 Solution

For $\mathrm{Im}\lambda <c$, close $C_2$ upwards. The pole at $\lambda$ is outside this contour, while ${\alpha }_{+}$ is inside; the pole ${\alpha }_{-}$ is also outside. The residue theorem and the minus sign in the definition of $J$ giveThis rational function is holomorphic throughout $\mathrm{Im}\lambda <c$, since its only pole lies above $c$. In particular the formula includes $\lambda ={\alpha }_{-}$: the original line-integrand has no singularity on $C_2$ there, and no evaluation of $g({\alpha }_{-})$ is needed. Closing downwards would require a decay hypothesis in the lower half-plane that has not been given; closing upwards avoids that unjustified step.