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

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.

The rational factors give four simple poles at $z=\pm i,\pm 2i$; none is canceled by the numerator. The square root has branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$ in the integrand.

For this part, choose $h(z)=\sqrt{z^2-1}$ analytic off $[-1,1]$ and satisfying $h(z)\sim z$ at infinity. Its values are $h(x)=\sqrt{x^2-1}$ for $x>1$, $h(x)=-\sqrt{x^2-1}$ for $x<-1$, andThus the chosen branch cut is $[-1,1]$, and the function is analytic elsewhere apart from the four poles. Infinity is not an intrinsic branch point: $g(z)=O(z^{-2})$ on this branch.

Another legitimate convention applies the positive-axis square root of part (a) to $w=z^2-1$; it places the cuts on $(-\infty ,-1]$ and $[1,\infty )$. Its upper-half-plane values agree with the branch used here, so the contour calculation below and the positive real square root in the required integral are unchanged. The branch choice must be stated because a square root alone does not specify a single-valued complex function.

Use the branch from part (i) and close the real-axis path with a large upper semicircle, indenting clockwise around $-1$ and $1$. On the segment between them, take the upper-bank value of the branch cut. This is an instance of square-root endpoint integrals by an upper-half-plane contour.

The contributions can be controlled separately. Since $g(z)=O(z^{-2})$, the large arc contributes $O(R^{-1})\to 0$. Near either branch point, $g=O({\epsilon }^{-1/2})$, so an arc of length $O(\epsilon )$ contributes $O(\sqrt{\epsilon })\to 0$. On the upper bank of $(-1,1)$,which is odd in $x$, so its integral is zero. On the exterior real intervals $h(-x)=-h(x)$, while the rational numerator is odd. Hence $g(-x)=g(x)$ and the two exterior contributions sum to $2J$.

The enclosed residues use $h(i)=i\sqrt{2}$ and $h(2i)=i\sqrt{5}$:The residue theorem gives $2J=2\pi i[-i/(3\sqrt{2})-i/(6\sqrt{5})]$. ThereforeAs an independent check, $u=\sqrt{x^2-1}$ converts the integral towhich yields the same value. This check also confirms the branch signs and the factor of two in the contour argument.