= Solution
Parameterize the second contour by $\zeta=-is$, $s\geq0$. Then
$$
\widetilde F(z)=-i\int_0^\infty\frac{e^{2izs}}{1+is^3}\,ds.
$$
The exponential decays when $\operatorname{Im}z>0$, uniformly on compact subsets of that half-plane. Differentiation under the integral there proves <holomorphy>. Thus the maximal open sector of convergence and <holomorphy> in this representation is \b[$\boxed{\alpha=0,\quad\beta=\pi}$]. The boundary integrals themselves can converge, but that does not enlarge this open analytic sector.
For $0<\arg z<\pi/2$, rotate the original contour clockwise through the fourth quadrant. On a large quarter-circle the integrand is $O(R^{-3})$ times a decaying exponential, so the arc integral tends to zero. The only enclosed pole is $\zeta_0=e^{-i\pi/3}$, and its <residue> is $e^{-2z\zeta_0}/(3\zeta_0^2)$. The clockwise orientation gives, by the <residue theorem>,
$$
F(z)-\widetilde F(z)=-2\pi i\frac{e^{-2z\zeta_0}}{3\zeta_0^2}
=-\frac{2\pi i}{3}e^{2\pi i/3}\exp(-2z e^{-i\pi/3}).
$$
Consequently the desired <analytic continuation> for $\pi/2\leq\arg z<\pi$ is
$$
\boxed{F_{\mathrm{cont}}(z)=\widetilde F(z)-\frac{2\pi i}{3}e^{2\pi i/3}\exp(-2z e^{-i\pi/3}).}
$$
Both terms are <holomorphic> in the upper half-plane and the expression agrees with the original function in the overlap, so it is an <analytic continuation>. The residue term is essential: the pole lies between the two rays.
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