Finding $\alpha$-stable distributions by (probably) Levy's continuity theorem Let $X_1,...,X_n$ be iid random variables uniformly distributed on $[-1,1]$. Now let's put $$Y_n = \frac{sgn X_n}{|X_n|^{1/\alpha}}, \, n= 1,2,...,$$ with a set value $\alpha \in (0,2)$.
The goal is to:
$\textbf{(a)}$ Show that $Z_n := \frac{Y_1 + ... + Y_n}{n^{1/\alpha}} $ converges by distribution.
$\textbf{(b)}$ Find the characteristic function of the limit.
And that would mean that the limiting distribution is stable as it has its own domain of attraction consisting of $Z_n$.
My best guess is to do this by showing that the characteristic function of $Z_n$ converges to a certain function that's continuous at $t=0$ and then using the Levy's continuity theorem (https://en.wikipedia.org/wiki/L%C3%A9vy's_continuity_theorem), but I was sadly unable to do so.
Do You perhaps have other ideas that would answer this problem or maybe it's indeed solvable by Levy's theorem? If so, how?
Edit:
I've calculated that the characteristic function of $Z_n$ should be: $$\phi_{Z_n}(u) = \left[ \frac{1}{2} \int_0^1 exp(iu \frac{-1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx + \frac{1}{2} \int_0^1 exp(iu \frac{1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx \right]^n,$$ which is $$\left[ \int_0^1 cos \left( \frac{ u}{(xn)^{1/\alpha}} \right) dx \right]^n$$
Edit 2:
I've found an answered question that shows what the limit should be, but still doesn't have what I need. Symmetric alpha stable distributions with $X_1+X_2+\cdots+X_n \stackrel{d}{=} n^{1/\alpha}X$ as definition
 A: Problem
Let $ X_1,X_2,\cdots $ be a sequence of iid random variables uniformly distributed on $ [-1,1] $.
Now let's put
\begin{equation*} Y_n=\frac{\mathrm{sgn}X_n}{|X_n|^{1/\alpha}}, \qquad n\ge 1, \end{equation*}
with a set value $ \alpha\in(0,2) $.
(a) Show that $ Z_n=\frac{Y_1+\cdots+Y_n}{n^{1/\alpha}} $ converges by distribution.
(b) Find the characteristic function of the limit.
And that would mean that the limiting distribution is stable as it has its own domain of attribution consisting of $ Z_n $
Answer  Let
$$ \phi_n(t)=\mathsf{E}\Big[\exp\Big(\frac{itY_1}{n^{1/\alpha}}\Big)\Big]
=\int_{0}^{1}\cos\Big(\frac{t}{(xn)^{1/\alpha}}\Big)dx. $$
Then
$$ \phi_{Z_n}(t)=\mathsf{E}[\exp(itZ_n)]=(\phi_n(t))^n. $$
Meanwhile,
\begin{align*}
 n[1-\phi_n(t)]&=n\int_0^1\Big[1-\cos\Big(\frac{t}{(nx)^{1/\alpha}}\Big)\Big]\,dx\\
 &=\alpha |t|^\alpha\int_{|t|/n^{1/\alpha}}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\
 &\to  C(\alpha) |t|^\alpha, \qquad \text{as}\quad n\to\infty.
\end{align*}
where(cf. Sato, Lévy Processes and Infinitively Divisible Distributions, Cambridge University Press, 1999, Lemma 14.1 p.84,
or Y. S. Chow &  H. Teicher, Probability Theory, 3rd ed., Springer Verlag, 1997, p.469--.)
\begin{align*}
 C(\alpha)&=\alpha\int_{0}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\
 &=\begin{cases}
  \cos\Big(\dfrac{\alpha\pi}{2}\Big)\Gamma(1-\alpha),& \alpha \in (0,1)\cup(1,2)\\
  \quad\dfrac{\pi}{2}, &\alpha=1.
 \end{cases}
\end{align*}
Then
$$ \lim_{n\to\infty}\phi_{Z_n}(t)=\lim_{n\to\infty}[1-(1-\phi_n(t))]^n=\exp[-C(\alpha) |t|^\alpha]. $$
