Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 14A / Solution

The Dirichlet series for the Riemann zeta function converges for $\Re z>1$. In the Hankel integral representation the branch is a branch of $t^{z-1}$, so the argument restriction concerns $t$, not the parameter $z$. Use the principal branch cut along the negative real axis and a positive Hankel loop with a fixed small indentation around zero. Its integral $I(z)$ is an entire function of $z$, since the tails have exponential decay and the fixed contour avoids zero. Multiplication by $\Gamma (1-z)/(2\pi i)$ gives the meromorphic continuation of $\zeta$: it is directly defined away from positive integers, has a pole at $1$, and the apparent singularities at $2,3,\dots$ are interpreted by limits.

For a check in the initial half-plane, collapsing the indentation for $\Re z>1$ givesExpansion of $(e^r-1{)}^{-1}$ into exponentials justifies the last equality there, and the Gamma reflection formula recovers the printed integral representation.

Take the upper semicircle contour with the straight segment oriented from left to right, making the total orientation positive. Its enclosed poles are $t=2\pi ij$, $j=1,\dots ,N$; each has residue $-(2\pi ij{)}^{z-1}$. ThusThe reverse orientation changes the sign.

Initially take $\Re z<0$, so the series on the right converges as $N\to \infty$ and the large arcs can be discarded under the given assumption. Denote by $L_{+}$ and $L_{-}$ the integrals along horizontal lines just above and below the real axis, both from left to right. The residue computation and its lower-half-plane clockwise counterpart giveContour deformation about the negative-axis branch cut gives $I=L_{-}-L_{+}$. HenceMultiplying by $\Gamma (1-z)/(2\pi i)$ yieldsIt extends by meromorphic continuation to all $z\ne 1$, with removable products evaluated by limits where individual factors diverge. The derivation does not assert ordinary convergence of the original positive-term series outside $\Re z>1$.