# $L_2$ bounds for tails of $\zeta(s)$ on a vertical line

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)?

• I'm afraid that elementary approach gives something worse than convexity. – H A Helfgott Nov 4 '17 at 11:48
• But I still get my green, right? – GH from MO Nov 4 '17 at 17:46
• 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. – H A Helfgott Nov 4 '17 at 17:55

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 $\sigma$.)

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.

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