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

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.

Write $s=\sigma +it$ fixed, with $\xi (s)\ne 0$. The zeros satisfy $0<\beta <1$. For $|\gamma |>2(|t|+1)$,The previous bound gives at most $O(2^{j}j)$ zeros in each dyadic ordinate band $2^j\le |\gamma |<2^{j+1}$, so its total majorant is $O(j{2}^{-j})$. That series converges. There are only finitely many zeros in the remaining bounded bands, and none has the forbidden denominator zero at the specified nonzero point of $\xi$. Thus the real logarithmic derivative sum converges absolutely. This proves the absolute convergence of the real xi logarithmic derivative without claiming absolute convergence of the unpaired complex sums of $1/(s-\rho )$.

Evaluate the supplied real logarithmic derivative at $2+it$. Since $1<2-\beta <2$, its positive summand is bounded above and below by constant multiples of $(1+|t-\gamma {|}^2{)}^{-1}$. On the other hand, differentiating the defining xi expression givesAt real part two the last term is bounded by the absolutely convergent series $\sum \Lambda (n)n^{-2}$; the gamma logarithmic derivative is $O(\log t)$. ThereforeFor $\gamma \in [T,T+1]$, use $t=T$: each term in the displayed sum is at least $1/2$. Hence the number of zeros in that unit ordinate interval, counted with multiplicities, is $O(\log T)$. These are the local zeta zero-count bound and the corresponding smoothed bound.

Under the Riemann hypothesis, every $\beta =1/2$, so for $\sigma >1/2$ the absolutely convergent formula givesThe zero set is nonempty: otherwise Hadamard factorization would make $\xi$ an exponential of a linear polynomial, and its functional symmetry would force it to be constant, contrary to gamma growth on the positive real axis. Thus the inequality is strict in the open right half-plane. Continuity at $\sigma =1/2$, including at boundary zeros, proves the claimed increasing modulus on the closed half-line.

Conversely, suppose the modulus is nondecreasing for every fixed $t$. If a zero $\beta +i\gamma$ had $\beta >1/2$, then nonnegativity and monotonicity would force $|\xi (\sigma +i\gamma )|=0$ throughout $1/2\le \sigma \le \beta$. The identity theorem would make $\xi$ identically zero, a contradiction. A zero left of the line reflects to one right of the line by the functional equation and complex conjugation symmetry. Hence every zero lies on the critical line. This proves the xi modulus criterion for the Riemann hypothesis. The two following roman headers refer to supplied asymptotic assumptions, not further questions, and require no Solution sections.