I'm trying to prove Pòlya's theorem, but got stuck at the very first step (I was given a plan of the proof)
This step is to proof, that $\phi_a(x) = (1 - a|x|) \cdot I(|x| \leqslant \frac{1}{a})$ is characteristic.
I used Lévy theorem and found, that if $\phi_a$ is characteristic function of $\xi$, then $p_\xi(t) = \frac{a - a\cos(\frac{t}{a})}{\pi t^2}$. Now i need to proof, that $\int_{-\infty}^{\infty} \left [ \cfrac{e^{ixt} \cdot (a - a \cos(\frac{t}{a}))}{\pi t^2} \right ]dt = \phi_a(x)$, or, equally, $\int_{-\infty}^{\infty} \left [ \cfrac{\cos(xt) \cdot (a - a \cos(\frac{t}{a}))}{\pi t^2} \right ]dt = \phi_a(x)$
How to compute the value of this integral or proof the original statement ($\phi_a(x)$ is characteristic) in any other way?
Let us do the case $a=1$. Define for $x\neq 0$, $$I(c):=\int_{-\infty}^{\infty} \cfrac{\cos(xt) }{\pi t^2} \mathrm dt.$$ Then using the expansion of $\cos(a\pm b)$, we have $$ \int_{-\infty}^{+\infty} \left [ \cfrac{\cos(xt) \cdot (1 - \cos t)}{\pi t^2} \right ]\mathrm dt =I(x)-\frac{I(x+1)+I(x-1)}2. $$ Now, to compute $I(x)$, use the substitution $s=xt$ to get that $I(x)$ is $|x|\cdot I(1)$.