Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 5 / b / Solution

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.