In lecture, we had the following corollary (without proof, unfortunately): If $ A \in (0,2) $ and $X$ is a random variable (real-valued) with the following characteristic:
$$ \mathbb P(X > x) = \mathbb P(X < -x) = \frac{x^{-A}}{2} \text{ for any } x \in [1, \infty).$$
Then, the following is valid as $t \to 0$:
$1-\phi(t)\sim C|t|^A $, where $\phi$ is the characteristic function and $\displaystyle C = \int_0^\infty \frac{1-\cos(u)}{u^{A+1}} \, du$.
Why is this correct? Can somebody show me?
Note that $X$ is symmetric hence $1-\phi(t)=E[1-\cos(tX)]$, that is, $$ 1-\phi(t)=\int_1^{+\infty}(1-\cos(tx))\frac{A\mathrm dx}{x^{A+1}}=A|t|^A\int_{|t|}^{+\infty}\frac{1-\cos(u)}{u^{A+1}}\mathrm du, $$ thanks to the change of variable $u=|t|x$. The last integral converges when $t\to0$ because $A\lt2$ and at $+\infty$ because $A\gt0$ hence the result holds with $$ C=A\int_{0}^{+\infty}\frac{1-\cos(u)}{u^{A+1}}\mathrm du. $$