18
$\begingroup$

What sort of bounds (explicit of preference) can one give for $$\int_T^{2 T} \frac{dt}{|\zeta(1+i t)|^2} \;\;\;\;\;?$$ Some obvious points:

  • One can give a pointwise bound $\frac{1}{|\zeta(1+ it)|} \leq C \log t$ (with $C\leq 42.9$ for $t\geq 2$) and deduce a bound of the form $\leq K T (\log T)^2$ on the integral above. If possible, we would like to do better (asymptotically and also as far as the constants involved are concerned).
  • To know that the integral is always finite, one needs to know that $\zeta(1+it)\ne 0$ for all $t$. Thus, it is not enought to just apply a mean-value theorem.
$\endgroup$
13
$\begingroup$

Problems like this are classical (as noted in Terry's answer), and have been considered more recently with attention to uniformity in the moments. To give a quick indication, one can show that $$ \zeta(1+it) \approx \prod_{p\le y} \Big( 1-\frac{1}{p^{1+it}}\Big)^{-1} $$ for all but a set of values $T\le t\le 2T$ of small measure, provided $y$ is bigger than some power of $\log T$. Such ideas go back to Littlewood (on RH, which can then be removed via zero-density results). See Lemmas 2.1 and 2.2 in Extreme values of $|\zeta(1+it)|$ for example. From this one can work out uniform moments of $\zeta(1+it)$ (positive, negative, complex ...) and understand its distribution. Look in the cited paper, and also work of Lamzouri.

The result is $$ \frac 1T \int_T^{2T} |\zeta(1+it)|^{-2} dt \sim \sum_{n=1}^{\infty} \frac{\mu(n)^2}{n^2} = \frac{\zeta(2)}{\zeta(4)}, $$ as one would expect.

An alternative (which may be useful if one wishes to get explicit results, and avoid zero-density estimates) is to use a (very) long approximation to $1/\zeta(1+it)$. For example, first show that $$ \frac{1}{\zeta(1+it)} \approx \sum_{n\le \exp((\log T)^3)} \frac{\mu(n)}{n^{1+it}}. $$ This would need an explicit zero free region -- I would guess something off-the-shelf from the work of Kadiri et al should be good, and you can make the sum even longer. Usually such a long approximation would be useless, but on the $1$-line you can do whatever you want. Now one can show (easiest to smooth and unsmooth) that $$ \int_T^{2T} \frac{dt}{|\zeta(1+it)|^2} = \sum_{n\le \exp((\log T)^3) } (T+O(n)) \frac{\mu(n)^2}{n^2} = \frac{\zeta(2)}{\zeta(4)} T + O((\log T)^3). $$ It shouldn't be too bad to make this explicit, but not something I'd care to do!

$\endgroup$
3
  • $\begingroup$ Thanks! Is the use of zero-density results essential? $\endgroup$ Sep 10 at 20:58
  • $\begingroup$ Thanks again! Excuse me if I am not seeing something, but why do you need to go as far up as $\exp((\log T)^3)$? $\endgroup$ Sep 11 at 1:02
  • 3
    $\begingroup$ I'm thinking of just using Perron's formula and shifting contours inside the zero free region. At this point $x=\exp((\log T)^3)$, we'd save $x^{-c/\log T}$ which is extremely big. Obviously you can take a smaller stopping point and save less. My point here is that where you stop almost doesn't matter in the mean-square. So go out very far, so that any calculations needed for the approximation become pretty much trivial. $\endgroup$
    – Lucia
    Sep 11 at 1:21
13
$\begingroup$

This expression is $\asymp T$. I will sketch a proof of the upper bound of $O(T)$; the lower bound is proven similarly.

