Integral inequality. Useful for optimization Let us introduce the following notation:
\begin{eqnarray}
&& f_\alpha (z)=\frac{1}{2\pi}\int_{-\infty}^\infty \cos(tz)e^{-|t|^\alpha} \, dt \\
&& F_{\alpha}(x)=\int_{-\infty}^x f_{\alpha}(z)\,dz
\end{eqnarray}
with $\alpha \in (1,2]$.
I am trying to prove mathematically my empirical result that if $\alpha_1<\alpha_2$ then
$$\int_{-\infty}^x F_{\alpha_1}(z)dz\geq\int_{-\infty}^x F_{\alpha_2}(z)dz, \quad \forall x \in \mathbb{R}\quad \forall \alpha_1,\alpha_2 \in (1,2]$$
which according to Ruszczynski and Dencheva (2003) is equivalent to
$$\int_{-\infty}^x (x-z)f_{\alpha_1} (z) \, dz\geq\int_{-\infty}^x (x-z) f_{\alpha_2} (z)  dz, \quad \forall x \in \mathbb{R}\quad \forall \alpha_1,\alpha_2 \in (1,2].$$
To solve the problem, it is enough to prove one of them. 
When solving the first inequality, we end up with tripple integrals while we will end up with double integrals if we solve the second inequality.
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PS1. We denote $\int_{-\infty}^x (x-z)f_{\alpha}(z)\,dz$ with $F_{\alpha}^{(2)}(x)$, i.e.
$$F_{\alpha}^{(2)}(x)=\int_{-\infty}^x F_{\alpha}(z)\,dz=\int_{-\infty}^x (x-z)f_{\alpha}(z)\,dz$$
PS2. $f_{\alpha}(x)$ and $F_{\alpha}(x)$ are the density and distribution function of a stable random variable $X \sim S_{\alpha}(1,0,0)$ respectively. See 
this link: https://en.wikipedia.org/wiki/Stable_distribution
 A: Some of my ideas and people who advise me:
1)
Let us use empirical distribution functions. We know by Glivenko–Cantelli theorem that empirical distribution functions converge to distribution functions. 
So let's represent $$F_{\alpha}(z;n)=\frac{1}{n}\sum_{i=1}^{n}I(X^{(i)}_{\alpha}<z) \mbox{  and   }F_{\alpha}(z;n)\rightarrow_{\mathbb{P}}F_{\alpha}(z)\mbox{ as }n\rightarrow \infty$$
So if $\alpha_1<\alpha_2$ then we get
$$ F_{\alpha_1}^{(2)}(x;n)=\int_{-\infty}^{x}F_{\alpha_1}(z;n)dz=\int_{-\infty}^{x} \frac{1}{n}\sum_{i=1}^{n}I(X^{(i)}_{\alpha_1}<z)dz=\frac{1}{n}\sum_{i=1}^{n}\int_{-\infty}^{x}I(X^{(i)}_{\alpha_1}<z)dz $$
Hence the logic might be based on the fact that it is more likely for the realizations of the random variables with smaller $\alpha$ to get into the tail of the distribution and then to use some convergence properties.
2) We can calculate the characteristic function $\psi(t;z)$ of $(z-X)^+$. However, what we obtained is
$$\frac{\partial \psi(0;x)}{\partial t}=\int_{-\infty}^{x}(x-z)f_{\alpha}(z)dz,$$
which is not very helpful.
3) Since we have a cosine in the integrand, we can calculate $$2\int_{-\infty}^x\int_0^\infty \cos(tz)\left[ (x-z)\left( e^{-t^{\alpha_1}}-e^{-t^{\alpha_2}}  \right)  \right] \, dt \, dz$$ in regions where it is positive and where it is negative and if the positive part outweighs the negative one, then the proof is ready. So in $\mathbb{R}^2$ we determine regions where cos(tz) is nonnegative and where it is positive.
4) We can use Taylor exapansion. The Tailor expansion of the integrand can be conducted in WolframAlpha.com by the key expression: "taylor expansion of (x-z)cos(zt)*(exp(-t^a)-exp(-t^b))"
A: Another empirical observation:
The function $y_{\alpha}(x)=F_{\alpha}^{(2)}(x)-x^+$ is an even function in $x$ (i.e. $y_{\alpha}(x)=y_{\alpha}(-x)$ for $\forall \alpha \in (1,2]$) where
$$F_{\alpha}^{(2)}(x)=\int_{-\infty}^x F_{\alpha}(z)\,dz=\int_{-\infty}^x (x-z)f_{\alpha}(z)\,dz$$
This feature can be quite easily proven because $f_{\alpha}(x)$ is symmetric in $x$.
