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It is known that the function $ \frac1\pi\arg\zeta\left(\frac12+iT\right) $ is continuous, except when T is the imaginary part of a Zeta zero.

In that case the jump of this function can only be large at Zeta zeros with high multiplicity. Is that correct ?

Cristian Dumitrescu
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1 Answers1

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It depends how you define it, many different branches..

  • Under the RH there is a branch of $\text{arg } \zeta(s)$ which is continuous on $\Re(s)\in (1/2,1)$,

letting $$\text{arg }\zeta(1/2+it)=\lim_{\epsilon\to 0^+}\text{arg }\zeta(1/2+it+\epsilon)$$ then yes there is a $+ k\pi$ jump at each non-trivial zero of multiplicity $k$.

  • If you don't believe in the RH then there is a branch of $\text{arg } \zeta(s)$ continuous on $\Re(s) > 1$ and on every horizontal strip without zeros,

in which case $\lim_{\epsilon\to 0^+}\text{arg }\zeta(1/2+it+\epsilon)$ has a $+ k\pi$ jump at each zero of multiplicity $k$ on $\Re(s)=1/2$ and a $+2k\pi$ jump at each $t$ such that there are $k$ zeros (counted with multiplicity) on $\Im(s)=t,\Re(s) > 1/2$.

reuns
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