Is Fourier transform of a $L^{1}$ integrable function is $L^{1}$ integrable?

Question

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 $x= -1, 1$.

Let $g:\mathbb R \to \mathbb R$ such that $$g(x): = |f(x)|.$$

As $g\in L^{1}(\mathbb R)$, we get $\hat{g} \in C_{0}(\mathbb R)$.

My questions are: (1) Can we compute the Fourier transform of $g$ ? (2) Is it true that $\int_{\mathbb R} |\hat {g} (\xi)| d\xi = \infty $, that is, $\hat{g} \notin L^{1} (\mathbb R)$ ?

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Fourier transform of $f$ :

By complex analysis method one can compute the Fourier transform of cardinal sine function or sinc function : $$\int_ {\mathbb R}\frac{\sin (ax)}{ax} e^{-2\pi i x \xi} dx = \frac{\pi}{a}; (|\xi|\leq \frac{a}{2\pi}); (*).$$

Note that, $$\frac{\sin (\pi x )}{x (1-x^{2})}= \pi \frac{\sin (\pi x)}{\pi x }+ \frac{\pi}{2} \frac{\sin (\pi (x+1))}{\pi (x+1)}+\frac{\pi}{2}\frac{\sin (\pi (x-1))}{\pi (x-1))} .$$

We define $f_{1}:\mathbb R \to \mathbb R$ such that $f_{1}(x)= \frac{\sin (\pi x)}{\pi x}$ for $x\not = 0$ and f(0)=1. Also we define, $f_{1}, f_{2}:\mathbb R \to \mathbb R$ such that $f_{1}(x):= f(x+1)$ and $f_{2}(x):= f(x-1)$.

By $(*)$ we get,

$$ \hat{f_{1}(\xi)}=\begin{cases} 1 & \text{if} \ |\xi| < \frac{1}{2},\\ \frac{1}{2} & \text{if} \ |\xi| = \frac{1}{2}, \\ 0, & \textbf{if} \ |\xi|> \frac{1}{2}. \end {cases}$$

By translation property of Fourier transform we get, $$ \hat{f_{2}(\xi)}=\begin{cases} e^{2\pi i \xi} & \text{if} \ |\xi| < \frac{1}{2},\\ \frac{e^{2\pi i \xi}}{2} & \text{if} \ |\xi| = \frac{1}{2}, \\ 0, & \text {if} \ |\xi|> \frac{1}{2}; \end {cases}$$ and $$ \hat{f_{3}(\xi)}=\begin{cases} e^{-2\pi i \xi} & \text{if} \ |\xi| < \frac{1}{2},\\ \frac{e^{-2\pi i \xi}}{2} & \text{if} \ |\xi| = \frac{1}{2}, \\ 0, & \text {if} \ |\xi|> \frac{1}{2}. \end {cases}$$

Now, by using the linearity of the Fourier transform, we get, $\hat{f}(\xi)= \pi \hat{f}_{1}(\xi)+ \frac{\pi}{2}\hat{f}_{2}(\xi ) + \frac{\pi}{2} \hat{f}_{3}(\xi);$ next by putting above values, $$ \hat{f(\xi)}=\begin{cases} \pi (1 + \cos (2 \pi \xi)) & \text{if} \ |\xi| < \frac{1}{2},\\ 0, & \text {if} \ |\xi|\ge \frac{1}{2}. \end {cases}$$

Clearly, $\hat{f} \in L^{1} (\mathbb R)$.

Fourier transform of $|\frac{\sin x}{x}|$ :

Fourier transform of $\left|\frac{\sin x}{x}\right|$

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Thanks to Math fraternity;-)

Answer 1

By coincidence, I happened to see this post: Sobolev spaces and integrability of Fourier transforms

It seems to me that your function is in $W^{1,2}$, hence it's Fourier transform is in $L^1$.

To see that $g'$ is in $L^2$, you can see that its derivative is discontinuous at the integers (except $0, \pm 1$), but they are simple discontinuities in that the left and right limits exist. Also, the derivative decays like $1/|x|^3$ as $x \to \infty$.