About the number of zeros of the zeta function?

Question

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.

Answer 1

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.