Fourier transform of $(1/\sqrt{x}) \mathbb{1}_{[0,1]}$

Question

I struggle with the following: Given the function $f(x)=\frac{1}{\sqrt{x}}\mathbb{1}_{[0,1]}$, I want to prove that its fourier transform $\hat{f}\notin L^1(\mathbb{R}) \cup L^2(\mathbb{R}) \cup C_0(\mathbb{R})$. I can calulate the following (given I didn't miscalculate, I used the transformation $x=t^2$) $$\hat{f}(\omega) = \frac{2}{\sqrt{2\pi}} \int_0^1 \mathrm{e}^{-\mathrm{i}\omega t^2} \mathrm{d}t$$ Is it possible to evaluate this integral further or can I deduce from this form that the transformed function is not an element in the given spaces? Any help would be appreciated!

Answer 1

Summarizing the hints that I gave you in the comments:

You don't have $\widehat{f} \in C_0$ (the space of continuous, compactly supported functions), since $f$ has compact support, which implies that $\widehat{f}$ is analytic. Thus, if $\widehat{f}$ had compact support, we would have $\widehat{f} \equiv 0$, and then $f = 0$ (almost everywhere) by Fourier inversion.

Next, you have $f \in L^1 \setminus L^2$, so that Plancherel's theorem implies that $\widehat{f} \notin L^2$ as well.

Finally, if you had $\widehat{f} \in L^1$, then by Fourier inversion you would get $f(x) = \widehat{\widehat{f}}(-x)$ almost everywhere, which implies (why?!) that $f$ is almost everywhere equal to a continuous function. This is impossible (why precisely?).