# 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...

• Does your course include distributions? If not, you'll have to ask your teacher what is meant with this exercise... – Hans Lundmark Jun 5 '18 at 18:57
• It does include distributions, but I don't see how does it change things. – Waddles Jun 5 '18 at 21:17
• Well, since $f$ is not an $L^1$ function, the Fourier transform isn't defined by the usual integral. Instead, $\mathcal{F}(f)$ is going to be a distribution. – Hans Lundmark Jun 6 '18 at 7:47
• I think I realize now. I have to think of $f$ as a distribution so the Fourier transform is defined as \begin{align*} \langle \mathcal{F}(f), \phi\rangle=\langle \overset{\sim}{f}, \mathcal{F}(\phi) \rangle \end{align*} and calculate integral $\int_{R} f(-\xi) \mathcal{F}(\phi)(\xi)d\xi$, where $\overset{\sim}{f(x)}=f(-x)$. But doesn't the problem with the convergence of integrals stay? – Waddles Jun 6 '18 at 10:54
• I understood your comment, but I don't know how should expression $\langle \overset{\sim}{f}, \cdot\rangle$ be defined if not as an integral, for $f(x)=|x^2-1|$. I know Dirac delta isn't a regular distribution, but other than that I don't know any or how would they be defined. – Waddles Jun 6 '18 at 12:53

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 $$(|x^2 - 1|, e^{-2 \pi i \xi x}) = -\frac {\delta''(\xi)} {4 \pi^2} - \delta(\xi) + \frac {\sin 2 \pi \xi - 2 \pi \xi \cos 2 \pi \xi} {\pi^3 \xi^3}.$$