 Solution

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.