8
$\begingroup$

Let $0<\sigma\leq 1$. Let $T$ be large. How can we give good explicit $L^2$ bounds on the tails of $\zeta(\sigma+it)$? That is, we want to bound the quantity $$\int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$ Morally, we should expect something about as good as $1/T$ for $\Re(s)\geq 1/2$, and presumably $1/T^{2\sigma}$ for $0<\Re(s)\leq 1/2$ (or maybe I ate a $\log$). In particular, we should certainly expect to do better than what we can get from convexity bounds on individual values of $\zeta(s)$. (Such bounds are, incidentally, explicit (Backlund)).


How to go about this? One inviting approach is to use the fact that the Mellin transform is an isometry. (That is, if $F$ is the Mellin transform of $f$, then $\frac{1}{2\pi} \int_{-\infty} |F(\sigma+it)|^2 dt=\int_0^\infty |f(x)|^2 x^ {2\sigma-1}$.) Recall that $\zeta(s)/s$ is the Mellin transform of $x\mapsto \lfloor 1/x\rfloor$. An obvious (even crass) sub-approach would be to let $$g(x) = \frac{1}{2 \pi i} \int_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} x^{-s} ds,$$ and then note that the Mellin transform of $\lfloor 1/x\rfloor - g(x)$, evaluated at $\sigma+it$, equals $0$ when $|t|< T$ and $\zeta(s)/s$ when $|t|>T$. (I'm papering over the fact that we should check that there is no pole at $s=1$, so that we can analytically continue the Mellin transform to the left of $\Re s=1$.) It follows, by isometry, the quantity in the first displayed equation above (the integral over the tails) equals $$\int_0^\infty |\lfloor 1/x\rfloor - g(x)|\cdot x^{2\sigma-1} dx.$$ We can write down $g(x)$ explicitly, though it is not completely trivial; if I understand correctly (see Theorem 5.2 in Montgomery-Vaughan), $$g(x) = \frac{1}{\pi} \sum_{n\geq 1} \mathrm{si}(T \log n x),$$ where $\mathrm{si}$ is the sine integral function. Of course the problem here is that $g(x)$ is not a particularly pleasant function to work with.

My instinct is to work instead with a function of the form $$\frac{1}{2\pi i} \int_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} \phi(s) x^{-s} ds,$$ where $\phi(s)$ is close to $0$ for $s = \sigma+i t$, $|t|\geq T$, and is close to $1$ on most of the segment $\lbrack \sigma-iT,\sigma+iT\rbrack$. I haven't hit on the right choice of weight $\phi$, though. It should give some sort of advantage over $g$ above, or else it is pointless.

Any other approaches? Is this all known (explicitly)?

$\endgroup$
  • $\begingroup$ I'm afraid that elementary approach gives something worse than convexity. $\endgroup$ – H A Helfgott Nov 4 '17 at 11:48
  • $\begingroup$ But I still get my green, right? $\endgroup$ – GH from MO Nov 4 '17 at 17:46
  • 1
    $\begingroup$ Yes! I just did! I just hope this will not halt the discussion. If it does, I might take it back and invite you for ice cream instead. $\endgroup$ – H A Helfgott Nov 4 '17 at 17:55
9
$\begingroup$

Let $\sigma>0$ be fixed, and let $T\geq 2$ be a parameter.

By Theorem 4.11 in Titchmarsh: The theory of the Riemann zeta-function, $$ \zeta(\sigma+it)=\sum_{n\leq T}n^{-\sigma-it}+O(T^{-\sigma}),\qquad T<|t|\leq 2T.$$ It follows that $$ \int_{\sigma+iT}^{\sigma+2T}|\zeta(s)|^2\,ds\ll T^{1-2\sigma}+\int_T^{2T}\left|\sum_{n\leq T}n^{-\sigma-it}\right|^2\,dt.$$ We estimate the integral on the right hand side by Theorem 9.1 in Iwaniec-Kowalski: Analytic number theory. We obtain $$ \int_{\sigma+iT}^{\sigma+2iT}|\zeta(s)|^2\,ds\ll T\sum_{n\leq T}n^{-2\sigma}\ll\begin{cases}T,&\sigma>1/2;\\T\log T,&\sigma=1/2;\\T^{2-2\sigma},&\sigma<1/2.\end{cases}$$ Finally, applying a dyadic decomposition in the original integral and using the last bound, we conclude $$ \int_{\sigma+iT}^{\sigma+i\infty}\frac{|\zeta(s)|^2}{|s|^2}\,ds\ll\begin{cases}T^{-1},&\sigma>1/2;\\T^{-1}\log T,&\sigma=1/2;\\T^{-2\sigma},&\sigma<1/2.\end{cases}$$ As you expected. (Note: implied constants depend on $\sigma$.)

