I'm reading Amplitudes and the Riemann Zeta Function, which recently appeared in Physical Review Letters. It's received some publicity, including my own campus' PR operation. From the abstract (adapting the notation)
"Physical properties of scattering amplitudes are mapped to the
Riemann zeta function. Specifically, a closed-form amplitude is
constructed, describing the tree-level exchange of a tower with masses
$m_n^2=\gamma_n^2$, where $\zeta(1/2+i\gamma_n)=0$. Requiring real
masses corresponds to the Riemann hypothesis…"
NB: I'm skeptical that this will lead to any progress on the Riemann Hypothesis, and not interested in that aspect.
I'm trying to determine the new (if any) mathematical content. The author develops the identity
$$-4+\frac{\pi^2}{8}+G+\frac{\zeta^{\prime\prime}(1/2)}{2\zeta(1/2)}-\frac{1}{8}\left(C+\frac{\pi}{2}+\log8\pi\right)^2
=2\sum_{n=1}^\infty\frac{1}{\gamma_n^2}.$$
Here $G$ is the Catalan constant and $C$ denotes the Euler constant. (I can check this via the logarithmic derivative of the Hadamard product for $\Xi(\sqrt{s})$, and differentiating the functional equation for $\zeta(s)$ twice.)
More generally, with
$$\zeta_n(s)=\frac{\zeta^{(n)}(s)}{\zeta(s)}, \qquad\zeta_n^k=\zeta_n(1/2)^k,$$
he has identities for $2\sum_{n=1}^\infty 1/\gamma_n^{2k}$ for odd $k$, for example
$$
2\sum_{n=1}^\infty \frac{1}{\gamma_n^6}=-128+\frac{1}{7680}\Psi^{(5)}(1/4)-\zeta_1^6+3\zeta_1^4\zeta_2-\frac{9}{4}\zeta_1^2\zeta_2^2+\frac{1}{4}\zeta_2^3-\zeta_1^3\zeta_3+\zeta_1\zeta_2\zeta_3-\frac{1}{12}\zeta_3^2+\frac{1}{4}\zeta_1^2\zeta_4-\frac{1}{8}\zeta_2\zeta_4-\frac{1}{20}\zeta_1\zeta_5+\frac{1}{120}\zeta_6
$$
The author writes
"[These] can be proven exactly albeit laboriously, without appeal to
our amplitude, using repeated differentiation of the functional
equation and the Hadamard product form of the zeta function, as well
as various polygamma identities… what is remarkable is that our
amplitude construction allows for much simpler, physical derivations
of these identities."
Are such identities new? Are they interesting? Regarding the latter, the author in an online talk describes such sums as moments of the $\{\gamma_n\}$, and they look like moments to me.
Best Answer
Thanks to @reuns for the answer in the comments. I've asked him to post as an answer, and I will accept it if he does. Meanwhile, his comments encouraged me to look again, and here is another approach, quite easy (Lemma 4.11 in Equivalents of the Riemann Hypothesis vol. II)
Let $\Xi(s)=\xi(1/2+is)$, so $\Xi(-s)=\Xi(s)$. Let $\gamma_n$ be the zeros with positive real part, so the Hadamard product looks like $$ \Xi(s)=\xi(1/2)\prod_{n=1}^\infty\left(1-\frac{s^2}{\gamma_n^2}\right) $$ Expanding $\Xi(s)$ as a power series about $s=0$ $$ \sum_{k=0}^\infty \frac{(-1)^k\xi^{(2k)}(1/2)}{2k!}s^{2k}=\Xi(s)=\xi(1/2)\left(1-s^2\left(\sum_n \gamma_n^{-2}\right)+s^4\left(\sum_{m,n}\gamma_m^{-2}\gamma_n^{-2}\right)-\ldots\right) $$ Equating coefficients and the Newton Identities gives the moments as polynomials in the $\xi^{(2k)}(1/2)$, for example $$ \sum_n\frac{1}{\gamma_n^6}=\frac{\xi^{(6)}(1/2)}{240\,\xi(1/2)}-\frac{\xi^{(2)}(1/2)\xi^{(4)}(1/2)}{16\,\xi(1/2)}+\frac{\xi^{(2)}(1/2)^3}{8\,\xi(1/2)}. $$