Is the Fourier transform of the function $f$ which agrees with $1_{[-1.1]}|x|^\alpha$ on $[-1,1]$ and then decays very fast to zero to become a compactly supported continuous function, is in $L^1(\mathbb R)$, where $\alpha\in(1,2)$? My guess is that the answer is true. If I can show that the identity $\widehat{D^\alpha f}(\zeta)=(2\pi i\zeta)^\alpha\hat{f}(\zeta)$ holds in some sense and we have $D^\alpha(1_{[-1,1]}|x|^\alpha)=\alpha!|x|1_{[-1,1]}$ in some sense. Then $\hat{f}$ would have enough decay at infinity to make it in $L^1(\mathbb R)?$ May be fractional derivatives and their interaction with the Fourier transform might be helpful. But I could not find any good reference (easily readable) which can make the above argument rigorous? For my purpose if one such $f$ exists for which the Fourier transform is integrable is enough?