 Solution

Parameterize the second contour by $\zeta =-is$, $s\ge 0$. ThenThe exponential decays when $\mathrm{Im}z>0$, 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 $\boxed{\alpha =0,\beta =\pi }$. 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,Consequently the desired analytic continuation for $\pi /2\le \arg z<\pi$ isBoth 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.