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For what $n \geq 0, n \in \mathbb{Z}$ it is true that $f(t) = e^{-|t|^n}$ is a characteristic function?

I tried to use Bochner's theorem, but got stuck with it. $e^{-|z_j - z_k|^n} \geq 0$ for all n, but I have no idea how to evaluate the whole sum. (Determining if something is a characteristic function)

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sooobus
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    For $a$ real, the function $\varphi_a(t)=e^{-|t|^a}$ defines a characteristic function if and only if $0<a\leqslant2$. Thus the answer to your question is: $n=1$ or $n=2$. To show that no $n>2$ yields a characteristic function, choose any $a>2$, then it happens that $$\lim_{s\to0}s^{-2}\lim_{t\to0}|t|^{-2a}\left|\begin{matrix}\varphi_a(0)&\varphi_a(t)&\varphi_a(t+ts)\\varphi_a(-t)&\varphi_a(0)&\varphi_a(ts)\\varphi_a(-t-ts)&\varphi_a(-ts)&\varphi_a(0)\end{matrix}\right|=-a^2<0$$ hence $\varphi_a$ is not a characteristic function (why?). My guess is ... – Did Dec 07 '16 at 00:27
  • ... that shortcuts to this are available when $n=3$ or $n=4$ or... – Did Dec 07 '16 at 00:27

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