Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 22 / 3 / Solution

Use the Jacobi theta function $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$, $t>0$. Its transformation law follows directly from the Gaussian Fourier transform. With the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(u)e^{-2\pi i\xi u}du$,For example, differentiating the transform and using integration by parts gives ${\hat{f}}^{\prime }(\xi )=-(2\pi \xi /t)\hat{f}(\xi )$; its value at zero is the Gaussian integral $t^{-1/2}$. Periodize the Gaussian. Its Fourier coefficients are these transforms at integer frequencies, and both series converge absolutely, so evaluating its Fourier series at zero proves the Poisson summation formula here:

For $\Re s>1$, integrating the exponentially convergent series termwise gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $u=1/t$ in the lower part, using the transformation just proved. The contribution of $t^{-1/2}-1$ is $1/(s-1)-1/s$. We obtain the pole-subtracted theta integral for the completed zeta functionThe integral is entire: $\Theta (t)-1$ decays exponentially, uniformly dominating the integrand and all its $s$ derivatives on compact sets. Both the rational term and the integral are invariant under $s\mapsto 1-s$. This proves the meromorphic identityEquivalently the Riemann xi functionis entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication formulas for the Gamma function turn the same identity intounderstood by meromorphic continuation at removable exceptional points. This establishes the functional equation of the Riemann zeta function, rather than assuming it.

If $N(T)$ counts zeros $\rho =\beta +i\gamma$ in the critical strip with $0<\gamma \le T$, including zero multiplicity, the Riemann–von Mangoldt formula isAt a zero ordinate one may instead use the conventional half-weight boundary count; that only alters the stated error. In particular $N(T)\sim T\log T/(2\pi )$.

To invert without assuming simple zeros, fix $0<\epsilon <1$ and set $T_n^{\pm }=(1\pm \epsilon )2\pi n/\log n$. Then $\log T_n^{\pm }/\log n\to 1$, soFor large $n$, $N(T_n^{-})<n<N(T_n^{+})$, and hence $T_n^{-}<{\gamma }_n\le T_n^{+}$. Let $\epsilon \downarrow 0$. The asymptotic inversion of the zeta zero count givesAs in the zero-counting formula, the sequence lists ordinates with zero multiplicity; “increasing” means nondecreasing here. The argument does not require the Riemann hypothesis or simplicity of zeros.