I am trying to understand the behaviour of $$\int^\infty_{-\infty}\frac{\xi(1-it)}{\xi(1+it)}h(t)\frac{dt}{t}$$ where $h$ is a Schwartz function on $\mathbb R$, and $\xi(s)$ the completed Riemann zeta function. Clearly it is the quotient of zeta functions that is the most difficult to study.

One knows certain things about $\zeta(1+it)$, for example the pole at $t=0$ and the nonvanishing for all $t$. (See this question, Also the paper referenced in the answer to this question.) But what can we say about this quotient, at least on average, say?


1 Answer 1


For zeta itself, there is a clear result, that eventually the argument of zeta on the line $\Re(s)=1$ becomes very regular. Some more sophisticated things around this are in

D. Hejhal, "On a result of G. Polya concerning the Riemann $\xi$-function", J. D'analyse Math. 55 (1990), 60-95.

Also, (5.14) of that paper recalls the consequence $$ \log \zeta(1+iu) - \log \zeta(1+it) \;=\; O\Big({\log t \over \log\log t}\Big)\cdot (u-t) $$ for $u\ge t$ from [Titchmarsh~1986]. In an earlier edition, this was on page 98. In the edition revised by Heath-Brown, this is (5,17.4) on page 112.

For more generaly automorphic $L$-functions, somewhere in Iwaniec-Kowalski they show that similar results follow for self-adjoint (data for) $L$-functions assuming sufficient GRH's ... but/and it is not clear to me whether/how the delicate-but-elementary arguments for zeta and maybe Dirichlet $L$-functions over $\mathbb Q$ would generalize...

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    $\begingroup$ +1 Thank you for the very nice reference! $\endgroup$
    – Tian An
    Sep 25, 2015 at 23:41
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    $\begingroup$ In what sense use you the word regular? "... the argument of zeta of the line $\sigma=1$ becomes very regular". Of course it is a real analytic function. But, for $0<t<T$ there are $cT$ points where it is $=0$. But it is not bounded. The probability to be greater than $2\pi$ being very small, to be $\ge 4\pi$ almost incredibly small, and so on. I will not call this behavior "regular". $\endgroup$
    – juan
    Sep 26, 2015 at 10:30
  • $\begingroup$ @juan You mean $\sigma=\frac12$ I believe? $\endgroup$ Sep 26, 2015 at 11:57
  • $\begingroup$ @ მამუკა ჯიბლაძე @paul garrett No, I mean $\sigma=1$, I speak of the function $\arg \zeta(1+i t)$. This is real analytic vanishes at $c T$ points (think on the x-ray, but this can be proved ). Also we know how to compute the probabilities I speak about. $\endgroup$
    – juan
    Sep 26, 2015 at 12:12
  • $\begingroup$ Oh sorry I just did not notice the word "argument" :D $\endgroup$ Sep 26, 2015 at 12:22

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