This question loosely builds on the second part of this one.

Take the Riemann $\xi$-function: $\xi(s) =\frac12 s\,(s-1) \,\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)\, \zeta(s)$. Numerical evidence suggests that for all $a \in \mathbb{R}$, the zeros of:

$$g(s,a):=\xi(a+s)\pm \xi(a+1-s)$$

all reside on the critical line $\Re(s)=\frac12$.

The graph below illustrates the claim for $-2<a<2$ (and $\pm=+$). Each line shows $\Im(s)$ of a zero for $g(s,a)$ at $\Re(s)=\frac12$. When $a=0$ the zeros are equal to the non-trivial zeros $\rho$ of $\zeta(s)$.

enter image description here

The highlighted part $(a\ge 1)$ shows the domain where $\displaystyle \sum_{n=1}^\infty\frac{1}{n^{a+s}}$ and $\displaystyle \sum_{n=1}^\infty\frac{1}{n^{a+1-s}}$ as well as their Euler products both do converge whilst using values of $s$ with $\Re(s)<1$.

Let's take $a=1$, $\pm=-$ and $0 < \Re(s) <1$, then for each $\xi(1+s) - \xi(2-s)=0$ the following equation should hold:


or in terms of the primes from the Euler products:

$$\prod_{p=prime} \left( \dfrac{p^{s+1}- p^{2s-1}}{p^{s+1}-1} \right)= \frac{2^s\,\pi^{s-1}\,\cos\left(\frac{\pi\,s}{2}\right)\,\Gamma(3-s)}{s\,(s+1)}$$

Note that the real part of the exponent in the term $p^{2s-1}$ only becomes $0$ when $\Re(s)=\frac12$. It gets negative for $s < \frac12$. I realise this is quite a broad question, but is there anything more to say about the influence of this term on the location of the zeros in the highlighted domain?

Additional note:

The term $p^{2s-1}$ is independent of the choice of $a$; the LHS always is: $\displaystyle \prod_{p=prime} \left( \dfrac{p^{s+a}- p^{2s-1}}{p^{s+a}-1} \right)= yyy$.


2 Answers 2


This problem (and more general ones) was solved completely and unconditionally (i.e., without any RH) by P.R. Taylor before the second world war, but I don't remember the reference, it should be easy to Google. Download for example the paper by McPhedran and Poulton "The Riemann Hypothesis for Symmetrised combinations of the Zeta Function"

  • $\begingroup$ The Taylor piece is 1945, Quarterly J. of Math. Oxford, which is indeed visible in the McPhedran-Poulton arXiv paper. Slightly surprisingly, it has been overlooked in many papers in the last few decades that in fact do similar things. Taylor explicitly addressed $\xi(s-{1\over 2})/\xi(s+{1\over 2})$, but shifts (in one direction) not involving RH, or (in the opposite direction) assuming it, would follow from the same reasoning. $\endgroup$ Commented Dec 26, 2016 at 23:59

The equation doesn’t hold, because: $$\lim\limits_{s\downarrow 0} s\zeta(s+1)=1\ne\frac{\pi}{3}=\lim\limits_{s\downarrow 0} \frac{\zeta(2-s)2^s\pi^{s-1}\cos(\frac{\pi s}{2})\Gamma(3-s)}{s(s+1)}$$ and $$\lim\limits_{s\uparrow 1} \frac{1}{(1-s)\zeta(2-s)}=1\ne\frac{3}{\pi}=\lim\limits_{s\uparrow 1} \frac{2^s\pi^{s-1}\Gamma(3-s)}{\zeta(s+1)s(s+1)}\frac{\cos(\frac{\pi s}{2})}{1-s}$$

  • $\begingroup$ I am not claiming that the formula holds for all $s$. As I stated in the question, it only works for complex $s$ where $\xi(1+s) - \xi(2-s)=0$ and $0 < \Re(s) <1$ (the latter to ensure the series and their associated Euler products converge). The conjecture is that zeros only occur at points when $\Re(s)=\frac12$. $\endgroup$
    – Agno
    Commented Apr 30, 2016 at 10:02
  • $\begingroup$ O.k.. Because of the symmetry it seems to be so, but I cannot proof it. $\endgroup$
    – user90369
    Commented Apr 30, 2016 at 12:58

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