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23:06
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Q: $\int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$ where $\zeta$ is the Riemann zeta function

Ashyln Brooke$$I=\int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$$ where $\zeta$ is the Riemann zeta function. It is known by Balazard Saias and Yor Paper that I is integrable and $0\leq I<\infty$. Since by Schwarz Reflection principle, integrand is an even function, so, $$I= 2\i...

I don't understand your question. What does it mean $I$ integrable? Everything you need is proven in the Balazard papers, why don't you follow them. $\frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$ is locally integrable (because $\zeta$ is meromorphic), if it is integrable then $I=\lim_{\epsilon\to 0}I_\epsilon$ trivially. Is it integrable?
user501743
@reuns Please give me an idea about how $I=lim_{\epsilon \to 0}I_{\epsilon}$ Trivially?
user501743
@Conrad Thanks for your edit. Please answer any of the two questions.
Did you read my comment?
user501743
@reuns So $\frac{log|\zeta(1/2+it)|}{1/4+t^2}$ is integrable on $[t_n+\epsilon,t_{n+1}-\epsilon]$. Can you shed some light on this?
user501743
23:06
@reuns Yes I read it but I could not understand how the limit is trivial.
user501743
@Conrad I know I am lacking somewhere. Please give a minute hint. Please.
user501743
@reuns Can you please give me a hint as to why the limit is equal to I? How is it locally integrable?
user501743
@Conrad log|x| is integrable near 0.
user501743
@Conrad Because it is defined near 0 but not at 0 possibly.
user501743
@Conrad I have worked hard on typing this question as I am new to latex. Please answer. I am totally confused. Is what I have written in the question true? If yes then please tell me the reason. A small hint will be enough. Thanks
user501743
23:06
@Conrad log|x| is locally integrable on $\mathbb{R}$. Please now answer. Please.
user501743
@Conrad . How can we pass on the limit ? Do we use dominated convergence theorem?
user501743
@Conrad $\int_{-1}^{1} log|x| dx = 2\int_{0}^{1} log x dx$
user501743
@Conrad$2\int_0^1 log x dx = x log x - x $ then as x $\to $ 0 it tends to a finite limit.
user501743
@Conrad x logx $\to 0$ as x$\to 0$
user501743
@Conrad I think I have answered your question. Now please answer Conrad. Please.
user501743
23:06
@Conrad Sorry I don't know this answer.
user501743
@Conrad please give some an answer as to how we get $\lim_{\epsilon \to 0} I_{\epsilon}= I$
user501743
@Conrad Is my question 1 and question 2 both true?
user501743
@reuns How is $\frac{log|\zeta(1/2+it)|}{1/4+t^2}$ locally integrable?
user501743
@Conrad How is $\frac{log|\zeta(1/2+it)|}{1/4+t^2}$ locally integrable?
user501743
@Conrad I am just expecting an answer from your part. Please answer.
user501743
23:06
@Conrad Please answer. Please
user501743
@Conrad Just one answer will suffice. Please don't leave me confused. I won't be able to sleep.
user501743
@Conrad Thanks. What is $n$? And z is any complex number?
user501743
@Conrad Thanks again. And reuns answered that $\zeta$ is meromorphic so $\frac{log|\zeta(1/2+it)|}{1/4+t^2}$ is locally integrable. How does this hold?
user501743
@Conrad It is an analytic function everywhere in its domain except for a set of poles
user501743
At point 1 simple pole
user501743
23:06
@Conrad So how is homlomorphic related to the locally integrability of the function? We also have a $log |\zeta(1/2+it)|$ which is $-\infty$ for infinitely many t
user501743
@Conrad Please write as an answer.
user501743
@Conrad Ok then please tell me that can I say that $log|\zeta(1/2+it)|/(1/4+t^2)$ is integrable in $[0,\infty)$ so it is locally integrable in every compact set $[t_n+\epsilon,t_{n+1}-\epsilon]$. So $ I_{\epsilon} \to I$?
