Proving that the characteristic function of $f(x)=\frac{1-\cos(x)}{\pi x^2}$ is $\max(1-|t|,0)$

The inversion formula states that $f(x) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}e^{-itx}\phi(t)\,dt$ where $f(x)$ is a probability density function of a random variable $X$ and $\phi(t)$ is the characteristic function of $X$.

Since $\phi(t) = \max(1-|t|,0)$ is non-zero only for $-1 < t < 1$, we have:

$f(x) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}e^{-itx}\phi(t)\,dt = \dfrac{1}{2\pi}\displaystyle\int_{-1}^{1}e^{-itx}(1-|t|)\,dt$

Simply split up the integral over $[-1,1]$ into an integral over $[-1,0]$ and an integral over $[0,1]$, and then use integration by parts for both.