What is the inverse Fourier transform of the japanese bracket $(1+4\pi^2|\xi|^2)^{-s/2}, s>0$ in $\mathbb{R}^d$ and how does it decay?

Question

I would like to compute and understand properties of the inverse Fourier transform of $(1+4\pi^2|\xi|^2)^{-s/2}:=\langle 2\pi\xi\rangle^{-s}$ on $\mathbb{R}^d$. This function and its inverse Fourier transform plays an important role in proving the Sobolev embedding theorem and analyzing the Sobolev spaces $H^s(\mathbb{R}^d)=\{u: \int_{\mathbb{R}^d}|\hat{u}(\xi)|^2 \langle \xi\rangle^{s}\,d\xi<\infty\}$.

Answer 1

The first step to understanding properties of the tempered distribution $$G_s(x)=\int_{\mathbb{R}^d} e(\xi\cdot x) \langle \xi\rangle^{-s}\,d\xi\in\mathcal{S}'(\mathbb{R}^d)$$ with $e(x)\equiv\exp(2\pi ix)$ is to notice that it will decay superpolynomially by virtue of the fact $\langle \xi\rangle^{-s}\in C^\infty(\mathbb{R}^d)$ is smooth. The next step is to use the following identity $$A^{-s/2}=\frac{1}{\Gamma(s/2)}\int_0^\infty t^{s/2-1}e^{-At}\,dt$$ valid for any positive $A, s$ to write $$(1+4\pi^2|\xi|^2)^{-s/2}=\frac{1}{\Gamma(s/2)}\int_0^\infty t^{s/2-1}e^{-t} e^{-\pi|\xi|^2(2\sqrt{\pi t})^2}\,dt.$$ We have hence written $\langle 2\pi\xi\rangle^{-s}$ as a linear superposition of Gaussians. This is the key step of the proof, as we can now just use the fact that $e^{-\pi|\xi|^2}$ is its own Fourier transform to find that $$G_s(x)=\frac{2^{-d}\pi^{-d/2}}{\Gamma(s/2)}\int_0^\infty e^{-t} e^{-|x|^2/4t}t^{(s-d)/2-1}\,dt$$ Our first result is that miraculously $$G_s(x)>0\quad \text{for all } x\in\mathbb{R}^d, \text{for all } d \tag{1}. $$ We now show that $G_s(x)$ has exponential decay for large values of $|x|$. Let $|x|\ge 2$. We simultaneously have that $$\frac{|x|^2}{4t} +t \ge \frac{1}{t}+t$$ and $$\frac{|x|^2}{4t}+t \ge |x| $$ for all $t>0$. Hence, we have that $$\frac{|x|^2}{4t} +t\ge \frac{1}{2t}+\frac{t}{2} + |x|/2,$$ leading to $$ G_s(x) \le \frac{2^{-d}\pi^{-d/2}}{\Gamma(s/2)}e^{-|x|/2}\int_0^\infty e^{-t/2 -1/2t} t^{(s-d)/2-1}\,dt\lesssim_{s,d}e^{-|x|/2}\tag{2}. $$ and hence exponential decay of $G_s$ (notice the integral is convergent at both ends).

Much of my answer I owe to Classical and Modern Fourier Analysis by Loukas Grafakos