= Solution
Write $r_+=\operatorname{Res}_{z=\alpha_+}g(z)$. For $\operatorname{Im}\lambda>-c$ and $\lambda\ne\alpha_+$, close $C_1$ upwards as before. The <residue theorem> includes the poles at $\lambda$ and $\alpha_+$, and the assumed decay in the upper region makes the arc contribution vanish. Thus
$$
H(\lambda)=g(\lambda)+\frac{r_+}{\alpha_+-\lambda}=g(\lambda)-\frac{r_+}{\lambda-\alpha_+}.
$$
It is essential to include the apparently exceptional point $\lambda=\alpha_+$. The <Laurent series> there is $g(z)=r_+/(z-\alpha_+)+g_0(z)$ with $g_0$ <holomorphic>. At that point the integrand is $r_+/(z-\alpha_+)^2+g_0(z)/(z-\alpha_+)$, whose <residue> is $g_0(\alpha_+)$. Hence
$$
\boxed{H(\lambda)=g(\lambda)-\frac{r_+}{\lambda-\alpha_+},\quad H(\alpha_+)=g_0(\alpha_+)}.
$$
The displayed difference is interpreted by its <removable singularity>. It therefore defines a <holomorphic> function at every point of the required half-plane, including $\alpha_+$.
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