Consider a function $f\in L^1(\mathbb{R})$, $f\geq0$, and let $\hat{f}$ be its Fourier transform, \begin{equation} \hat{f}(\omega)=\int_{\mathbb{R}}f(x)\,e^{-i\omega x}\,\mathrm{d}x. \end{equation} We know that, informally, the "more rapidly decaying" $f$ is, the "more regular" $\hat{f}$ will be. For example, suppose \begin{equation} \int_{\mathbb{R}}xf(x)\,\mathrm{d}x<+\infty: \end{equation} then the first derivative of $\hat{f}$ exists and is simply given by \begin{equation} \hat{f}'(\omega)=-i\int_{\mathbb{R}}xf(x)\,e^{-i\omega x}\,\mathrm{d}x, \end{equation} which is finite; in particular, $\hat{f}'(0)=-i\int_{\mathbb{R}}xf(x)\,\mathrm{d}x$.
Now suppose that, instead, $$\int xf(x)\,\mathrm{d}x=+\infty.$$ The argument above then fails and, in general, $\hat{f}'(0)$ does not exists. Still, the left and right derivatives of $\hat{f}$ at $\omega=0$ may exist and be finite: a notorious example is the Cauchy distribution, $f(x)=(x^2+1)^{-1}$, whose Fourier transform is $\hat{f}(\omega)=\pi e^{-|\omega|}$.
However, now let us make an additional assumption: suppose that $f$ is supported in a semi-infinite interval, e.g. the half-line $\mathbb{R}_+$. I wonder whether, in this case, necessarily $\hat{f}$ cannot admit finite left and right derivatives st $\omega=0$.
My guess arises from the following argument. Supposing $f$ to be supported on $\mathbb{R}_+$, it is easy to show that $\hat{f}$ automatically admits an extension to the lower complex half-plane: \begin{equation} \mathbb{C}^-\cup\mathbb{R}\ni\zeta\mapsto\hat{f}(\zeta)=\int_{\mathbb{R}_+}f(\zeta)e^{-i x\zeta}\,\mathrm{d}x \end{equation} which is analytic in $\mathbb{C}^-$ and continuous up to the real axis. In particular, given $\eta>0$, \begin{equation} \hat{f}'(-i\eta)=-i\int_{\mathbb{R}_+}xf(x)\,e^{-\eta x}\,\mathrm{d}x, \end{equation} and thus, by monotone convergence, $|\hat{f}'(-i\eta)|$ diverges as $\eta\to0$, "suggesting" that $\hat{f}$, on the real line, cannot admit finite left and right derivatives.
I would like to know whether my guess is true (and whether the argument above can be perfected to provide a proof of it) or whether a counterexample can be found.
Thanks in advance.