 Solution

For $0\le \alpha <1$, symmetry gives $\mathbb{E}{Y}_1=0$ andbecause the tail integrand is $O(x^{\alpha -2})$. The strong law of large numbers for independent identically distributed integrable variables therefore gives $S_n/n\to 0$ 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 isFor $u>0$, 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 $\mathbb{E}{e}^{iu{S}_n/n}=\phi (u/n{)}^n=e^{-|u|}$. Thus $S_n/n$ is exactly standard Cauchy for every $n$. Consequently

For $0\le \alpha <1/2$, the variance is finite and equals $\mathbb{E}{Y}_1^2=2m(2\alpha )$, using the integral notation in the question. The central limit theorem for independent identically distributed variables with finite variance givesIn 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.