Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 4

Let $f$ be meromorphic on and inside a positively oriented simple closed contour $C$, with finitely many isolated poles $z_k$ inside and none on $C$. The residue theorem statesThe positive orientation is counterclockwise. A clockwise contour reverses the sign, and the residue is the coefficient of $(z-z_k{)}^{-1}$ in the local Laurent series.

The denominator vanishes at the four roots of $z^4=-1$, namely $z_k=e^{i(2k+1)\pi /4}$ for $k=0,1,2,3$. They are simple poles, since $(1+z^4{)}^{\prime }=4z^3$ is nonzero at each root and the numerator does not vanish there. The simple-pole residue formula givesExplicitly,There are no branch points or other finite singularities. At infinity the function is $O(z^{-2})$, so infinity is removable rather than a pole.

Close the interval $[-R,R]$ with an upper semicircle, traversed counterclockwise. On its large arc,so the estimation lemma bounds the arc integral by $\pi R^3/(R^4-1)\to 0$. The enclosed residues are those at $(1+i)/\sqrt{2}$ and $(-1+i)/\sqrt{2}$, whose sum is $-i/(2\sqrt{2})$. The residue theorem therefore givesThe positive sign agrees with the nonnegative real integrand.

Choose a branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, placing its branch cut down the negative imaginary axis. Thus $\mathrm{Log}x=\ln x$ for $x>0$, while $\mathrm{Log}(-x)=\ln x+i\pi$ on the negative real axis for $x>0$.

Integrate $\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small clockwise semicircular indentation above zero. The large arc contributes $O(\ln R/R)\to 0$, and the small arc contributes $O(\epsilon |\ln \epsilon |)\to 0$. The only enclosed pole is $ia$, withWriting $I(a)$ for the required integral, the negative real segment contributes $I(a)+i\pi {\int }_0^{\infty }dx/(x^2+a^2)$. Hence the residue theorem givesThe imaginary terms cancel, leavingThe integral converges at both endpoints because $|\ln x|$ is integrable near zero and $\ln x/x^2$ is integrable at infinity. As a real-variable check, scaling $x=au$ reduces the result to the vanishing first logarithmic moment at $a=1$, one of the logarithmic moments of the Cauchy kernel.