Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S1 / 2

A branch point obstructs a single-valued analytic continuation around a closed circuit. Continuing $\log z=\log |z|+i\theta$ once around zero increases its value by $2\pi i$. A branch cut removes a curve joining branch points or extending to infinity so that a continuous argument can be chosen on the remainder. Zero and infinity are logarithmic branch points; the positive-axis cut used here selects $0<\arg z<2\pi$. The cut is a choice of representation, not a line of poles of the multivalued logarithm.

The large circle contributes $O((\log R{)}^n/R)$ and the small circle contributes $O(ϵ|\log ϵ{|}^n)$, so both disappear. The enclosed simple poles are $i$ and $-i$, with logarithms $i\pi /2$ and $3i\pi /2$. Their residues are $(i\pi /2{)}^n/(2i)$ and $-(3i\pi /2{)}^n/(2i)$. The positive orientation of the keyhole therefore givesThe upper bank runs outward and the lower bank inward. Their logarithms differ by $2\pi i$; keeping that orientation is essential for the sign.

Taking the difference of the two bank integrals at $n=2$ givesThe $I_j$ are real and convergent. Comparing real and imaginary parts proves $\boxed{I_0=\pi /2,I_1=0}$. Independently, $x\mapsto 1/x$ changes the sign of each odd logarithmic moment and preserves the measure, so every odd $I_j$ vanishes.

At $n=3$ the bank difference is $-6\pi i{I}_2+12{\pi }^2{I}_1+8{\pi }^{3}i{I}_0$, while the residue result is $13i{\pi }^4/4$. Substitution of $I_0,I_1$ yields $I_2={\pi }^3/8$.

At $n=5$, the odd moments vanish and the remaining relation isThusAs an independent generating-function check, $F(s)={\int }_0^{\infty }{x}^s/(1+x^2)dx=\pi /[2\cos (\pi s/2)]$ for $|\mathrm{Re}s|<1$. Differentiating twice and four times at zero gives the same moments.