Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 5

Laurent's theorem states that a function holomorphic on $R<|z-a|<S$ has a unique expansionconverging uniformly on every compact subannulus. Its coefficients arewith counterclockwise orientation. The value is independent of the chosen intermediate radius. The nonnegative powers form a power series convergent inside the outer radius, and the negative powers form the principal part of a Laurent series.

The residue $\mathrm{res}(f;a)$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series at the isolated singularity. Equivalently it is $(2\pi i{)}^{-1}$ times the integral around a sufficiently small counterclockwise circle enclosing $a$.

The residue theorem states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour and its interior except for finitely many isolated singularities $a_j$ inside, thenNo singularity may lie on the integration contour. A more general contour weights each residue by its winding number.

For $z=x+iy$, $e^{iz}=e^{-y}{e}^{ix}$, so $|e^{iz}|=e^{-y}$. Thereforethe closed upper half-plane. Equality holds on the real axis.

The Taylor expansions giveand hence $f(z)=b/z+iab+O(z)$. Since $b>0$, zero is a simple pole with residue $b$. It is not a removable singularity, despite the vanishing numerator, because the denominator has a zero of order two.

Use the upper half-disc of radius $R$, with a clockwise semicircular indentation of radius $\epsilon$ above zero. The integrand is holomorphic inside that indented contour, so its total integral is zero. On the large arc,because $a+b>0$, $a-b>0$ and $\mathrm{Im}z\ge 0$. Its integral has modulus at most $\pi /R$ and vanishes as $R\to \infty$.

On the small clockwise arc, $f(z)=b/z+O(1)$, so the integral tends to $-i\pi b$. Therefore the remaining real-axis integral satisfiesThe imaginary part is even and is an ordinary absolutely convergent integral: near zero its integrand tends to $ab$, and at infinity it is $O(x^{-2})$. Taking imaginary parts givesThe complex integral needs the Cauchy principal value because its real part has a $b/x$ singularity; the requested sine-product integral does not. This sine-product integral with inverse-square weight is an application of the upper-half-plane indentation rule.