Is the Fourier transform of $e^{-|x|^n}$ positive? Let
$$\Phi(x) = \int_{\mathbf{R}^n} e^{-|y|^n +i (x,y)} dy.$$
Is $\Phi$ positive everywhere in $\mathbf{R}^n$?
Could someone helps me answer this question or gives a reference for it? Thanks. 
 A: Here is a full characterization.
Theorem. The function $e^{-|x|^a}$ is positive definite for $0\le a \le 2$ and is not positive definite for $a>2$. Thus, its Fourier transform is positive and non positive for the said ranges.
Proof. The claim $a=0$ is trivial. Assume $0<a<2$. Then, it can be shown that
\begin{equation*}
-|x|^a = C_a\int_{-\infty}^\infty \frac{\cos (xt) - 1}{|t|^{a+1}}dt,
\end{equation*}
where $C_a$ is a constant depending on $a$. Since $\cos(xt)$ is a positive definite function, we see that $-|x|^a$ is conditionally positive definite (because of the $-1$ term). Hence, $\exp(-|x|^a)$ is positive definite, and consequently by Bochner's theorem, it's FT is positive.
For $a>2$, it is easy to construct numerical examples where the associated function is not positive definite, and hence its FT is not positive. Carlo Beenakker's answer gives an example. 
To obtain a formal proof of this, here's an outline by contradiction. In particular, suppose that for some $a > 2$, the kernel $e^{-|x-y|^a}$ is positive definite, $|x-y|^a$ is negative definite. Thus, by appealing to Schoenberg's theorem, it must be the case that $d(x,y) := |x-y|^{a/2}$ is a metric on $\mathbb{R}$. Choosing $x,y,z=(0,1,2)$ and comparing $d(x,y)=d(y,z)=1$ but $d(x,z)=2^{a/2}>2$, a contradiction to the triangle inequality. 
Reference. Chapter 5, Positive definite matrices, R. Bhatia. Princeton University Press, 2007.
A: no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have
$$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$
I checked that it becomes negative for $n=3$:
$$\Phi_3(x)=\frac{4\pi}{|x|}\int_0^\infty e^{-r^3}r\sin(|x|r)\,dr$$