user501743
@Conrad Then how $I_{\epsilon} \to I$?
user501743
@Conrad Please elaborate the theorem used in measure theory. I will read about it.
user501743
@Conrad Please tell about the theorem used to prove the limit.
user501743
23:06
@Conrad Please write a rough proof as an answer. I will be obliged to give your answer a check mark.
user501743
@Conrad Thank you so much for patiently answering. Please write as an answer. A short answer is enough.
user501743
@Conrad Please answer it about a holomorphic function. I will do the rest Conrad. Please. 1 or 2 lines answer will suffice.
user501743
@Conrad Give me a one line answer in comment as to how the limit is equal to I.
user501743
@Conrad Please answer. I have got 7 upvotes but not a single answer. Please answer Sir.
user501743
@Conrad I am sorry. Please answer my last question? Are my both questions true? Please reply. Thanks a lot.
23:06
your question are not well-posed (hint $\liminf (t_{n+1}-t_n) = 0$, so for fixed $\epsilon$ there is a large $n(\epsilon)$ for which $[t_n+\epsilon,t_{n+1}-\epsilon]$ becomes ill defined at least for some $n$ and I hoped after all this discussion you would get that, how to modify them to be well posed (one way is to define $I_{\epsilon}$ only for $n < n(\epsilon)$) and how to answer them (local integrability to ensure $I_{\epsilon}$ exists and global integrability to insure the limit exists;
user501743
@Conrad I did not get you. So how should i redefine my n and $\epsilon$?
do you know that $\liminf (t_{n+1}-t_n) = 0$?
user501743
@Conrad Why are we talking about $ lim \ inf(t_{n+1}-t_n)$?
because it means that if you fix $\epsilon >0$, the interval $[t_n+\epsilon,t_{n+1}-\epsilon]$ becomes ill defined for infinitely many large $n$
user501743
@Conrad Why is it ill defined?
23:06
beacuse $(t_{n+1}-t_n) < \epsilon$ for infinitely many large $n$
user501743
@Conrad So what should be the interval instead of $[t_n+\epsilon,t_{n+1}-\epsilon]$?
user501743
@Conrad $t_n$ and $t_{n+1}$ are consecutive zeros of zeta on the critical line then how is $t_{n+1}-t_n<\epsilon$?
not sure about the goal of the question; the way one solves this (usually) is to restrict to $n(\epsilon)$ for which $(t_{n+1}-t_n) < \epsilon, n \le n(\epsilon)$
do you know how many zeroes (unconditionally proven by Selberg and improved by others) does RZ have on the critical line up to level $T$?
user501743
@Conrad There are finitely many zeros upto the height T.
of course but that is general theory of holomorphic functions; for RZ much more precise results are known
user501743
23:06
@Conrad It is some constant times T log T
correct - which means that the average spacing of the zeroes is what?
user501743
@Conrad I am sorry. I don't know this one.
let's spell it out (a rough heuristic argument that can be made rigorous); up to $T$ we have $cT\log T$ zeroes, and up to $T+1$ we have $c(T+1)\log (T+1)$ zeroes, so how many roughly are between $T$ and $T+1$?
user501743
@Conrad Is it $c(1+\frac{log T}{T})$?
user501743
@metamorphy Please point out the error in the question.
23:06
The answer is $c(T+1)\log (T+1)-cT \log T=c\log (T+1)+O(1/T)$ hence the average spacing is $c/\log T$ which is also (roughly) $c/\log t_n \to 0$ so the spacings go to zero at least for a subsequence, hence the OP is ill formulated as it stands;
user501743
@Conrad I got it. Thank you so much. Please tell me how my $[t_n+\epsilon,t_{n+1}-\epsilon]$ becomes ill defined if $t_{n+1}-t_{n}<\epsilon$?
