# Show that $X_n$ follow somewhat a CLT

Let $X_i$ be i.i.d. $f_X(x)=\dfrac{1}{|x|^3}1_{\{|x|>1\}}$. Show that $\dfrac{\sum_{i=1}^nX_i}{\sqrt{n\log n}}\to N(0,1)$.

I realized that $Var(X_i)=\infty$ and $E(X_i)=0$. I cannot apply Lindeberg or Lyapounov CLT in any way, so I tried to use characteristic functions.

I observe that $$\lim_{n\to\infty}\varphi_{S_n}\left(\dfrac{t}{\sqrt{n\log n}}\right)=\exp\left[\lim_{n\to\infty}-\left (\int_1^\infty\dfrac{4n}{x^3}\sin^2 \left(\dfrac{tx}{2\sqrt{n\log n}}\right)dx\right)\right]$$

I now want to show that this limit is equal to $\exp(-t^2/2)$. How can I do that? I don't think I can apply Dominated Convergence here.

So essentially the task is to prove that the integral converges to $t^2/2$ as $n\to\infty$.

## 1 Answer

To use characteristic functions is definitely the way to go. We have, through integration by parts: $$\int \frac{\sin^2(t x)}{x^3}\,dx = t^2\text{Ci}(2tx)-\frac{t}{2x}\sin(2tx)-\frac{1}{2x^2}\sin^2(tx)\tag{1}$$ hence for large $n$s: $$\int_{1}^{+\infty}\frac{4n}{x^3}\,\sin^2\left(\frac{tx}{2\sqrt{n\log n}}\right)\,dx \approx \frac{1}{2\log(n)}\left[3t^2-2t^2\text{Ci}\left(\frac{t}{\sqrt{n\log n}}\right)\right]\tag{2}$$ and the problem boils down to computing: $$\lim_{n\to +\infty}\frac{3-2\,\text{Ci}\left(\frac{t}{\sqrt{n\log n}}\right)}{\log(n)}\tag{3}$$ that by De l'Hopital theorem equals: $$\lim_{n\to +\infty}\left(1+\frac{1}{\log n}\right)\cos\left(\frac{t}{\sqrt{n\log n}}\right) = 1.\tag{4}$$