Fourier transform vanishes on a half line

Question

Let $\phi \in L^1(\mathbb{R})$ and $\phi>0$. Prove that there is an $F$ with $|F|=\phi$ and the Fourier transform $\widehat{F}(x)$ vanishes on $x<0$ if and only if the following inequality \begin{equation} \int_{\mathbb{R}} \frac{\log \phi(x) dx}{1+x^2}>-\infty \end{equation} holds. My though is that the function \begin{equation} u(z)=\frac{1}{\pi} \int_{\mathbb{R}} \frac{y}{(x-t)^2+y^2} \log \phi(t) d t \end{equation} is harmonic in the upper half plane and consider $f=e^{u+iv}$, where $v$ is its harmonic conjugate. Since $f$ is analytic in the upper half plane, Fatou theorem for the upper half plane implies that \begin{equation} \lim_{y \to 0} |f(x+iy)|=\phi(x) \end{equation} holds a.e. on the real axis. However, I can go further and I don't know how to apply the given inequality in my qustion. Any suggestions?