Is the Fourier transform of an integrable function ( in $L^1(-\infty, \infty)$ ) also integrable? Put another way if: $f \in L^1(\mathbb{R})$ then is
$$ \hat{f}(t) = \int_{-\infty}^\infty e^{i xt} f(t) \, dt \stackrel{?}{\in} L^1(\mathbb{R})$$
It would seem to me perfectly reasonable the Fourier transform $\mathcal{F}$ takes $L^1$ to itself since it is an isometry on $L^2$.
No: if $f(x)=\chi_{[-1,1]}(x)$, then $$ \hat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi ix\xi}\;dx=\int_{-1}^1e^{-2\pi ix\xi}\;dx=2\frac{\sin(2\pi \xi)}{2\pi \xi}$$ if $\xi\neq 0$, and $\hat{f}(0)=2$.
One can show that the function $g(y)=\frac{\sin y}{y}$ is not absolutely integrable on $(0,\infty)$ (although interestingly the improper Riemann integral exists, see here) hence $\hat{f}$ is not in $L^1$.