Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 6 / Solution

Let $F(z)=e^{i\lambda z}/(z-i)$. For $\lambda >0$, close the interval $[-R,R]$ by the upper semicircle, counterclockwise. Its only enclosed pole is $z=i$, with residue $e^{-\lambda }$. The residue theorem givesOn $z=R{e}^{i\theta }$, $|e^{i\lambda z}|=e^{-\lambda R\sin \theta }$ and $|z-i|\ge R-1$. Hence the arc is bounded byby the supplied estimate, with parameter $\lambda R$.

For $\lambda <0$, close below, with clockwise orientation. The pole at $i$ lies outside, and the exponential decays there. Parametrizing the lower arc yields the same bound with $|\lambda |R$ in place of $\lambda R$. Its contribution again tends to zero, and no residue remains. Therefore

This is the contour-decay mechanism behind Jordan's lemma; the pole contributes only when the decaying closure encloses it.