Let $s>0$, show that $\mathscr{F}[(1+|x|^2)^{-s}]\in L^1(\mathbb{R}^d)$.
The original goal is to prove that $W^{s,p}(\mathbb{R}^d)\hookrightarrow L^p(\mathbb{R}^d)$ for all $s>0,1\le p\le \infty$, where the norm of $W^{s,p}$ is $||f||_{s,p}=||\mathscr{F}[(1+|\omega|^2)^s\hat f(\omega)]||_p$
For $x\in\mathbb{R}^d$, define $\displaystyle G(x)=\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt$. Since $t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}>0$, $0<G(x)\le\infty$ is well-defined.
By Tonelli's theorem $\displaystyle||G||_1=\int_{\mathbb{R}^d}dx\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt=\int_0^\infty t^{s-1-d/2}e^{-t}dt\int_{\mathbb{R}^d}e^{-\frac{|x|^2}{4t}}dx\\\displaystyle=(4\pi)^{d/2}\int_0^\infty t^{s-1}e^{-t}dt=(4\pi)^{d/2}\Gamma(s)<\infty$
So $G\in L^1(\mathbb{R}^d)$.
By Fubini's theorem $\displaystyle\hat G(\xi)=\int_{\mathbb{R}^d}G(x)e^{-2\pi i x\xi}dx=\int_0^\infty t^{s-1-d/2}e^{-t}dt\int_{\mathbb{R}^d}e^{-\frac{|x|^2}{4t}}e^{-2\pi i x\xi}dx\\\displaystyle=(4\pi)^{d/2}\int_0^\infty t^{s-1}e^{-t}e^{-4\pi^2 t|\xi|^2}dt=(4\pi)^{d/2}\int_0^\infty (1+4\pi^2 t|\xi|^2)^{-s}w^{s-1}e^{-w}dw=(4\pi)^{d/2}\Gamma(s)(1+4\pi^2 t|\xi|^2)^{-s}$.
By uniqueness of Fourier transform, $\displaystyle\mathscr{F}^{-1}[(1+4\pi^2 t|\xi|^2)^{-s}](x)=\frac{G(x)}{(4\pi)^{d/2}\Gamma(s)}\in L^1(\mathbb{R}^d)$