Central limit theorem for two-sided Pareto distribution

Question

I am trying to solve the following problem, which provides an example for a central limit theorem in spite of the fact that the variance is infinite.
Consider the two-sided Pareto distribution with density $$ f(x) = \begin{cases} \frac{1}{|x|^3}, \enspace &\textrm{for} |x| \geq 1 \\ 0, \enspace &\textrm{otherwise}. \end{cases} $$ (a) Prove that the characteristic function $$ \varphi(t) = 1 - t^2 \left( \log \frac{1}{|t|} + O(1) \right) \enspace \textrm{as} \enspace t \rightarrow 0$$ (b) Let $X_1, X_2, \dots$ be independent, identically distributed random variable with the density as above. Prove that $$ \frac{\sum_{k=1}^{n} X_k}{\sqrt{n \log n}} \overset{d}{\rightarrow} \mathcal{N}(0,1) \enspace \textrm{as} \enspace n \rightarrow \infty.$$ I have proven (b) using (a). But I have no idea how to prove (a). I cannot use the same method as in the proof of the central limit theorem, since the variance is not finite. And I tried to calculate the characteristic function but this was not beneficial.
Thank you very much for your help.

Answer 1

$$ \begin{align*} \varphi(t) &= \mathbb{E}\left(e^{itX}\right)\\ &= \int_{\mathbb{R}} e^{itx}f(x)dx = \int_{-\infty}^{-1} \dfrac{e^{itx}}{-x^3}dx + \int_1^\infty \dfrac{e^{itx}}{x^3}dx\\ &= \int_{1}^\infty \dfrac{e^{-itx}}{x^3}dx + \int_1^\infty \dfrac{e^{itx}}{x^3}dx \\ &= \int_1^\infty \dfrac{e^{itx} + e^{-itx}}{x^3}dx \end{align*} $$ Since $$ e^{itx} = \cos(tx) + i \sin(tx), \text{ and } e^{-itx} = \cos(tx) - i\sin(tx) $$ we have $$ \varphi(t) = 2 \int_1^\infty \dfrac{\cos(tx)}{x^3}dx $$ Now, notice that $$ \int_1^\infty \dfrac{1}{x^3}dx = \dfrac{1}{2} $$ Thus, $$ \varphi(t) = 1 - 2 \int_1^\infty \dfrac{1 - \cos(tx)}{x^3}dx $$ Since we only care about the behavior of $\varphi$ as $t \rightarrow 0$, we can assume that $0 < \vert t \vert < 1$. With this assumption, we split the integral into two: $$ \varphi(t) = 1 - 2 \int_1^{\frac{1}{\vert t \vert}} \dfrac{1 - \cos(tx)}{x^3}dx - 2 \int_{\frac{1}{\vert t \vert}}^\infty \dfrac{1 - \cos(tx)}{x^3}dx $$

The second integral, notice that $$ 0 \le \int_{\frac{1}{\vert t \vert}}^\infty \dfrac{1 - \cos(tx)}{x^3}dx \le \int_{\frac{1}{\vert t \vert}}^\infty \dfrac{2}{x^3}dx = t^2 $$ Thus, $$ \int_{\frac{1}{\vert t \vert}}^\infty \dfrac{1 - \cos(tx)}{x^3}dx = t^2 O(1) $$

Now, for the first integral, consider the Taylor expansion of $\cos$: $$ \cos(tx) = 1 - \dfrac{t^2x^2}{2} + t^4 O\left(x^4\right) $$

So, $$ \int_1^{\frac{1}{\vert t \vert}} \dfrac{1 - \cos(tx)}{x^3}dx = \int_1^{\frac{1}{\vert t \vert}} \left[\dfrac{t^2}{2x} + t^4O(x) \right]dx = \dfrac{t^2}{2} \ln\left(\dfrac{1}{\vert t \vert}\right) + t^2 O(1) $$

Therefore, $$ \varphi(t) = 1 - t^2\left(\ln\left(\dfrac{1}{\vert t \vert}\right) + O(1)\right), \text{ as } t \rightarrow 0 $$