How to prove this inequality about characteristic function?

Question

$X$ is a random variable and $\phi(t)$ is the characteristic function of $X$

Then prove: For every $\alpha>0$ it holds

$$\mathbb{P}\big(|X|\geq\frac{2}{\alpha}\big)\leq\frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-\phi(t))dt$$

Answer 1

Let $h(t) := 1-\frac{1}{\alpha t}$

As I stated in the comment we have:$$\frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-\phi(t))dt = \frac{1}{\alpha}\int_{-\alpha}^{\alpha}\mathbb{E}(1-e^{itX})dt = \mathbb{E}(\frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-e^{itX})dt)$$ where the last conversion holds true since $|1-e^{itX}| \le 2$.
By integrating the last term we get:
$$2\mathbb{E}(1-\frac{sin(\alpha X)}{\alpha X})$$
Because $sin$ is bounded we can estimate this from below by: $$2\mathbb{E}(1-\frac{1}{|\alpha X|})$$
By using Markov's inequality $h(a)P(X \ge a) \le E(h(X))$ we obtain:
$$\frac{1}{2}Pr(|X| \ge \frac{2}{\alpha}) =h(\frac{2}{\alpha})Pr(X \ge \frac{2}{\alpha}) \le\mathbb{E}(1-\frac{1}{|\alpha X|})$$

Therefore we proved that $$Pr\big(|X|\geq\frac{2}{\alpha}\big)\leq\frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-\phi(t))dt$$ holds true.