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I am learning about to zeta-function, I am a beginner. I am trying to find:

$$\lim_{t \to \infty}\zeta(\frac{1}{2} + it)$$

From the book Theory of the Riemann zeta-function-clarendon by Titchmarsh in the theorem 8.12 I know:
If $\frac{1}{2} \leq \sigma < 1$ the $|\zeta(\sigma + it)|> e^{\log^{\alpha}y}$ with $\alpha < 1- \sigma $ and for some indefinitely large values of $t$

For my case if $0<\alpha<\frac{1}{2}$ then since $ |\zeta(1/2 + it)|> e^{\log^{\alpha}t}$ for some indefinitely large values of $t$ since $t \to \infty$ then I don't know if that's enough to conclude that $$\lim_{t \to \infty}\zeta(\frac{1}{2} + it) $$ does not converge.

Please any help is good.

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    $\begingroup$ What do you mean by $|\zeta(1/2 + it)|> \infty$? How can something be larger than infinity? For which values of $t$? We know that $|\zeta(1/2 + it)|$ has infinitely many zeros, so Titchmarsh's inequality cannot hold for all $t$. What is $t$ in Titchmarsh's inequality? You ahve $x$ and $y$ on the left-hand side. Please revise your question carefuly and address these issues. $\endgroup$
    – Gary
    Nov 7, 2021 at 22:55
  • $\begingroup$ Thanks you Mr Gary, I tryed to write better my doubt. $\endgroup$
    –  Maëlys
    Nov 7, 2021 at 23:39
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    $\begingroup$ The theorem still makes no sense. What is the relation of $x$ and $y$ to $t$? As I said $\zeta(1/2+i t)$ is $0$ infinitely often as $t \to +\infty$, so you cannot bound it from below by a positive expression of $t$. $\endgroup$
    – Gary
    Nov 7, 2021 at 23:39
  • $\begingroup$ It's on page 204 of the second edition, for the record. $\endgroup$ Nov 7, 2021 at 23:46
  • $\begingroup$ I corrected the statement. You replaces $\sigma$ and $t$ from the book by $x$ and $y$ for unknown reasons in certain places. Titchmarsh talks about $\limsup_{t\to +\infty}$. The limit does not exist, see the answer below. $\endgroup$
    – Gary
    Nov 7, 2021 at 23:46

2 Answers 2

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As @Gary comments, there is some ill-definedness and muddle in the statements of the question, and the supposedly-cited facts.

One very clear point is that it is known by now that there are infinitely-many zeros on the critical line (Levinson... Conrey... showed that at least 2/5 (?) or so are on-the line...), but/and away from zeros zeta grows. So there's no actual limit of $\zeta({1\over 2}+it)$ as $t\to \infty$.

Naturally, with or without RH, things are even more chaotic to the right of $\Re(s)={1\over 2}$. For example, Voronin's universality theorem overwhelmingly indicates that there are no elementary asymptotics on any vertical line to the right of $\Re(s)={1\over 2}$. Overkill, yes, but really decisive.

But/and, perhaps your true question can be more refined...?

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By Titchmarsh's theorem, $$ \mathop {\lim \sup }\limits_{t \to + \infty } \,\left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right| = + \infty . $$ We also know that there is an infinite number of zeros along the critical line (with an accumulation point at infinity), so $$ \mathop {\lim \inf }\limits_{t \to + \infty } \,\left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right| = 0. $$ Since the two are not equal, the limit $\lim_{t \to + \infty } \left| {\zeta\! \left( {\tfrac{1}{2} + it} \right)} \right|$ does not exist, and hence the limit $\lim_{t \to + \infty } \zeta\! \left( {\tfrac{1}{2} + it} \right)$ does not exist.

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