= Solution
For $0\le\alpha<1$, symmetry gives $\mathbb EY_1=0$ and
$$
\mathbb E|Y_1|=2\int_0^\infty\frac{x^\alpha}{\pi(1+x^2)}\,dx<\infty,
$$
because the tail integrand is $O(x^{\alpha-2})$. The <strong law of large numbers> for independent identically distributed integrable variables therefore gives $S_n/n\to0$ almost surely and hence in distribution. At $\alpha=0$, $Y_1$ is an equally likely sign, which is included in this argument.
For $\alpha=1$, $Y_1=X_1$ has the standard <Cauchy distribution>. Its <characteristic function> is
$$
\varphi(u)=\frac1\pi\int_{\mathbb R}\frac{e^{iux}}{1+x^2}\,dx=e^{-|u|}.
$$
For $u>0$, close a contour in the upper half-plane; the pole at $i$ has residue $e^{-u}/(2i)$ and the semicircle <integral> vanishes. For $u<0$, symmetry gives the same value with $|u|$, and $u=0$ gives one. Independence now yields $\mathbb Ee^{iuS_n/n}=\varphi(u/n)^n=e^{-|u|}$. Thus $S_n/n$ is exactly standard Cauchy for every $n$. Consequently
$$
\boxed{S_n/n\Rightarrow\delta_0\ (0\le\alpha<1),\qquad
S_n/n\Rightarrow\operatorname{Cauchy}(0,1)\ (\alpha=1).}
$$
For $0\le\alpha<1/2$, the <variance> is finite and equals $\mathbb EY_1^2=2m(2\alpha)$, using the <integral> notation in the question. The <central limit theorem> for independent identically distributed variables with finite <variance> gives
$$
\boxed{S_n/\sqrt n\Rightarrow N(0,2m(2\alpha)).}
$$
In particular $m(0)=1/2$ gives <variance> one at $\alpha=0$. The invoked implications from almost-sure to distributional convergence and the characteristic-function uniqueness theorem are the standard probability results used above.
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