Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 4 / 17B (Complex Methods)

Set $M_R={\max }_{0\le \theta \le \pi }|f(R{e}^{i\theta })|$. The complex limit at infinity gives $M_R\to 0$. Parametrizing the semicircle and using $|e^{i\lambda R{e}^{i\theta }}|=e^{-\lambda R\sin \theta }$ yieldsBy symmetry and the supplied sine bound,The arc integral is therefore at most $(\pi /\lambda )M_R$, tending to zero. This proves the required vanishing, the relevant form of Jordan's lemma. No holomorphicity of $f$ is required just for this arc estimate.

Integrate $e^{iz}/z$ around the positively oriented indented upper-half-plane contour. The outer semicircle runs from $R$ to $-R$; the inner semicircle runs clockwise from $-r$ to $r$, excluding zero. The integrand is holomorphic inside this region, so Cauchy's integral theorem gives total integral zero.

The outer arc tends to zero by part (a) with $f(z)=1/z$ and $\lambda =1$. On the inner arc $z=r{e}^{i\theta }$, $\theta$ decreases from $\pi$ to zero, andThe two real segments combine asPassing to the limits gives $2iI-i\pi =0$, and hence the Dirichlet integralThis is an ordinary conditionally convergent improper integral: the integrand tends to one at zero, and integration by parts gives a tail bounded by $2/A$ beyond $A$. The clockwise indentation sign is essential.