Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 13A (Further Complex Methods)

Take the Hankel contour about the negative real axis: it starts at $-\infty$ just below the cut, circles the origin counterclockwise, and returns to $-\infty$ just above the cut. The notation $(0+)$ records this return side of the branch point. Choose the branch $-\pi <\arg t<\pi$ for $t^{z-1}$.

For $\mathrm{Re}z>1$, collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula $\Gamma (z)\Gamma (1-z)=\pi /\sin (\pi z)$, recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on $z$. The apparent poles of $\Gamma (1-z)$ are cancelled by zeros of the contour integral except at $z=1$. It therefore gives the analytic continuation of the Riemann zeta function to $\mathbb{C}\setminus \{1\}$.

For $n\ne 0$, the denominator has derivativeHence the simple-pole residue is

Apply the Hankel representation with $z$ replaced by $1-z$:Enlarge $H$ to a modified Hankel contour that encloses the simple poles $t=2\pi in$, $n\in \mathbb{Z}\setminus \{0\}$. In a strip where the deformation and residue sum converge, the outer pieces vanish and the orientation gives the Hankel integral as minus $2\pi i$ times the sum of the enclosed residues. Part (b) givesSubstitution therefore yieldsBoth sides are meromorphic functions of $z$, so the identity theorem extends the equality from the initial strip to every point where either side is defined.