Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 23 / 3 / a / Solution

The Riemann xi function is the entire functionwith $\xi (s)=\xi (1-s)$. Its Hadamard factorization iswhere the Nontrivial zeros of the Riemann zeta function are repeated by multiplicity and the factors are canonical genus-one factors.

Here is the growth estimate needed for the Jensen zero-count bound. For $|s|\le 2T$, functional symmetry reduces to $\Re s\ge 1/2$. Euler summation truncated at $T^2$ bounds $(s-1)\zeta (s)$ by a fixed power of $T$, uniformly in that region; the multiplication cancels the pole at one. The logarithmic gamma estimate bounds $\log |\Gamma (s/2)|$ by $O(T\log T)$, including the bounded small-$s$ part separately. The remaining elementary factors obey the same bound. ThusFor a zero with $|\rho |\le T$, its contribution in Jensen's formula on radius $2T$ is at least $\log 2$. ConsequentlyIf a zero lies on the integration circle, use nearby radii and continuity of the zero-count estimate. Hence $n_{\xi }(T)=O(T\log T)$ for $T>2$. The growth also gives order at most one and justifies the stated Hadamard factorization; the zero-count bound gives convergence of its genus-one factors.

The printed logarithmic Stirling hint drops the term $-\frac{1}{2}\log z$. The correct expansion is $(z-\frac{1}{2})\log z-z+\frac{1}{2}\log (2\pi )+O(|z{|}^{-1})$ in a fixed sector. Its consequence $\log |\Gamma (z)|=O(|z|\log (2+|z|))$ is all that the argument needs.