Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 7

The Cauchy integral formula for derivatives, with $z_0=0$, givesThe derivative cycles of sine and cosine give ${\sin }^{(n)}(0)=\sin (n\pi /2)$ and ${\cos }^{(n)}(0)=\cos (n\pi /2)$. ThereforeIn particular the sine integral vanishes for even $n$, and the cosine integral vanishes for odd $n$. For the remaining cases their signs alternate. The radius is irrelevant provided $R>0$, since the numerator is entire and the origin is enclosed once counterclockwise.

Factor the denominator as $(z-z_0)(z+z_0)$. Since $z_0\ne 0$, its zeros are distinct, and the potential simple poles and their residues areFor example, the first follows by taking ${\lim }_{z\to z_0}(z-z_0)f(z)$. If the numerator vanishes at either denominator zero, the singularity there is removable and the corresponding residue is zero. Thus the data identify the possible poles, not an unconditional pole at a point canceled by $g$.

If $f$ is analytic on and inside a positively oriented simple closed contour except for finitely many isolated poles $z_j$ strictly inside it, then the residue theorem statesThere must be no singularity on the contour. More generally each residue is multiplied by the contour's winding number about its point.

Let $n\ge 2$ be even and integrate $f(z)=1/(1+z^n)$ around an upper-half-plane semicircle of radius $R>1$ with the real segment traversed from $-R$ to $R$. There are no real poles, and the $n/2$ upper-half-plane poles areEach is simple. Since $z_j^n=-1$,On the semicircle the integral magnitude is at most $\pi R/(R^n-1)$, which tends to zero. The real integral is absolutely convergent because its tail is of order $|x{|}^{-n}$. ThusEvaluate the finite geometric series, writing $\theta =\pi /n$:ConsequentlyFor $n=2$ this reduces to $\pi$, as a check on the contour orientation and residue signs.