Take the Hadamard product for the Riemann $\xi$-function ($\rho$ is a non-trivial zero of $\zeta(s)$):

$$\xi(s) =\frac12\, s\,(s-1) \,\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)\, \zeta(s) =\frac12\,\prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$

Firstly strip out the factor $\frac12$ on both sides and then find the solution for:

$$s\,(s-1) \,\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)=1$$ This yields the unique value of $\mu=3.171242735975879847\dots$ and therefore:

$$\zeta(\mu) = \prod_\rho \left(1- \frac{\mu}{\rho} \right) \left(1- \frac{\mu}{1-\rho} \right)$$ Hence, since $\mu \gt 1$ this gives a direct relation between the products of primes and $\rho$'s:

$$\prod_{p \in \mathbb{P}} \left( \dfrac{1}{1-p^{-\mu}} \right)=\prod_\rho \left(1- \frac{\mu}{\rho} \right) \left(1- \frac{\mu}{1-\rho} \right)$$

Is there anything known about $\mu$? (I tried link to known constants on Wolfram, but no success)

  • 1
    $\begingroup$ I've never heard anything about this constant before. $\endgroup$ – Greg Martin Apr 16 '15 at 21:59
  • 1
    $\begingroup$ I don't see why the value of $\mu$ would be useful or important in any way. $\endgroup$ – Peter Humphries Apr 17 '15 at 17:21

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