Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 6

On the unit circle, put $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\cos \theta =(z+z^{-1})/2$, soWrite $r=2-\sqrt{3}$ and $s=2+\sqrt{3}$. Only $r$ lies inside the unit circle; the rational function $h(z)=z/[(z-r{)}^2(z-s{)}^2]$ has a double pole there. Its residue isBy the residue theorem,

Choose the branch $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi /2<\arg z<3\pi /2$. It is holomorphic off the downward imaginary ray and takes values $\log x$ on the positive real axis and $\log x+i\pi$ at $-x$ for $x>0$.

Take $R>2$ and $0<\epsilon <2$. The positively oriented contour consists of the real segments $[-R,-\epsilon ]$ and $[\epsilon ,R]$, the small upper semicircle from $-\epsilon$ to $\epsilon$ traversed clockwise, and the large upper semicircle from $R$ to $-R$ traversed counterclockwise. For $F(z)=\mathrm{Log}z/(4+z^2)$, the only enclosed pole is $2i$, withThe integral over the large arc is $O(\log R/R)$, since its length is $\pi R$ and $|4+z^2|\ge R^2-4$. The small arc contributes $O(\epsilon |\log \epsilon |)$, since its length is $\pi \epsilon$ and $|4+z^2|\ge 4-{\epsilon }^2$. Both tend to zero.

Let $J={\int }_0^{\infty }\log x/(4+x^2)dx$; it converges because $|\log x|$ is integrable near zero and $\log x/x^2$ is integrable at infinity. The two real segments, after substituting $z=-x$ on the negative one, tend toThe residue theorem therefore givesTaking real parts yields