This question was suggested when trying to find an explicit example of a continuous function with compact support in $\mathbb{R}$ whose Fourier transform is not integrable. The existence of such a function was proved by an abstract argument in this [MO question](https://mathoverflow.net/questions/3764/does-there-exist-a-continuous-function-of-compact-support-with-fourier-transform), but no explicit example was given. It is clear that such a function cannot be differentiable everywhere.


The functions $(1-x^2)^{\alpha}\\_{+}$ for $0<\alpha<1$ look like reasonable candidates, since they have a singularity at $\xi=\pm1$ and converge as $\alpha\to0$ to the characteristic function of $[-1,1]$, whose Fourier transform is not integrable. However $\hat f\_{\alpha}(\xi)=O(|\xi|^{-(1+\alpha)})$ as $|\xi|\to\infty$, and hence is integrable.


For any $\alpha,\beta>0$, the function
$$g\_{\alpha,\beta}(x)=(1-\beta\log(1-x^2))^{-\alpha}\hbox{ if }|x|<1,\quad 0 \hbox{ if } |x|\ge1$$
is more singular than any $f\_{\alpha}$. So my questions are:


* Is $\hat g\_{\alpha,\beta}$ is integrable?
* What is the asymptotic behaviour of $\hat g\_{\alpha,\beta}$ at infinity?


I suspect, based on numerical calculations, that the answer to the first question is no, at least for sufficiently small $\alpha$ and $\beta$.



Is the Fourier transform of 1/(1-log(1-x^2)) (supported in [-1,1]) integrable?
If one performs a smooth dyadic decomposition of $g\_{\alpha,\beta}$ around the singularities $x = \pm 1$ (i.e. using smooth partitions of unity to decompose $g\_{\alpha,\beta}$ into pieces that are localised in the region $1-|x| \sim 2^{-n}$ for $n \geq 0$), and then takes the Fourier transform of these pieces, one soon arrives at the conclusion that $\hat g\_{\alpha,\beta}$ decays like $\frac{1}{|\xi| \log^\alpha |\xi|}$ (times something like $\sin \xi$) as $\xi \to \infty$ (and one can then back up this intuition with stationary phase, or use some localised form of the Plancherel theorem for an $L^2$ averaged result, which should be enough for the application at hand). So absolute integrability should fail for $\alpha \leq 1$.


