Fourier transform of $\lvert x^2-1\rvert $ One of the exercises of my assignment was to determine the Fourier transform of function 
$$f(x)=\lvert x^2-1\rvert$$
The domain wasn't specified. First I was puzzled since $f$ isn't a $L^1$ function. 
If I were to calculate 
$$\mathcal{F}(f)(\xi)=\int_\mathbb{R} \lvert x^2-1\rvert e^{-2\pi i x \xi}\,dx$$ I would calculate it separately on $\langle-\infty, -1], \langle-1,1\rangle$ and $[1, +\infty \rangle$ but in case of real domain, it doesn't converge (first and third integral). 
In case of complex domain, separate parts converge depending on the imaginary part of $\xi$, but not at the same time so the whole integral diverges.
Am I missing something here? I would have said at first that the Fourier transform of this function isn't defined, but why would it be an exercise then...

The Fourier transform of a distribution is defined as
$$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]).$$
That is, the action of $\mathcal F[f]$ on $\phi$ is given by the rhs of this identity, where we know that $\mathcal F[\phi]$ is well-defined and is again a valid test function.
We can find the transforms of $1$ and of $x^2$ by finding the transform of $\delta^{(n)}$ and also can find the transform of $|x^2 - 1| - (x^2 - 1)$ directly to get
$$(|x^2 - 1|, e^{-2 \pi i \xi x}) =
-\frac {\delta''(\xi)} {4 \pi^2} - \delta(\xi) +
 \frac {\sin 2 \pi \xi - 2 \pi \xi \cos 2 \pi \xi} {\pi^3 \xi^3}.$$
