Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 13A / b / iii / Solution

Let $g$ be a <meromorphic function> with simple poles $\alpha_+$ above $c$ and $\alpha_-$ below $-c$, and $zg(z)\to0$ uniformly at infinity in $\operatorname{Im}z\ge-c$. 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.

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 Incoming links: Half-plane decomposition by Cauchy integrals