Let $f:\mathbb R \to \mathbb R$ such that $$f(x)= \frac{\sin \pi x}{x (x^{2}-1)}$$ for $x\in \mathbb R - \{ 0, -1, 1 \}$ and $f(x):= \pi $ for $x=0$ and $f(x)=-\frac{\pi}{2}$ for $x= -1, 1$.
Clearly, $f\in C_{0}(\mathbb R) $ and note that $\int_{\mathbb R} |f(x)| (1+|x|) dx < \infty.$
My questions are:
(1) How to compute the Fourier transform $\hat{f}$ of the $f$ (as defined above), that is, for $\xi \in \mathbb R$, $\hat {f}(\xi)= \int_{\mathbb R} f(x) e^{-2\pi i x \xi} dx = $ ? Or, even without computation can we conclude that that, $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{f}(\xi)| d\xi < \infty $ ?
(2) Define $g:\mathbb R \to \mathbb R$ such that $g(x):= |f(x)|$ , for $x\in \mathbb R$. Is it true that $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{g}(\xi)| d\xi < \infty $ ?
(Motivation:
Let $a, b \in \mathbb R$ such that $ab> 1$ ; put
$$L^{1}_{a}(\mathbb R)= \{ f:\mathbb R\to \mathbb C \ \text {measurable} : ||(1+|x|)^{a}f||_{L^{1}(\mathbb R)}< \infty \},$$ and $$FL^{1}_{b}(\mathbb R)= \{ f:\mathbb R \to \mathbb C \ \text {measurable} : ||(1+|w|)^{b} \hat {f}||_{L^{1}(\mathbb R)}< \infty \}.$$ We consider a Foureier-Lebsgue space, $$FL:=L^{1}_{a}(\mathbb R) \cap FL^{1}_{b}(\mathbb R).$$ The question is: If $f\in FL$. Can we expect $|f|\in FL$ ?; so, my motivation here is that above example serve here as a counter example. )
Thanks to math fraternity;-)
Use partial fractions, and the fact that you know the Fourier transform of $\frac{\sin(\pi x)}x$, and also if you know the Fourier transform of $f(x)$, then it is easy to obtain the Fourier transforms of $f(x+1)$ and $f(x-1)$. I think you will get something compactly supported with Lipschitz derivative.
$g = |f|$ is harder. We will outline a proof $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{g}(\xi)| d\xi = \infty $. Since the Fourier transform of $f$ is very nice, we will instead work with $h = \frac12(f+g) = f_+$. It is sufficient to show $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{h}(\xi)| d\xi = \infty $.
Find a smooth bump function $\phi$, supported on $[-3,3]$, such that $\phi(x) h(x) = \sin(\pi x) I_{|x|<2}$. We can compute $\widehat{\phi h}(\xi)$ explicitly, seeing that it is something like an oscillating function that decays like $|\xi|^{-2}$. Hence $\int_{\mathbb R} (1+|\xi|)^{2} |\widehat{\phi h}(\xi)| d\xi = \infty $.
However, $\widehat{\phi h} = \hat\phi*\hat h$, where $*$ denotes convolution. Since $\hat\phi$ is localized, we can make some approximations when $\xi$ is large, and see that $\int_{\mathbb R} (1+|\xi|)^{2} |\widehat{\phi h}(\xi)| d\xi \le C \int_{\mathbb R} (1+|\xi|)^{2} |\hat{h}(\xi)| d\xi $. I hope that you are satisfied with proving this last bit yourself, because it is probably going to be a bit fiddly.