Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 4 / d

Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives