Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 2

For $\Re s>1$, the absolutely convergent Dirichlet seriesdefines the Riemann zeta function. To continue it, use partial summation in Stieltjes form:Since $0\le \{x\}<1$, the final integral converges locally uniformly for $\Re s>0$ and is holomorphic there. The displayed expression is consequently a meromorphic function on that half-plane, with its only pole at $s=1$. Since $s/(s-1)$ has residue one there, so does $\zeta$. Agreement in $\Re s>1$ makes this continuation unique by the identity theorem.

The functional equation isEquivalently,

For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for $\Re s>1$,Split the integral at one, substitute $u\mapsto 1/u$ in the lower half, and use the theta transformation. The result isThe integral is an entire function of $s$ because $\Theta (u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto 1-s$, proving both the analytic continuation and the Functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.

Write the second form of the functional equation asFor $-2\le \sigma \le 2$, the elementary exponential formula for the sine gives, uniformly for $t\ge 4$,The stated Stirling formula givesuniformly on the same strip. The bounded factors $2^{\sigma }{\pi }^{\sigma -1}$ and the cancelling exponentials therefore show thatTaking absolute values in the functional equation proves the Vertical-strip factor in the Riemann zeta functional equation: