Show that $|\phi_{X}(t)-1-t^2\log|t||\le3t^2$ Let $f(x)=|x|^{-3}1_{|x|\ge1}$ be the density function of a random variable $X$
and $\phi_{X}(t)=E[e^{itX}]$. Show that $\forall t\in[-1,1] $ $$|\phi_{X}(t)-1-t^2\log|t||\le3t^2$$
I couldn't think of anything apart from using Taylor's theorem. Any ideas?
 A: A first step would be to write $\phi_X$ in integral form. The bound is obviously true for $t = 0$. Since the density $f$ is symmetric, $\phi_X$ is even. Without loss of generality, assume that $t \in (0,1)$. Then:
$$\phi_X (t) = \int_1^{+ \infty} \frac{e^{itx}+e^{-itx}}{x^3} \ dx = 2t^2 \int_1^{+ \infty} \frac{\cos (tx)}{(tx)^3} t \ dx = 2t^2 \int_t^{+ \infty} \frac{\cos (u)}{u^3} \ du.$$
Now, the integral diverges as $t$ goes to $0$. So we use Taylor-Lagrange inequality on the cosine, which will yields nice main terms. Since this implies working close to $0$, we have to choose a cutoff: close to $0$, we work with with Taylor's approximation of the cosine, while far from $0$, we keep the cosine alone. I will choose this cutoff to be $1$, which incidentally, is also the choice made in the exercise.
$$\phi_X (t) = 2t^2 \int_t^1 \frac{\cos (u)}{u^3} \ du + 2t^2 \int_1^{+ \infty} \frac{\cos (u)}{u^3} \ du = (1) + (2).$$
I'll let you control $(2)$. On $(1)$, write:
$$2t^2 \int_t^1 \frac{\cos (u)}{u^3} \ du = 2t^2 \int_t^1 \frac{1-u^2/2+g(u)}{u^3} \ du,$$
use Taylor-Lagrange inequality to bound $g(u)$, and compute the integral. The bound I got was slightly better than what was asked.
