= Solution
For $\operatorname{Im}\lambda<c$, 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 $J$ give
$$
\boxed{J(\lambda)=-\frac{r_+}{\alpha_+-\lambda}=\frac{r_+}{\lambda-\alpha_+}}.
$$
This rational function is <holomorphic> throughout $\operatorname{Im}\lambda<c$, 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.
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