Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Analysis / 5 / Solution

The residue at $a$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series. A simple pole means $f(z)=c_{-1}/(z-a)+h(z)$ near $a$, with $c_{-1}\ne 0$ and $h$ holomorphic across $a$.

For the specified positively oriented arc, direct parametrization gives ${\int }_{{\gamma }_{ϵ}}(z-a{)}^{-1}dz=i(\beta -\alpha )$. The remaining integral is bounded by $ϵ(\beta -\alpha )\sup |h|$, which tends to zero because $h$ is bounded near $a$. ThereforeThe simple-pole hypothesis matters. For $f(z)=(z-a{)}^{-2}$, with $\alpha =0$, $\beta =\pi$, the residue is zero butFor the particular function, expanding the exponential givesThus $\boxed{\mathrm{res}(f;0)=1/2}$, and the pole is simple. On the large upper semicircle, $|e^{iz}|\le 1$, givingClose the real-axis path from $-R$ to $-ϵ$ and $ϵ$ to $R$ by this upper arc and a clockwise upper indentation at zero. The function is holomorphic in the indented region, so the sum of all four integrals is zero. The clockwise indentation tends to $-i\pi /2$. For real $x$,The real part is odd and cancels on the symmetric real pieces; the imaginary part is even. Its integral converges since $(x-\sin x)/x^3\to 1/6$ at zero and is $O(x^{-2})$ at infinity. Letting $ϵ\to 0$ and $R\to \infty$ therefore yields $2iI-i\pi /2=0$, where $I$ is the requested real integral. Hence