Given a characteristic function $\varphi(t) = \frac{2-2 \cos t}{t^2} $ and I have to use the inversion theorem to determine the value of this integral: $2 \int_0^{\infty} \frac{1- \cos t}{\pi t^2}~dt$.
My idea was to use the inversion theorem like this:
$\int_{\mathbb{R}}| \varphi(t)| dt = 2 \pi < \infty $ (Wolfram Alpha) and $2 \int_0^{\infty} \frac{1-\cos t}{\pi t^2} ~dt = \int_{-\infty}^{\infty} \frac{1-\cos t}{\pi t^2} ~dt$ because cos is an even function.
The inversion theorem says: "If $\int_{\mathbb{R}}| \varphi(t)| dt < \infty $ then the probability measure $\mu$ has density $f(x) = \frac{1}{2\pi} \int e^{-itx} \varphi(t) dt$"
And Wolfram Alpha says that the value of this integral should be 1. So I wanted to use the inversion theorem to show that $\frac{1-\cos t}{\pi t^2} $ is a density, hence integrating over $\mathbb{R}$ should imply that the value of the integral is 1. But how do I do this? Is this idea even correct?
Let us consider triangular PDF $$ f(x)=(1-|x|)\mathbb 1_{x\in[-1,1]}. $$ It is very simple to find its CF and get $$\tag{1}\label{1}\varphi(t) = \int_{-1}^1 e^{itx}(1-|x|)\,dx=\frac{2-2 \cos t}{t^2}.$$
I suppose that the fact is known that the function $f(t)=(1-|t|)\mathbb 1_{t\in[-1,1]}$ is a characteristic function. It can be proved by Polya's criterion. This function is absolutely integrable on $\mathbb R$, and the inversion theorem can be applied.
We get the PDF that corresponds to the CF $f(t)=(1-|t|)\mathbb 1_{t\in[-1,1]}$ (compare with (\ref{1})): $$ g(x)=\frac{1}{2\pi}\int_{\mathbb R}e^{-itx}f(t)\,dt = \frac{1}{2\pi}\int_{-1}^1 e^{-itx}(1-|t|)\,dt = \frac{2-2 \cos x}{2\pi x^2}=\frac{1- \cos x}{\pi x^2}. $$ We proved that the function $\frac{1- \cos x}{\pi x^2}$ is a PDF. So, it integrates to $1$.