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Let $N(T)$ denote the number of zeros $b=α+iβ$ (counted with multiplicity) of the Riemann zeta function $ζ(s)$ for which $0<β<T$. The functional equation and the argument principle implies that $$N(T)=(T/(2π))log(T/(2πe))+(7/8)+(1/π)argζ((1/2)+iT)+O((1/T))$$ If $T=20$, the we get $$N(20)= 0.99967<1$$ which means that there is no zeros in this region. But we know that there exist only one zero in this region. How this could be possible.

Safwane
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  • N(T) is formula (132) on page 38 in the paper "A theory for the zeros of Riemann ζ and other L-functions" by Guilherme Franca and André LeClair, found here: https://arxiv.org/abs/1407.4358 – Mats Granvik Jan 10 '17 at 18:32
  • @MatsGranvik: So, my calculations are wrong. – Safwane Jan 10 '17 at 18:41
  • No I did not say that. I was just surprised to see this formula of yours without reference to Franca LeClair. I thought initially that Franca LeClair had arrived at this formula but your wording says otherwise. – Mats Granvik Jan 10 '17 at 18:47
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    E. James, whether they are right or not I think ignores the main issue: you have an $O(T^{-1})$ term: without a bound on the constant involved, $N(20)$ could conceivably be quite large, only in the limit can you ignore it. – Adam Hughes Jan 10 '17 at 18:48
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    @MatsGranvik: The formula is due to Riemann in 1851 – Safwane Jan 10 '17 at 18:49
  • @AdamHughes: Yes you are right. THE VALUE WITH THE THERM IS >1. – Safwane Jan 10 '17 at 18:55

1 Answers1

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You made a mistake when computing $\text{arg}\ \zeta(1/2+it)$ which in this context is defined like this :

$\log \zeta(s)$ is analytic on $Re(s) > 1$, choose $\log \zeta(2) = 0$, and for some $t$ such that $\zeta(s)$ has no zeros $Im(s) = t$, continue $\log \zeta(s)$ analytically from $2+it$ to $1/2+it$. Finally set $\text{arg}\ \zeta(1/2+it) = \text{Im}\ \log \zeta(1/2+it)$.

In other words, our branch of $\log \zeta(s)$ is continous on horizontal lines $(it-\infty,it+i\infty)$ where $\zeta(s)$ has no zeros.

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