Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 4 / a

Choose $0<\arg z<2\pi$, so $\sqrt{z}=\sqrt{R}{e}^{i\theta /2}$ on the arc. Its clockwise orientation givesBecause $f$ is an analytic function at zero, $f(R{e}^{i\theta })=f(0)+O(R)$ uniformly on the small arc. Integration then givesThereforeThe estimation lemma also gives $|I(R)|\le \pi \sqrt{R}{\max }_C|f|$. Hence a sufficient large-arc condition isFor example $f(z)=O(|z{|}^{-1/2-\eta })$ uniformly on the arc, with $\eta >0$, suffices. Under the stronger literal assumption that $f$ is analytic at infinity, its Laurent expansion is $f(z)=f(\infty )+O(1/z)$; the necessary and sufficient condition is then $f(\infty )=0$. A nonzero constant term would instead give the growing contribution $2\sqrt{2R}f(\infty )$.

For $x>0$, the upper-bank argument tends to zero, whereas the lower-bank argument tends to $2\pi$. ThusThe analytic function $f$ has identical boundary values across this part of the real axis. Hence the branch cut introduces a sign change:This compares integrand values. When two contour segments traverse opposite banks in opposite directions, their orientation supplies an additional minus sign.