If $f$ has an integrable (weak) derivative of order $n$, then the map $\xi\mapsto \xi^n\widehat f(\xi)$ is in $C_0$, the space of continuous functions which go to $0$ at infinity. This is because we expressed $\xi^n\widehat f(\xi)$ as a Fourier transform of an integrable function.
We deduce that $\widehat f$ decays at least like $|\xi|^{-n}$.
Conversely, a good decay of Fourier transform used with inverse Fourier transform and differentiation under the integral gives smoothness of the function. Furthermore, we can "read" on the Fourier transform how smooth is the function.
Note that we can write
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\frac{1}{2i}\int_{-\infty}^\infty\frac{e^{ik}-e^{-ik}}{k}e^{ikx}\,dk \tag1$$
Observe that for each of the principal value integrals
$$I_{\pm}(x)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{\pm ik}}{k}e^{ikx}\,dk\right)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x\pm 1)}}{k}\,dk\right)$$
the integrand has a pole at $k=0$. To evaluate these integrals, we analyze the contour integrals
$$\begin{align}
\oint_{C}\frac{e^{iz(x\pm 1)}}{z}\,dz&=\int_{-R}^{-\epsilon} \frac{e^{ik(x\pm 1)}}{k}\,dk+\int_{\epsilon}^{R} \frac{e^{ik(x\pm 1)}}{k}\,dk\\\\
&+\int_{\text{sgn}(x\pm 1)\pi}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\
&+\int_0^{\text{sgn}(x\pm 1)\pi}\frac{e^{i R e^{i\phi}}}{R e^{i\phi}}\,iR e^{i\phi}\,d\phi \tag 2
\end{align}$$
As $R\to \infty$, the last integral on the right-hand side of $(2)$ tends to $0$. As $\epsilon\to 0$, the third integral on the right-hand side of $(2)$ tends to $-\text{sgn}(x\pm 1)\pi$. Therefore, we have
$$\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x+ 1)}-e^{ik(x-1)}}{k}\,dk\right)=\text{sgn}(x+1)\pi-\text{sgn}(x-1)\pi=\begin{cases}2\pi &,|x|<1\\\\0&,|x|>1\\\\\pi&,|x|=1\end{cases} \tag 3$$
Substituting $(3)$ into $(1)$ yields
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$
Best Answer
We know that $f \in L^1(\mathbb R)$, then $\hat{f}$ is bounded and continuous. If $\hat{f} \in L^1(\mathbb{R})$ then almost everywhere $$f(x)=\dfrac{1}{2\pi}\int_{\mathbb R}e^{i\xi x}\hat{f}(\xi)\ \mathrm{d}\xi$$ This means that $f$ is equal almost everywhere to a continuous function $g$. We find a contradiction around $x=0$ because we know that there exists $\delta > 0$ such that for $x \in [-\delta, \delta], \ g(x) \in [g(0)-0.1, g(0)+0.1]$. Therefore for almost every $x \in [-\delta, \delta], \ f(x) \in [g(0)-0.1, g(0)+0.1]$, which is not the case.