Zeros of Fox-Wright Function This question is stimulate by the previous two question here and here.   We are interested in studying the following special case of Fox-Wright function
\begin{align}
 \Psi_{1,1}  \left[ \begin{array}{l} (1/k,2/k) \\ (1/2,1)\end{array} ; -x^2\right], x\in \mathbb{R}, k\in (1,\infty.
\end{align} 
Where    Fox-Write function $\Psi_{1,1}  \left[ \begin{array}{l} (a,A) \\ (b,B)\end{array} ; z\right]$ is defined as
\begin{align}
\Psi_{1,1}  \left[ \begin{array}{l} (a,A) \\ (b,B)\end{array} ; z\right] =\sum_{n=0}^\infty \frac{\Gamma(a+An)}{\Gamma(b+Bn)} \frac{z^n}{n!}.
\end{align}
The reason this is interesting is because of the following Fourier cosine:
\begin{align}
\Psi_{1,1}  \left[ \begin{array}{l} (1/k,2/k) \\ (1/2,1)\end{array} ; -x^2\right]= \int_0^\infty \cos(xt) e^{-t^k} dt
\end{align}
see this question for more details. 
Goal We are interest in finding  conditions on $k$ such that   $ \Psi_{1,1}  \left[ \begin{array}{l} (1/k,2/k) \\ (1/2,1)\end{array} ; -x^2\right]$ has no zeros. 
The current conjecture is that there are no zeros for $k \le 2$ and at least one zero for $k>2$.  Thanks for any ideas you might have. 
Edit: Perhaps there is a connection to the notion of Stable Distribution in probability. See comments below.
 A: I'll sketch an answer.
I use the notation:
$$
\psi_k(z) = \sum_{n=0}^\infty \frac{\Gamma(\frac{1}{k} + \frac{2n}{k})}{\Gamma(\frac12 + n)} \cdot \frac{z^n}{n!},
$$
and
$$
\phi_k(z) = \int_{0}^\infty \cos(z t)e^{-t^k} \, \rm{d} t.
$$
First of all, the correct identity seems to be:
$$ \phi_k(x) = \frac{\sqrt{\pi}}{k} \psi_k\left(-\left(\frac{x}{2}\right)^2\right).$$
Elementary cases ($\phi_k > 0$):
$$k=1\rm{:}\,\, \phi_1(x) = \frac{1}{1+x^2}, \quad k=2\rm{:}\,\, \phi_2(x) = \frac{\sqrt{\pi}}{2} e^{-\frac{x^2}{4}}. $$
The case $k>2$:
The function $e^{-t^k}$ is differentiable at $t=0$ (from the right, at least twice), so integrating by parts twice, we have $\phi_k(x) = O(x^{-2})$. Thus, $\phi_k(x)$ is integrable (it is clearly bounded). Since $\cos(z) = \frac12 (e^{iz} + e^{-iz})$, we can write
$$ \phi_k(z) = \frac12 \int_{-\infty}^\infty e^{izt} e^{-|t|^k} \, \rm{d} t.$$ We might think of $\frac{1}{\pi} \phi_k(x)$ as the "density" of a random variable $X$, whose characteristic function is $\varphi_k(t) = e^{-|t|^k}$. If $\varphi_k(t)$ was a characteristic function, then since it is at least twice differentiable at $t=0$, we would have $\mathbb{E} [X^2] = -\varphi_k^{(2)} (0) = 0$. Thus $X$ is a trivial random variable, and the contradiction implies that $\frac{1}{\pi} \phi_k$ is not a density.
The case $k<2$:
I'm not sure if any easy proof is available. An indirect way to prove $\phi_k > 0$ is to show
$$ \varphi_k(t) = \int_0^\infty e^{-t^2 y} \, \rm{d} \mu_k(y) ,\quad t \in \mathbb{R}, \tag{1} \label{eq:sample} $$
for some non-negative Radon measure $\mu_k$. Then we have,
$$ \phi_k(x) = \frac12 \int_{-\infty}^\infty e^{i x t} \varphi_k(t) \, \rm{d} t = \frac12 \int_0^\infty \left[ \int_{-\infty}^\infty e^{i x t - t^2 y} \, \rm{d} t \right] \, \rm{d} \mu_k(y) = \frac{\sqrt{\pi}}{2} \int_0^\infty y^{-\frac12} e^{-\frac{x^2}{4y}} \, \rm{d} \mu_k(y). $$
However, it is not straightforward to find the representation \eqref{eq:sample}. The details can be found in the paper "Stable laws of probability and completely monotone functions" by S. Bochner.
