The Riemann xi function has its genus-one Hadamard product over the nontrivial zeros of zeta

By [L1], xi is entire of finite order $1$. Applying [L2] with $\rho =1$ therefore yields constants $A,B\in \mathbb{C}$ such that $\xi (s)=e^{A+Bs}{\prod }_{{\rho }^{\mathrm{\prime }}}{E}_1(s\mathrm{/}{\rho }^{\mathrm{\prime }}),$ where the product runs over the zeros ${\rho }^{\mathrm{\prime }}$ of xi, counted with multiplicity.

By [L5] and [L4], $\xi (s)=\frac{1}{2}+\frac{1}{2}s(s-1)H(s)\mathrm{.}$ Hence $\xi (0)=\xi (1)=1\mathrm{/}2$, so neither $0$ nor $1$ is a zero of xi. Now fix $m\ge 1$. The symmetry from [L4] and [L5] gives $\xi (-2m)=\xi (1+2m)\mathrm{.}$ Since $1+2m>1$, [L6] gives $\zeta (1+2m)\mathrm{\ne }0$, and [L7] shows that $\mathrm{\Gamma }((1+2m)\mathrm{/}2)$ is finite because $(1+2m)\mathrm{/}2$ is not a nonpositive integer. The scalar factor $\frac{1}{2}(1+2m)(2m)$ is also nonzero, so [L5] gives $\xi (1+2m)\mathrm{\ne }0$, hence $\xi (-2m)\mathrm{\ne }0$. Together with [L3], this shows that the zeros of xi are exactly the nontrivial zeros of zeta.

Replacing the zero set $\{{\rho }^{\mathrm{\prime }}\}$ in step 1.1 by the nontrivial zeros of zeta from step 1.2 gives the announced product. The phrase "genus-one canonical sense" is exactly the convergence prescription supplied by [L2] for an order-one entire function.