Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 4

Let $f=u+iv$ be analytic on a neighbourhood of a bounded planar region $\Omega$ and its piecewise smooth boundary $C$, oriented with $\Omega$ on the left. The Cauchy integral theorem statesFor a region with holes, $C$ includes the clockwise inner boundaries as well as the counterclockwise outer one. In particular, the integral around a closed contour bounding a region throughout which $f$ is analytic is zero.

To prove this classical form, expand the line integral and apply Green's theorem to its real and imaginary parts:Both integrands vanish by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$. Analyticity supplies the continuous first partial derivatives needed by this Green's theorem proof. The requirement that the enclosed region contain no singularities is essential.

Put $s=1/n$, so $0<s<1$. For the complex exponentiation $z^s$, choose $-\pi /2<\arg z<3\pi /2$: the branch cut is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed simple pole is $z=ia$, with residueThe large arc is $O(R^{s-1})$ and the small arc is $O({ϵ}^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z{|}^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The residue theorem givesSince $1+e^{i\pi s}=2e^{i\pi s/2}\cos (\pi s/2)$,The endpoint estimates also prove convergence. More generally the same calculation is the fractional-power integral from an upper-half-plane contour, valid for $-1<s<1$.

Set $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\tan \theta =(z^2-1)/[i(z^2+1)]$, so the requested integral isThe denominator factors asFor $0<a<1$, its two inner roots are $z=\pm r$, $r=\sqrt{(1-a)/(1+a)}<1$, and the other two roots have modulus $1/r>1$. There is also a simple pole at zero. Their residues areThe latter values follow by differentiating the quadratic product, or from the supplied hint. The sum of the enclosed residues isThe residue theorem therefore givesAt the angles where the tangent is undefined, the original reciprocal has a removable zero limit. Its continuous extension gives the same integral; the transformed rational integrand has no pole on the unit circle.