Let $f(x)=(1+|x|^2)^{-a}$, with $x\in \Bbb R^n$ and $a>0$. Show that $f(x)$ is a constant multiple of $g(x)$. Let $f(x)=(1+|x|^2)^{-a}$, with $x\in \Bbb R^n$ and $a>0$. Show that $f(x)$ is a constant multiple of $$g(x)=\int_0^\infty t^{\alpha -1}e^{-t(1+|x|^2)}~\mathrm{d}t$$Use this fact to conclude that $\hat{f}$ is a positive function.
How would I do this proof?
The change of variables $t=u/(1+|x|^2)$ in the integral defining $g$ shows that $$ g(x)=\frac{1}{(1+|x|^2)^a}\int_0^\infty u^{a-1}e^{-u}du=\Gamma(a)f(x) $$ This implies that $$\eqalign{ \hat{f}(w)&=\frac{1}{\Gamma(a)}\int_0^\infty u^{a-1}e^{-u}\left(\int_{\mathbb{R}^n} e^{-u |x|^2+i\langle x,w\rangle}dx \right) du\cr &=\frac{1}{\Gamma(a)}\int_0^\infty u^{a-1}e^{-u}\prod_{k=1}^n\left(\int_{\mathbb{R}} e^{-u x_k^2+i x_kw_k }dx_k \right) du\cr &=\frac{1}{\Gamma(a)}\int_0^\infty u^{a-1}e^{-u} \prod_{k=1}^n\left(\sqrt{\frac{\pi}{u}}e^{-w_k^2/{4u}}\right)du\cr &=\frac{\pi^{n/2}}{\Gamma(a)}\int_0^\infty u^{a-1-n/2}e^{-u-|w|^2/(4u)}du } $$ which is clearly positive.
