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

Write $r_{+}={\mathrm{Res}}_{z={\alpha }_{+}}g(z)$. For $\mathrm{Im}\lambda >-c$ 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. ThusIt 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 $g_0({\alpha }_{+})$. HenceThe displayed difference is interpreted by its removable singularity. It therefore defines a holomorphic function at every point of the required half-plane, including ${\alpha }_{+}$.

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.

In the open strip $-c<\mathrm{Im}\lambda <c$, neither pole of $g$ is present. The two evaluations therefore giveBoth summands are holomorphic on this strip. This is a half-plane decomposition by Cauchy integrals; the pole subtraction in $H$ is what makes its extension holomorphic even at the upper pole of $g$.