Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Analysis / 5

Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\epsilon$ about zero removed. The real segments run from $-R$ to $-\epsilon$ and $\epsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented contour.

The large arc contribution tends to zero: its modulus is at most ${\int }_0^{\pi }{e}^{-R\sin \theta }d\theta$, which tends to zero by dominated convergence. On the small arc $z=\epsilon e^{i\theta }$, $\theta$ decreases from $\pi$ to zero, so its integral tends to ${\int }_{\pi }^{0}id\theta =-i\pi$. The contour theorem consequently givesTaking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at zero after assigning its limiting value one, so the inner indentation can be removed. Therefore

Let $\omega =e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument zero to $2\pi /3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi /3}$, with residue $1/(3z_0^2)=1/(3\omega )$. The outward real radial segment contributes $I_R={\int }_0^{R}dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x{)}^3=x^3$.

On the circular arc, $|1+z^3|\ge R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to 0$. Thus the residue theorem yieldsSince $\omega (1-\omega )=\omega -{\omega }^2=i\sqrt{3}$,The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.