Hence we can convcentrate on only $\mathbb{R}^-$ when proving this inequality.
A: Lemma 1
The function $y_{\alpha}(x)=F_{\alpha}^{(2)}(x)-x^+$ is even.
Proof:
Without loss of generality, let us assume that $h>0$ and prove that $y_{\alpha}(h)=y_{\alpha}(-h)$. So if we denote $\int_{x}^{\infty}f_{\alpha}(z)dz$ with $F^*_{\alpha}(x)$, then
\begin{eqnarray*}
&& y_{\alpha}(h)=F_{\alpha}^{(2)}(h)-h=\\
&&=\int_{-\infty}^{-h}F_{\alpha}(z)dz+\int_{-h}^0F_{\alpha}(z)dz+\int_{0}^{h}F_{\alpha}(z)dz-h=\\
&& = y_{\alpha}(-h)+\int_{-h}^0F_{\alpha}(z)dz+\int_{0}^{h}(1-F_{\alpha}^*(z))dz-h=\\
&& y_{\alpha}(-h)+\int_{-h}^0F_{\alpha}(z)dz-\int_{0}^{h}F_{\alpha}^*(z)dz=y_{\alpha}(-h).
\end{eqnarray*}
Lemma 2
The function $$\rho_{\alpha_1,\alpha_2}(x)=\frac{1}{\pi}\int_{-\infty}^x\int_0^\infty \cos(tz)\left[ (x-z)\left( e^{-t^{\alpha_1}}-e^{-t^{\alpha_2}}  \right)  \right] \, dt \, dz$$ is even.
Proof: It can be easily established that 
\begin{eqnarray*}
&& \rho_{\alpha_1,\alpha_2}(x)=\frac{1}{\pi}\int_{-\infty}^x\int_0^\infty \cos(tz)\left[ (x-z)\left( e^{-t^{\alpha_1}}-e^{-t^{\alpha_2}}  \right)  \right] \, dt \, dz=\\
&&=F_{\alpha_1}^{(2)}(x)-F_{\alpha_2}^{(2)}(x)=\left(F_{\alpha_1}^{(2)}(x)-x^+  \right)-\left(F_{\alpha_2}^{(2)}(x)-x^+  \right)=\\
&& y_{\alpha_1}(x)-y_{\alpha_2}(x),
\end{eqnarray*}
which proves the statement.
Lemma 3
For all $\alpha_1,\alpha_2\in (1,2]$, and for all $x\in\mathbb{R}$ if $F_{\alpha_1}(x)\geq F_{\alpha_2}(x)$, then $F_{\alpha_1}(-x)\geq F_{\alpha_2}(-x)$.
Proof:  Let us assume that $x>0$. Then
\begin{eqnarray*}
&& F_{\alpha_1}(x)- F_{\alpha_2}(x)=\left(y_{\alpha_1}(x)-x^+\right)-\left(y_{\alpha_2}(x)-x^+\right)=y_{\alpha}(x)-y_{\alpha}(x)=\\
&&=y_{\alpha}(-x)-y_{\alpha}(-x)=F_{\alpha_1}(-x)- F_{\alpha_2}(-x)
\end{eqnarray*}
which implies that $F_{\alpha_1}(x)$ is larger or equal $F_{\alpha_2}(x)$ if and only if $F_{\alpha_1}(-x)\geq F_{\alpha_2}(-x)$.
Theorem
If $\alpha_1,\alpha_2 \in (1,2]$, $\alpha_1>\alpha_2$, and $X_i\sim S_{\alpha_i}(1,0,0)$, $i=1,2$, then 
\begin{eqnarray*}
F^{(2)}_{\alpha_1}(x)\leq F^{(2)}_{\alpha_2}(x)\quad\forall x\in \mathbb{R}.
\end{eqnarray*}
Proof:
By lemma 3, it suffices to prove the statement in $\mathbb{R}^+$. (To be continued)
