I have deleted a previous, now obsolete question on the same topic.

Take the well-known Riemann integral:

$$\displaystyle \pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) =\int_1^{\infty} \left({x}^{\frac{s}{2}-1} + {x}^{-\frac{s}{2}-\frac12}\right)\,\sum_{n=1}^{\infty}e^{-\pi\,n^2\,x}\, \text{d}x - \frac{1}{s\,(1-s)}$$

and move $\frac{1}{s\,(1-s)}$ to the RHS to make both sides entire and call it $\widehat{\xi}(s)$:

$$\displaystyle \widehat{\xi}(s)=\pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) + \frac{1}{s\,(1-s)} =\int_1^{\infty} \left({x}^{\frac{s}{2}-1} + {x}^{-\frac{s}{2}-\frac12}\right)\,\sum_{n=1}^{\infty}e^{-\pi\,n^2\,x}\, \text{d}x$$

The function $\widehat{\xi}(s) = \widehat{\xi}(1-s)$ induces an infinite number of symmetrical zero ($\mu$) foursomes. enter image description here

After calculating the first 100.000 $\mu$'s (i.e. 400.000 zeros), I strongly believe that $\widehat{\xi}(s)$ has the following Hadamard product ( ' denotes that also $\overline{\mu_n},1-\mu_n,1-\overline{\mu_n}$ should be included in the operation):

$$\displaystyle \widehat{\xi}(s) = \widehat{\xi}(0) \prod_{n=1}^{\infty}' \Bigl(1-\frac{s}{\mu_n}\Bigr)$$

Where $\widehat{\xi}(0)=\widehat{\xi}(1) = 1+\frac12\gamma -\ln(2)-\frac12\,\ln(\pi)$ which also equals $\sum_{ \rho }\Bigl( \frac{1}{\rho}+\frac{1}{1-\rho}\Bigr)$.

It then simply follows that:

$$\displaystyle \pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) + \frac{1}{s\,(1-s)} = \widehat{\xi}(0) \prod_{n=1}^{\infty}' \Bigl(1-\frac{s}{\mu_n}\Bigr)$$

This allows equating the logarithmic derivatives $\frac{\widehat{\xi}'}{\widehat{\xi}}(s)$ of both sides and injecting $s=1$ gives:

$$\displaystyle \sum_{n=1}^{\infty}'\Bigl(\frac{1}{\mu_n}\Bigr)=\scriptsize{\frac{2\ln(\pi)^2+(8 \ln(2)-4\gamma)\ln(\pi)+8\ln(2)^2-6\gamma^2-8\gamma\ln(2)+\pi^2-16\gamma_1-16}{16+8\gamma-16\ln(2)-8\ln(\pi)}}$$

Also, the equation allows to express $\zeta(s)$ in terms of the $\mu$-zeros:

$$\displaystyle \zeta(s) = \dfrac{\widehat{\xi}(0) \prod_{n=1}^{\infty}' \Bigl(1-\frac{s}{\mu_n}\Bigr)-\frac{1}{s\,(1-s)}}{\pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\,}$$

and the RHS induces a non-trivial zero $\rho$ when: $\displaystyle \widehat{\xi}(0) \prod_{n=1}^{\infty}' \Bigl(1-\frac{s}{\mu_n}\Bigr)=\frac{1}{s\,(1-s)}$.

Note that the zeros of $\widehat{\xi}(s)$ follow a surprisingly regular pattern. I believe this pattern originates from them all residing increasingly deeper in the domain $\Re(s)>1$. In this domain $\zeta(s)$ is well behaved and rapidly approaches $1$ when $\Re(s) \rightarrow \infty$. So, in $\pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s)$, the component $\zeta(s)$ will have a negligible influence on the location of the $\mu$'s. $\Gamma\left(\frac{s}{2}\right)$ can be expressed in terms of one of its Stirling approximation and the result follows that for $\Re(s) \rightarrow +\infty$, the function $\widehat{\xi}(s)$ will induce increasingly more accurate zeros $\mu$ and $\overline{\mu}$ when:

$$\displaystyle 2\,(s-1)\,\sqrt{\pi\,s}=\left(\frac{2\,e\,\pi}{s}\right)^{\frac{s}{2}}$$

Since $\widehat{\xi}(s) = \widehat{\xi}(1-s)$ the zeros $1-\mu,\overline{1-\mu}$ can be simply derived.


Assuming this Hadamard product can indeed be proven, does the apparent regular pattern of the $\mu$'s impose any constraint on where the non-trivial zeros $\rho$ of $\zeta(s)$ can reside in the critical strip (e.g. could it imply a zero-free region)?


1 Answer 1


There is no problem in proving the Hadamard product if you write it as $$\hat{\xi}(s)=\hat{\xi}(0)\prod_{\mu}\bigg(1-\frac{s}{\mu}\bigg)\bigg(1-\frac{s}{1-\mu}\bigg)$$

Also, your observation about behavior of $\mu$ is correct. However, you can't get anything about Riemann zeros out of this. To put it in simple terms, the reason for this is that you can replace your function by something like $$\pi^{-s/2}\Gamma(s/2)\zeta(s)\bigg[1+\frac{a}{(s-\rho)(s-\overline{\rho})}\bigg]+\frac{1}{s(1-s)},$$ (where $\zeta(\rho)=0$ is a zero you like), and you will still have the same pattern.

  • $\begingroup$ Thanks Alex. Makes sense. To proof the Hadamard product, I have tried (assuming $\hat{\xi}(s)$ has order $1$) to solve $A$ and $B$ from $\displaystyle \hat{\xi}(s)=e^{A+B\,s}\prod_{\mu}\bigg(1-\frac{s}{\mu}\bigg)\, e^{\frac{s}{\mu}}$ by equating the logarithmic derivatives of both sides. This way, I can derive $A$ and $B$, however I struggle to make the final step towards $\widehat{\xi}(0)=\widehat{\xi}(1) = 1+\frac12\gamma -\ln(2)-\frac12\,\ln(\pi)$ (i.e. getting rid of the e-powers). Would you be so kind to share (a sketch of) your proof in the answer? $\endgroup$
    – Agno
    May 11, 2014 at 17:23
  • $\begingroup$ This function does have order 1. The idea of my (very well known) proposal is to get rid of exponents by taking the roots in pairs. (By the way, I missed $\exp(Bs)$; I forgot that the function isn't actually even.) But you may proceed your way very well, too. The product your wrote in the comments converges because $\hat{\xi}$ has order 1 (and other good properties). So, I do not understand what's exactly the problem. $\endgroup$ May 12, 2014 at 11:27
  • $\begingroup$ I may try and help you more if you explain it. $\endgroup$ May 12, 2014 at 11:32
  • $\begingroup$ Thanks Alex. No need for further help, I got it now :-) $\endgroup$
    – Agno
    May 12, 2014 at 21:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.