Let us introduce the following notation: \begin{eqnarray} f_\alpha (z)=\int_{-\infty}^\infty \cos(tz)e^{-|t|^\alpha} \, dt \end{eqnarray} with $\alpha \in (1,2]$.
I am trying to prove mathematically my empirical result that if $\alpha_1<\alpha_2$ then
$$\int_{-\infty}^x (x-z)f_{\alpha_1} (z) \, dz>\int_{-\infty}^x (x-z) f_{\alpha_2} (z) dz, \quad \forall x \in \mathbb{R}\quad \forall \alpha_1,\alpha_2 \in (1,2].$$
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My ideas and some stylized facts:
1) The integrand is a symmetric function therefore,
$$f_\alpha (z)=\int_{-\infty}^{\infty} \cos(tz)e^{-|t|^\alpha } \, dt = 2\int_0^\infty \cos(tz)e^{-|t|^\alpha} \, dt$$ 2) If $z=0$ then $$f_\alpha (0)=\int_{-\infty}^\infty e^{-|t|^\alpha} \, dt$$ This integral cannot be expressed in terms of elementary functions except for special cases for $\alpha$ when it is either 1 or 2. How ever it is evident that the smaller the $\alpha$, the larger is the value of $f_\alpha(0)$.
3) Asymptotically $f_\alpha(z)\sim \frac{c_\alpha}{z^{\alpha+1}}$ because for $\alpha\in (0,2]$, $\frac{1}{2\pi}f_\alpha(\cdot)$ is the density function of a stable distribution $S_\alpha(1,0,0)$.
4) If we substitute $f_\alpha (z)=\int_{-\infty}^\infty \cos(tz)e^{-|t|^\alpha} \, dt $ into $ \int_{-\infty}^x (x-z)f_\alpha (z) \, dz $, we get $$\int_{-\infty}^{x}\int_{-\infty}^{\infty}\cos(tz)\left[ (x-z)e^{-|t|^\alpha}\right] \, dt \, dz$$ by using the symmetry of the integrand, we get
$$2\int_{-\infty}^x\int_0^\infty \cos(tz)\left[ (x-z)e^{-|t|^{\alpha}}\right] \, dt \, dz$$ It is evident that the smaller the $\alpha$ the larger is the expression $\left[ (x-z)e^{-|t|^\alpha}\right]$, but the harmonic function is the biggest obstacle in establishing the inequality.