It is convenient to use the Gonek-Hughes-Keating explicit factorisation $$ \zeta(s) = \exp\left( \sum_n \frac{\Lambda(n)}{n^s \log n} \varphi(\frac{\log n}{\log R}) \right) \prod_\rho H( (s-\rho) \log R )$$ for any $R>1$ and bump function $\varphi: {\bf R} \to {\bf R}$ that equals $1$ near the origin, where the product is over non-trivial zeroes $\rho$ and $$ H(s) := \exp\left( \int_0^\infty \varphi'(u) E_1(su)\ du \right)$$ with $E_1(z) := \int_z^\infty \frac{e^{-t}}{t}\ dt$ the exponential integral; see equation (15) of this blog post of mine. This is an efficient way to control $\zeta$ in terms of small primes and nearby zeroes; one can view it is as an interpolant between the Euler product representation of $\zeta$ and the Hadamard factorisation. For our purposes, the only thing we need to know about $H$ is the bound $$ H(s) = \exp\left( O\left( \frac{e^{-c \mathrm{Re} s}}{(1 + |\mathrm{Im}(s)|)^{10}} \right) \right)$$ whenever $|s| \geq 1$ and $\mathrm{Re} s \geq 0$, where $c>0$ is a constant depending only on $\varphi$ (and the implied constant in the $O()$ notation also depends on $\varphi$). See equation (13) of the previously mentioned blog post. Informally, this says that $H$ is very close to $1$ on the right half-plane, as long as one stays away from the origin.

In particular, if $1+it$ stays at a distance at least $1/\log R$ from any zero, we have the estimate $$ \frac{1}{|\zeta(1+it)|^2} = \exp\left( 2\mathrm{Re} \sum_n \frac{\Lambda(n)}{n^s \log n} \varphi(\frac{\log n}{\log R}) + O( \sum_{\rho = \sigma+i\gamma} \frac{e^{-c (1 - \sigma)\log R }}{(1 + |\gamma-t| \log R)^{10}} ) \right).$$ Now, from zero density estimates we know that most of the $\asymp T \log T$ zeroes with imaginary part $\gamma \asymp T$ will lie in the range $\sigma \leq 3/4$ (say); for instance Ingham's estimate ((10.27) in Iwaniec-Kowalski) says that there are only $O( T^{3/5} \log^{O(1)} T)$ exceptions. Thus we can restrict attention to those $1+it$ that stay a distance at least $1/4$ from any zero, since the contribution of the exceptional $t$ is only $O( T^{3/5} \log^{O(1)} T)$ thanks to the uniform bound $1/\zeta(1+it) \ll \log T$ mentioned in the OP. If we now set $R := \exp(\sqrt{\log T})$ (say) then just from the Riemann-von Mangoldt formula we will have $$ \sum_{\rho = \sigma+i\gamma} \frac{e^{-c (1 - \sigma)\log R}}{(1 + |\gamma-t|\log R)^{10}} = o(1)$$ (in fact this expression will be $O( \log T / \exp(c \sqrt{\log T}) )$ for some $c>0$) and so we now have a clean asymptotic $$ \frac{1}{|\zeta(1+it)|^2} = (1+o(1)) \exp(X(t))$$ where $$ X(t) := 2 \mathrm{Re} \sum_n \frac{\Lambda(n)}{n^s \log n} \varphi(\frac{\log n}{\log R}).$$ From the triangle inequality and Mertens' theorem we have $X(t) \ll \log\log R$, but from central limit theorem heuristics we expect $X(t)$ to be normally distributed with mean zero and bounded variance, so the mean value of $\exp(X(t))$ should be bounded (cf., the proof of the Selberg central limit theorem). One can make this rigorous by upper bounding the moments $\int_T^{2T} |X(t)|^{2k}\ dt$ for $k$ up to $O(\log\log R)$ by the usual mean value theorem and then applying Chebyshev's inequality and optimising in $k$ to get Chernoff-type bounds for $X(t)$ which will give the claim.

With a bit more effort one should be able to get an asymptotic for the integral of the form $(c+o(1)) T$ for some explicit absolute constant $c>0$, basically arising from a central limit theorem for $\log |\zeta(1+it)|$ (which, now that I think about it, must already in the literature somewhere, probably in Laurincikas's book on limit theorems for the Riemann zeta function, for instance). Actually this method gives asymptotics for all fractional moments $\int_T^{2T} |\zeta(1+it)|^\alpha\ dt$ for any fixed $\alpha$ (either positive or negative).

