Proving that the characteristic function of $f(x)=\frac{1-\cos(x)}{\pi x^2}$ is $\max(1-|t|,0)$ How to prove the following?
$$\int_{\mathbb{R}}e^{itx}\frac{1-\cos(x)}{\pi x^2}dx=\max(1-|t|,0)$$
I would like a direct proof that does not involve the inversion formula. Thank you!
 A: 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. 
A: It could be nice to use the fact that the Fourier transform of the Fourier transform of a function gives you back the original function (up to constant that depends on the definition of Fourier transform) mirrored in the $y$-axis. Since your functions are even, you can neglect the last part. Just calculate the Fourier transform of $\text{max}\, (1-|t|,0)$. If you get $(1-\cos x)/(\pi x^2)$ (modulo some constant), you are done.
A: It is sufficient to show that for any $n\in\mathbb{R}^+$ we have:
$$ \int_{0}^{1}(1-x)\cos(\pi n x)\,dx = \frac{1-\cos(n\pi)}{\pi n^2}.\tag{1}$$
However, $(1)$ simply follows from integration by parts.
A: Let $\color{blue}{ g (t) := \max( 1 - |t|, 0 ) }$. Differentiating $g$ twice,
$$ g '' (t) = \delta(t+1) - 2 \delta(t) + \delta(t-1) $$
where $\delta$ denotes the Dirac delta. Arguably, the Fourier transform of a "train" of Dirac deltas is easier to compute.
$$ \mathfrak F \left\{ g '' \right\} = \frac{1}{2\pi} \left( e^{ix} - 2 + e^{-ix} \right) = \frac{2 \cos(x) - 2}{2\pi} = \frac{\cos(x) - 1}{\pi} $$
Since integrating twice in the $t$-domain corresponds to dividing twice by $ix$ in the $x$-domain,
$$ \mathfrak F \left\{ g \right\} = \frac{1}{(i x)^2} \mathfrak F \left\{ g '' \right\} = \color{blue}{\frac{1 - \cos(x)}{\pi x^2}}$$