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Thanks, GH! Let me have another go. I think the following is the right way to go about things, at least if one wants something self-contained and with good, explicit constants. (The latter more or less implies the former, given that almost all of the literature is non-explicit.)

We want to estimate $$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i \infty} \left|\frac{1}{s} -G(s)\right|^2 |\zeta(s)|^2 ds,$$ where $G(s)$ is the Mellin transform of a well-chosen function $g:\lbrack 0,\infty)\to \mathbb{R}$. It is easy to see that $G(s) \zeta(s)$ is the Mellin transform of $x\mapsto \sum_n g(n x)$. We will choose $g$ so that (a) $G(\sigma+it)$ is small for $|t|\geq T$, (b) the "physical-space" estimation we are about to do is easy.

By Plancherel, $$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i \infty} |1-G(s)|^2 \frac{|\zeta(s)|^2}{|s|^2} ds = \int_0^\infty |h(x)|^2 x^{2\sigma-1} dx,$$ where $$h(x) = \lfloor 1/x\rfloor - \sum_n g(n x).$$ (We will make sure that $G(1)=1$, so that there is no pole at $s=1$; in this way, the equation above will hold for $\Re(s)>0$, and not just for $\Re(s)>1$.)

First, let us show that $h(x)$ is bounded for all $x$. (This part is the same as what I had before.) By second-order Euler-Maclaurin, $$\sum_n g(n x) = \frac{1}{x} \int_0^\infty g(t) dt - \frac{g(0)}{2} - \frac{g'(0)}{12} + \textrm{err},$$ where $|\textrm{err}| \leq \frac{x}{24} \int_0^\infty |g''(t)| dt$. We will work with $g$ such that $g(0)=1$, $g'(0)=0$ and $\int_0^\infty g(t) dt = 1$. Then $$|h(x)| = \left|\lfloor 1/x\rfloor - \left(\frac{1}{x} - \frac{1}{2} + \text{err}\right)\right|\leq \frac{1}{2} + \textrm{err}$$ for all $x$.

We will now choose $g$ so that we can give a much better estimate for $x$ not too small. For starters, we will have $g(x)=1$ for $x\leq 1-\delta$ and $g(x)=0$ for $x\geq 1+\delta$. Then $$h(x) = 0$$ unless $n x \in \lbrack 1-\delta,1+\delta\rbrack$. Moreover, for each $x>2\delta$, there is at most one $n$ such that $n x$ is in that interval, since $\frac{1+\delta}{x} - \frac{1-\delta}{x} = \frac{2 \delta}{x} < 1$. Hence $$\begin{aligned}\int_0^\infty |h(x)|^2 x^{2\sigma-1} dx &\leq \int_0^{2\delta} c^2 x^{2\sigma-1} dx + \sum_{n\leq \frac{1}{2 \delta}} \int_{\frac{1-\delta}{n}}^{\frac{1}{n}} |1-g(n x)|^2 x^{2\sigma-1} dx \\ &+ \sum_{n\leq \frac{1+\delta}{2\delta}} \int_{\frac{1}{n}}^{\frac{1+\delta}{n}} |g(n x)|^2 x^{2\sigma-1} dx,\end{aligned}$$ where $c =1/2 + \frac{2\delta}{24} \int_0^\infty |g''(t)| dt$. Obviously $$\int_0^{2\delta} c^2 x^{2\sigma-1} dx = c^2 \frac{(2\delta)^{2\sigma}}{2\sigma}.$$ To estimate the two other integrals, we have to choose a convenient $g$ obeying our conditions. I will simply take $$g(x) = \begin{cases} 1 &\text{if $x\leq 1 - \delta$,}\\ \frac{1+\delta-x}{2\delta} &\text{if $1-\delta <x<1+\delta$,}\\0 &\text{if $x\geq 1 + \delta$.}\end{cases}$$ Hence $$\begin{aligned} \int_{\frac{1-\delta}{n}}^{\frac{1}{n}} |1-g(n x)|^2 x^{2\sigma-1} dx &+ \int_{\frac{1}{n}}^{\frac{1+\delta}{n}} |g(n x)|^2 x^{2\sigma-1} dx\\ &\leq \frac{2 \eta}{n^{2\sigma}} \int_1^{1+\delta} \left(\frac{1+\delta-t}{2\delta}\right)^2 dt = \frac{\delta \eta}{6 n^{2\sigma}},\end{aligned}$$ where $\eta=1$ if $1/2\leq \sigma\leq 1$ and $\eta = \frac{(1-\delta)^{2\sigma-1} + (1+\delta)^{2\sigma-1}}{2}$ if $0<\sigma\leq 1/2$. Now, for any $y\geq 0$, $$\sum_{n\leq y} \frac{1}{n^{2\sigma}} \leq \begin{cases} \zeta(2 \sigma)& \text{if $\sigma>1/2$,}\\ \log(2 y + 1) &\text{if $\sigma=1/2$,}\\ \frac{y^{1-2\sigma}}{1-2\sigma} & \text{if $\sigma<1/2$.}\end{cases}$$ Taking totals, we conclude that $$\int_0^\infty |h(x)|^2 x^{2\sigma-1} dx \leq c^2 \frac{(2\delta)^{2\sigma}}{2\sigma} + \begin{cases} \frac{\zeta(2\sigma)}{6} \delta& \text{if $\sigma>1/2$,}\\ \frac{\delta}{6} \cdot \log\left(\frac{1}{\delta} + 2\right) &\text{if $\sigma=1/2$,}\\ \frac{\eta (1+\delta)^{1-2\sigma}}{6\cdot 2^{1-2\sigma} (1-2\sigma)} \cdot \delta^{2\sigma} & \text{if $\sigma<1/2$.}\end{cases}$$