$\endgroup$
10
  • 2
    $\begingroup$ That would have been worth noting in the OP then :-). I've just added a remark here that the method described here in fact yields asymptotics for all fractional moments, not just the -2 moment; it seems difficult to replicate that level of generality with a Plancherel-based argument. $\endgroup$
    – Terry Tao
    Sep 10 at 16:22
  • 2
    $\begingroup$ Heuristically, $\zeta(1+it) = \prod_p (1-p^{-1-it})^{-1}$ should behave statistically like $\prod_p (1-p^{-1} e^{2\pi i\theta_p})^{-1}$ for iid random variables $\theta_p \in {\bf R}/{\bf Z}$. So the asymptotic constant $c$ should be $\prod_p \int_0^1 |1-p^{-1} e^{2\pi i\theta}|^{2}\ d\theta$. By Plancherel this simplifies to $\prod_p (1+\frac{1}{p^2}) = \zeta(2)/\zeta(4) = 15/\pi^2$. Perhaps there is a more elementary way to reach this asymptotic. $\endgroup$
    – Terry Tao
    Sep 10 at 17:03
  • 6
    $\begingroup$ I'm sure you are more than qualified to do so, Harald. $\endgroup$
    – Terry Tao
    Sep 10 at 17:15
  • 1
    $\begingroup$ Hah. I really mean that there are methods for which one can see that explicit work should be straightforward, and methods for which explicit work is most likely going to be a nightmare that self-respecting people do not put themselves through. I am still hoping for something in the first category. (People don't think of explicit work as "clean", but in fact it forces one to streamline.) $\endgroup$ Sep 10 at 17:19
  • 6
    $\begingroup$ Sounds like you can take it from here then. (If you were one of my graduate students, this would be the point where I would suggest that you work on it and report back in a week.) $\endgroup$
    – Terry Tao
    Sep 10 at 17:34
7
$\begingroup$

Balasubramanian, Ivić and Ramachandra ("An application of the Hooley-Huxley contour", Acta Arith. 65 (1993), no. 1, 45–51) prove the asymptotic result found by Lucia and Terry Tao. Their error term is $O(\log T)$, so slightly better than Lucia's error, but they make use of zero-density results.

More generally, their equation (1) gives an analogous result for $\int_{1}^{T} |\zeta(1+it)^k|^2 dt$ for any complex $k$. In fact, this is itself a special case of their equation (4), $$\int_{1}^{T} |F(1+it)|^2 dt = T\sum_{n \ge 1}|a_n|^2 n^{-2} + O\left( \log \log T + \sum_{n \le T^C} |a_n|^2 n^{-1}\right)$$ where $F(s) = \sum_{n \ge 1} a_n/n^s$ is a Dirichlet series 'arising' from L-functions (concretely, a product of powers of L-functions, their logarithms and their derivatives) and $C$ is a constant depending only on $F$.

Let me also mention that a quick conditional (on RH) proof of your required result is also given in Titchmarsh (2nd edition, p. 338).

$\endgroup$
2
  • 1
    $\begingroup$ Well, the result from Titchmarsh's book is stronger but requires RH (as the chapter title suggests) $\endgroup$
    – TravorLZH
    Nov 7 at 17:45
  • $\begingroup$ @TravorLiu Indeed, I've missed that, thanks. Answer is now modified. $\endgroup$ Nov 7 at 19:06
4
$\begingroup$

While @Lucia's answer is my favorite, I thought it might be worthwhile to sketch the Plancherel-based argument I alluded to in the above.

