 Half-plane decomposition by Cauchy integrals

Let $g$ be a meromorphic function with simple poles ${\alpha }_{+}$ above $c$ and ${\alpha }_{-}$ below $-c$, and $zg(z)\to 0$ uniformly at infinity in $\mathrm{Im}z\ge -c$. The left-to-right line integrals $H(\lambda )=(2\pi i{)}^{-1}{\int }_{\mathbb{R}-ic}g(z)/(z-\lambda )dz$ and $J(\lambda )=-(2\pi i{)}^{-1}{\int }_{\mathbb{R}+ic}g(z)/(z-\lambda )dz$ satisfy $H=g-r_{+}/(\lambda -{\alpha }_{+})$ above $-c$ and $J=r_{+}/(\lambda -{\alpha }_{+})$ below $c$, where $r_{+}$ is the upper residue. Upper semicircle closure proves both formulas by the residue theorem; subtracting the principal part makes $H$ holomorphic at ${\alpha }_{+}$ too. Consequently $H+J=g$ on the intervening strip. No lower-half-plane decay is needed.