Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S1 / 3

For a meromorphic function with no poles on a positively oriented closed contour, the contour integral equals $2\pi i$ times the sum of enclosed residues, weighted by winding number for a general contour. At a simple pole $a$, the residue is ${\lim }_{z\to a}(z-a)f(z)$. Analyticity on the contour and the absence of other interior singularities are necessary hypotheses.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos \theta =(z+z^{-1})/2$ and $\sin \theta =(z-z^{-1})/(2i)$. Rational expressions in these trigonometric functions become rational contour integrands on the unit circle. The method is especially useful when the resulting denominator has a small number of identifiable poles. A pole on the integration circle requires a separately specified principal value or indentation; the residue theorem does not automatically assign an ordinary integral there.

First remove the removable singularities introduced by tangent and secant: the integrand is $\cos \theta /(3+\cos \theta +2\sin \theta )$. The denominator stays positive since $3>\sqrt{5}$. With $z=e^{i\theta }$ the contour integrand isIts interior poles are zero and $r=-1/(1-2i)$; the other quadratic root is $-5/(1-2i)$ and lies outside. The residues are $1/[i(1+2i)]$ and $(r^2+1)/(4ir)$, whose sum is $i/10$. ConsequentlyAssigning arbitrary finite values at the isolated removable points does not change the integral.

Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.