First of all, let $\phi:[0,\infty)\to \mathbb{R}$ be such that $\phi(t)$ has fast decay as $t\to \infty$ and fast enough decay as $t\to 0^+$, where "fast enough" means "fast enough for the Mellin transform $\Phi(s)$ of $\phi$ to be defined for $\Re s\geq 0$." Assume as well that $\Phi(s)$ has fast decay as $|\Im s|\to \infty$, and that $\Phi(i t)$ does not vanish on $[1,2]$. It will be enough to show that $$\int_{1-i\infty}^{1+i\infty} \frac{|\Phi(s/T)|^2}{|\zeta(s)|^2} ds = O(T),$$ as the left side is bounded from below by $\gg \int_T^{2 T} \frac{dt}{|\zeta(1+ i t)|^2}$.

(It seems likely that one can obtain a bound of $\gg T$ on the original integral in the same way, but it should also be clear that this method will not yield an asymptotic for the original problem; at least I do not see how it could. At the same time, just carrying out the argument carefully probably gives an asymptotic $c T + o(T)$ rather than just an $O(T)$ in the above statement.)

We know that $\Phi(s/T)$ is the Mellin transform of $T \phi(x^T)$, and so $n^{-s} \Phi(s/T)$ is the Mellin transform of $x\mapsto T \phi((n x)^T)$. Thus, $\frac{\Phi(s/T)}{\zeta(s)}$ is the Mellin transform of $T F(x)$, where $F(x) = \sum_{n} \mu(n) \phi((n x)^T)$. By Plancherel, $$\frac{1}{2 \pi} \int_{1-i \infty}^{1+i \infty} \frac{|\Phi(s/T)|^2}{|\zeta(s)|^2} ds = T^2 \int_0^\infty |F(x)|^2 x dx.$$

We see that $F(x)$ is a smoothed sum of $\mu(n)$, essentially supported on an interval $I_T(x) = ((1-O(1)/T)/x,(1+O(1)/T)/x)$. For $x\gg 1$, $F(x)$ will be negligible (and in fact $0$, if $\phi$ is of compact support. For $1/K(T)\leq x \leq 1/T$ (where $K(T)$ will be chosen later), we use a trivial bound $|F(x)| = O(1/(T x))$, and obtain $$T^2 \int_{1/K(T)}^{1/T} |F(x)|^2 x dx \ll \int_{1/K(T)}^{1/T} \frac{dx}{x} = \log \frac{K(T)}{T}.$$ For $1/T\leq x\leq 1$, $F(x)$ will be non-negligible (and bounded by $|\phi|_\infty$) only a proportion $O(1/x T)$ of the time (whenever $I_T(x)$ happens to contain an integer), and so it should be simple to show that $$T^2 \int_{1/T}^{1} |F(x)|^2 x dx \ll T\int_{1/T}^1 dx = T,$$ which will be our main term. (We can probably get an asymptotic here without much work.)

It remains to consider $F(x)$ for $0<x\leq 1/K(T)$. We want $K(T)$ to be large enough for us to be able to get cancellation on a (smoothed) sum of $\mu(n)$ on an interval of length in the order of $K(T)/T$ centered around $T$. If we are willing to use short-interval results (which are of course completely fine, but rest on zero-free regions, at least if one wants to gain a factor of more than $\log K(T)$, as we need to) then we can set $K(T) = T^4$ (say). If we just want to use a result of Hadamard-de la Vallée-Poussin strength, then we can set $K(T) = \exp(C (\log T)^2)$, and obtain $$T^2 \int_0^{1/K(T)} |F(x)|^2 x dx = \int_0^{1/K(T)} \frac{1}{e^{c \sqrt{\log(1/x)}}} \frac{dx}{x} \ll 1.$$

Summing up, we get a bound of $O(T)$.

(In the last step, can also use weaker results than what follows from a classical zero-free region. For instance, there are explicit bounds on smooth sums of length $y$ that gain a factor of $C' (\log y)^2$ ($C'$ a constant, often large) with respect to the trivial bound. That is enough; we just need $K(T)$ to be smaller than $T$ (so that $\log K(T)/T$ does not overwhelm $T$ and we also need the gain over the trivial bound to be at least a little larger than $(\log y)$ (so that the integral $\int_0^{1/K(T)} |F(x)|^2 x dx$ converges).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.