Choose $f(x)=\frac{(\frac{1}{2} -x)}{log(x)}$ for $0 < x \leq \frac{1}{2}$ and $0$ otherwise. $$f(x)= \begin{cases} \dfrac{\frac12 -x}{\log(x)},&0<x\leq1/2\\ 0,&\text{otherwise} \end{cases}$$
Then $Im \int_{0}^{\infty} dk \ e^{-\varepsilon k} \int_{-\infty}^{\infty} dx \ f(x) e^{-ikx} = \int_{0}^{\frac{1}{2}} dx \ \frac{(\frac{1}{2} -x)}{log(x)} \cdot \frac{-x}{\varepsilon^{2}+x^{2}} \rightarrow \infty$ for $ \varepsilon \downarrow 0$ .$$\operatorname*{Im} \int_{0}^{\infty} dk \ e^{-\varepsilon k} \int_{-\infty}^{\infty} dx \ f(x) e^{-ikx} = \int_{0}^{\frac12} dx \ \frac{\frac12 -x}{\log(x)} \cdot \frac{-x}{\varepsilon^{2}+x^{2}} \to \infty\text{ for } \varepsilon \downarrow 0.$$
Therefore the Fourier transform of $f$ is not in $L^{1}$ .
