Proof of Convergence in Distribution with unbounded moment I posted the question here, but no one has provided an answer, so I am hoping I could get an answer here. Thanks very much!
Prove that given $\{X_n\}$ being a sequence of iid r.v's with density $|x|^{-3}$ outside $(-1,1)$, the following is true:
$$
\frac{X_1+X_2 + \dots +X_n}{\sqrt{n\log n}} \xrightarrow{\mathcal{D}}N(0,1).
$$
The original post has a 2 in the square root of the denominator. There should not be a 2. 

This is a proof by c.f. approach:
The c.f. of $X_i$ is
$$
 \phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx.
$$
        Hence, for $Y_n = (X_1+X_2+\dots+X_n)(\sqrt{n\log n})^{-1}$, we have
        \begin{align*}
  \phi_{Y_n}(t) =& \phi_i\left(\frac{t}{\sqrt{n\log n}}\right)^n\\
      =& \left(2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx\right)^n.\\
  \end{align*}
        We first consider the integral:
        \begin{align*}
   2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx =& 1 + 2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\
                       =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx \\
                       +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx,
  \end{align*}
        since for $x \in [1, \sqrt{n\log\log n}]$, ${\displaystyle \frac{tx}{\sqrt{n\log n}}} \to 0$ as $n \to \infty$. Hence, we can apply the Taylor expansion of the cosine term in the first integral around $0$. Then we have
        \begin{align*}
   2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}-\frac{t^2}{2n\log nx} + \left[\frac{t^4x}{24(n\log n)^2 }-\dots\right]dx \\
                       +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\
                       =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}-\frac{t^2}{2n\log nx}dx + o(1/n)\\
                       +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\
                       =& 1 -\frac{t^2\log( n\log\log n)}{2n\log n} + o(1/n)\\
                       +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\
  \end{align*}
        Now 
        \begin{align*}
   \int_{\sqrt{n\log\log n}}^{\infty}|\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}|dx \leq& \int_{\sqrt{n\log\log n}}^{\infty}\frac{2}{x^3}dx\\
                               =& \frac{1}{n\log\log n} \in o(1/n).
  \end{align*}
        Hence, 
        $$
  2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx = 1 -\frac{t^2\log( n\log\log n)}{2n\log n} + o(1/n).
  $$
        Let $n \to \infty$, we have
        $$
  \lim_{n \to \infty}\left(2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx\right)^n = \lim_{n \to \infty}\left(1 -\frac{t^2\log( n\log\log n)}{2n\log n}\right)^n = \lim_{n \to \infty}\left(1-\frac{t^2}{2n}\right)^n = e^{-t^2/2},
  $$
which completes the proof.