$t_n+\epsilon >t_{n+1}-\epsilon$
user501743
@Conrad Thanks a ton. So how to redefine so it works?
not sure of what you want to prove in the post
user501743
@Conrad Actually I am trying to define an interval such that $|\zeta(1/2+it)|$ is non zero. Please tell me an interval. This is my last question.
23:06
$[0, t_1-1/2]$ works where $t_1=14...$ is the first zero ordinate; not sure why does that help here
user501743
@Conrad and what for $[t_n,t_{n+1}]$ ? For $n\geq 1$
$[t_n+\epsilon_n,t_{n+1}-\epsilon_n]$ for any $\epsilon_n < \frac{t_{n+1}-t_n}{2}$ (in other words, there is no fixed $\epsilon$ that works for all $n$
user501743
@Conrad So $\epsilon_n$ is something less than $\frac{t_{n+1}-t_n}{2}$? And this holds for all $n\geq 0$
yes that would work; but note that $\liminf \epsilon_n=0$
user501743
@Conrad So if $0\leq \epsilon_{n}<\frac{t_{n+1}-t_n}{2}$. So as $n\to \infty$, $\epsilon_{n}$ $\to$ 0
user501743
23:06
@Conrad And then can we pass on the limit to arrive from $I_{\epsilon_n}$ to $I$?
user501743
@Conrad Thank you for your patience and for your help. I really admire you for this. Thank you a lot.
what limit? there are infinitely many $\epsilon_n$ so what does it mean to take a limit on them? as noted to prove that $I$ exists and is finite is trivial and doesn't require all these convolutions (truncate it at $N$, show that the remainder is negligible etc - ); the saddle part is to show the Balazard et others closed form and that is also not that deep btw
user501743
@Conrad I mean that $\lim_{n\to \infty} I_{\epsilon_{n}}=I$?
what does that mean? there is no $I_{\epsilon_n}$ but an $I_{(\epsilon_1, \epsilon_2,...\epsilon_n,...})$
user501743
@Conrad What do you mean by$ I_{(\epsilon_1,\epsilon_2,...)}$
23:06
if my understanding is correct you want to excise disjoint intervals around critical zeroes and their lengths depend on $n$; so in other words, the integral obtained would depend on an infinite sequence of different lengths and how you index it is up to you but then it becomes the question of what do you mean by taking a limit (as already at least a subsequence of the lengths go to zero with $n$ for a Fixed choice of the excisions); again not sure why try something like this when stopping at a finite $N$ works in defining $I$ and proving its finiteness and is the customary way done
user501743
@Conrad I need to know that how can we get $I$ from $I_{(\epsilon_1,\epsilon_2,...)}$
have no idea what the question above means; as noted I wouldn't approach $I$ this way since I am not sure how to make sense of a limiting process on infinite sequence
user501743
@Conrad I am asking that how to get the original integral I in terms of $I_(\epsilon_1,\epsilon_2,...)$
have no idea as noted since I do not know (and have no interest to think about) how to make sense of a limiting process of an infinite sequence of an infinite sequence (at least here where other techniques are available to deal with $I$)
user501743
@Conrad Ok. No problem. Thank you so much for your immense help and patience.
23:06
this last part was definitely more interesting and while not that deep, definitely more RZ like good question; local integrability at zeroes and estimates at infinity on the critical line are fairly trivial; still not clear about the purpose of the question
user501743
@Conrad The question was just an observance. You can edit the question so as to make it meaningful.
user501743
@Conrad I have edited the question $$I_{\epsilon}=2\sum_{n=0}^{\infty}\int_{t_n+\frac{\epsilon}{4}}^{t_{n+1}-\frac{\epsilon}{4}} \frac{log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt $$
user501743
@Conrad Please reply. Is now my integral making sense? $t_{n+1}-\frac{\epsilon}{4}> t_n+\frac{\epsilon}{4}$ which also holds when $t_{n+1}-t_n<\epsilon$

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