The Dirichlet $\beta$-function is defined for $\Re(s)>0$ as:

$$\beta(s) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}$$

It has the following Euler product (I used that Dirichlet character $\chi_{4}(p)=\sin\left(\frac{p \,\pi}{2}\right)$):

$$\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$$

Numerical evidence suggests that this Euler product also (slowly) converges for values $\Re(s)>\frac12$.

Does convergence in the domain $\frac12 < \Re(s) \le 1$ indeed occur? If so, could this be proven (I guess a proof would also imply that all complex zeros of $\beta(s)$ must reside on the critical line)?


Just to share that by multiplying the Leibniz formula for $\pi$ ($\beta(1)=\frac{\pi}{4}$) and this formula (35), gives the very elegant relationship:

$$\prod_p \bigg(\frac{p-\sin\left(\frac{p \,\pi}{2}\right)}{{p+\sin\left(\frac{p \,\pi}{2}\right)}} \bigg)=2$$

  • 1
    $\begingroup$ This is going to revolve around the convergence of the series $ \sum_p \chi_4(p) p^{-s} $ for $1/2<\Re(s)\leqslant 1$. I don't think we have enough information about the pattern of the primes modulo 4 to be able to say anything helpful. (Of course the nature of the beast is that such a relationship is equivalent to the function's Riemann hypothesis.) $\endgroup$
    – Chappers
    Jun 16, 2015 at 15:36
  • $\begingroup$ Chappers, you are probably right, although this is one of the very few Euler products that I know of that converges for $s=1$ ($\beta(1)$ is a well-known formula). Hence it could be an indication that further convergence exists. Numercal evidence does indeed suggest (very slow) convergence continues in the strip, however certainly not $\le \frac12$. $\endgroup$
    – Agno
    Jun 16, 2015 at 20:37
  • $\begingroup$ $\chi_4(2n+1) = (-1)^n$. Then $\sum_p \chi_4(p)p^{-s}$ and hence $\prod_p \frac{1}{1-\chi_4(p)p^{-s}}$ converge for $\Re(s) > \sigma$ iff $L(s,\chi_4)$ has no zeros for $\Re(s) > \sigma$ (the proof of $\Longleftarrow$ is very similar to the prime number theorem) $\endgroup$
    – reuns
    Nov 29, 2017 at 22:17
  • $\begingroup$ @reuns Thanks and clear. I have learned a lot since this question and wouldn't have asked it today. $\endgroup$
    – Agno
    Nov 30, 2017 at 14:03
  • $\begingroup$ Sure. To me this equivalence between the zeros and the abscissa of convergence is really the fundamental theorem of the theory of $\zeta(s), L(s,\chi)$, more than the PNT. $\endgroup$
    – reuns
    Nov 30, 2017 at 14:14

1 Answer 1


Since the product (over primes) for Dirichlet L-series is valid only for $\Re{(s)}>1$, as mentioned in all textbooks on this subject, the answer to Agno initial question is that not even the `easiest' case $s=1$ is proved for $\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$. I have investigated these products for $1/2 < \Re{(s)} \le 1$ for a while and also detected a slow numerical convergence, but this was just a conclusion on experimental computations, nothing such as a formal proof.

P.S.: Could someone (maybe Agno, himself, or Reuns) provide a reference in literature (or internet) for the formal proof of "the equivalence between the zeros and the abscissa of convergence" mentioned by Reuns?


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