Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 14E (Further Complex Methods)

Use the Hankel contour about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi <\arg t<\pi$. For $\mathrm{Re}s>0$, the small circle vanishes and the two banks giveThe gamma reflection formula then gives

Near zero,Subtracting any desired number of these terms makes the local contour integral converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma (1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in \mathbb{C}$.

At a nonpositive integer, the branch disappears and the collapsed Hankel contour extracts the residue at zero. For $s=0$,For $s=-2$, the coefficient of $t^{-1}$ inis $-1/4$; multiplication by $\Gamma (3)=2$ givesWhen $s=-2n-1$, the integrand is $t^{-2n-2}$ times an even power series, so it has no $t^{-1}$ coefficient. Hence

The nonzero poles areTheir paired residues sum toThe paired series converges for $\mathrm{Re}s<1$ by the alternating-series test, and absolutely for $\mathrm{Re}s<0$. ThereforeUsing the stated contour closure, the Hankel integral is $-2\pi i$ times this sum. Substitution into part (a) givesNow $1/\Gamma (1-s)=\Gamma (s)\sin (\pi s)/\pi$ and $\sin (\pi s)=2\sin (\pi s/2)\cos (\pi s/2)$, soIt was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the identity theorem therefore extends it to every $s\in \mathbb{C}$.