Now it just remains to estimate how much of the tail we captured. A quick calculation shows that we then have $$G(s) = \frac{(1+\delta)^{s+1} - (1-\delta)^{s+1}}{2\delta (s+1) s}.$$ Not unexpectedly, this is close to $1/s$ for $\Im(s)$ small. Even more to the point, for any $s$, $$|G(s)| \leq \frac{(1+\delta)+(1-\delta)}{2\delta |s+1| |s|} = \frac{1}{\delta |s+1| |s|},$$ and so $$|1-G(s) s|^2 \geq \left(1-\frac{1}{\delta |s+1|}\right)^2 \geq \left(1 - \frac{1}{\delta T}\right)^2$$ for every $s=\sigma+it$ with $|t|\geq T$. Hence $$\int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds \leq \frac{1}{\left(1 - \frac{1}{\delta T}\right)^2} \int_0^\infty |h(x)|^2 x^{2\sigma-1} dx .$$ Minimizing what will be the main term, we choose $\delta = 3/T$ if $\sigma\geq 1/2$, and $\delta = \frac{1+1/\sigma}{T}$ if $0<\sigma<1/2$. Then $\frac{1}{\left(1 - \frac{1}{\delta T}\right)^2}$ equals $9/4$ for $\sigma\geq 1/2$ and $(1+\sigma)^2$ for $0<\sigma<1/2$.

We conclude that $$\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma-iT} + \int_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds \leq \begin{cases} \frac{9 \zeta(2\sigma)}{8} \cdot \frac{1}{T} + \frac{9 c^2 6^{2\sigma}}{8\sigma} \cdot \frac{1}{T^{2\sigma}} & \text{if $\sigma>1/2$,}\\ \frac{9}{8} \frac{\log\left(\frac{T}{3}+2\right)+12 c^2}{T} &\text{if $\sigma=1/2$,}\\ \left(\frac{ \frac{1}{2} \left(\frac{1+\delta}{1-\delta}\right)^{1-2\sigma} + \frac{1}{2}}{6\cdot 2^{1-2\sigma} (1-2\sigma)} + c^2 \frac{2^{2\sigma}}{2\sigma}\right)\cdot \frac{(1+\sigma)^{2\sigma+2}}{\sigma^{2\sigma}}\cdot \frac{1}{T^{2\sigma}} & \text{if $\sigma<1/2$.}\end{cases}$$ Oh, by the way, $\int_0^\infty |g''(x)| = 1/\delta$, so $c = 7/12$.

No doubt the constants can still be improved, and any suggestions on how to simplify things further are very welcome. I think I can be happy now, though.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The coefficient for $\sigma<1/2$ looks like it can be simplified. Is there an easy way to do this? $\endgroup$ – H A Helfgott Nov 5 '17 at 21:16
  • 1
    $\begingroup$ Yes - once one does everything cleanly, the unpleasant bits in the coefficient for $\sigma<1/2$ magically disappear. $\endgroup$ – H A Helfgott Nov 8 '17 at 14:41

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.