is this finite?

Answer 1

Use polar coordinates in $\mathbb R^{d}$ (p, 176, Rudin's RCA):

$\int_{\mathbb R^d}\frac{1}{(1+\|x\|_2^d)^2}dx=C\int_0^{\infty} \frac {r^{d-1}} {(1+r^{d})^{2}}dr$ where $C=\sigma_{d-1}(S_{d-1})$ is a finite constant. The integral here is finite since $2d-(d-1)=d+1>1$.

[$C$ is $d-$ times the Lebesgue measure of the unit ball in $\mathbb R^{d}$].

Alternative proof: Let $B_r=\{x\in \mathbb R^{d}: 0<\|x\| \leq r\}$. Let $D_n=B_{n+1}\setminus B_n$. Note that $D_n$'s are disjoint and their union is $ \mathbb R^{d}$. Now $\int_{D_n}\frac 1 {(1+\|x\|^{d})^{2}}dx \leq \frac 1 {(1+n^{d})^{2}}m_d(D_n)$ where $m_d$ is Lebesgue measure on $\mathbb R^{d}$. But $m_d(D_n)=m_d(B_{n+1})-m_d(B_n)=C[(n+1)^{d}-n^{d}]$. Finally note that $\sum_n \frac {(n+1)^{d}-n^{d}} {(1+n^{d})^{2}} <\infty$.