Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 4

The only possible singularity of $f(z)/(z-z_0)$ inside the contour is at $z_0$, with residue $f(z_0)$. It is at most a simple pole; when $f(z_0)=0$, it is removable. Thus the residue theorem gives the Cauchy integral formula,for a counterclockwise simple contour containing $z_0$ once, with $f$ analytic on a neighbourhood of the contour and its interior. More generally the answer is $2\pi i\mathrm{Ind}(C,z_0)f(z_0)$, where the winding number accounts for multiple turns or reversed orientation. A clockwise simple contour has the opposite sign.

The integrand has simple poles at $z=ia$ and $z=-ia$. Only the first is inside the upper contour, and its residue isThe residue theorem therefore gives, for every $R>a$,On the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+a^2|\ge R^2-a^2$. The estimation lemma bounds the arc contribution by $\pi R/(R^2-a^2)\to 0$. Hence the real-axis integral has the same limit, and taking its real part givesThe imaginary part vanishes independently because $\sin x/(a^2+x^2)$ is odd. The real integral converges absolutely, since the denominator grows quadratically.

For $\epsilon \ne 0$, use $z=e^{i\theta }$. The interval traverses $|z|=1$ once counterclockwise, $\cos \theta =(z+z^{-1})/2$, and $d\theta =dz/(iz)$. ThusPut $d=\sqrt{1-{\epsilon }^2}>0$. The two simple poles areTheir moduli are $|\epsilon |/(1+d)<1$ and $(1+d)/|\epsilon |>1$, so exactly $z_{+}$ is enclosed, for either sign of $\epsilon$. Its residue isThe residue theorem yieldsThis also follows from the reciprocal affine sine integral after an angular shift. At $\epsilon =0$, the original integral is directly $2\pi$, agreeing with the boxed formula. The displayed two-pole formulas require $\epsilon \ne 0$: at zero the contour integrand reduces to $1/(iz)$ with a single pole. Thus the source's claim of two simple poles has this removable special-case exception.