9
$\begingroup$

Let $\alpha,\beta\in (0,1\rbrack$, $\alpha\ne \beta$. I wish to estimate $$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta}.$$ There is an obvious approach, namely, to estimate the inner sum first (the second- and third-order terms will be proportional to $\zeta(\beta)$ and $\zeta'(\beta)$; there is a connection with Ramanujan summation) and then input that estimate into the outer sum, which gets estimated in much the same way.

An improved version of the approach consists essentially in putting the longer of the two sums always inside, by means of a bit of combinatorial manipulation.

I have to wonder - is there a more elegant, less piecemeal approach, considering both sums in one go? If $\alpha$ were equal to $\beta$, the answer would be yes - we would get an estimate based on $(\zeta^2)'(s)$. Of course, that's precisely the case we aren't in.

(You may assume $\alpha = 2 \beta$, since that is the particular case I am most interested in.)

Update: what I get by a careful version of the above (continuous partition) and some pain is that the double sum equals $$\begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned}$$ plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$, where the implied constant is explicit (and fairly small). I do not know whether the error term ought to be of a lower order of magnitude. Let me see what I can do with Lucia's approach below.

$\endgroup$
9
  • 6
    $\begingroup$ Why not just start with $\frac{1}{2\pi i} \int_{(c)} \zeta(s+\alpha)\zeta(s+\beta)x^s/s^2 ds$ and compute residues? $\endgroup$ – Lucia Oct 31 '17 at 18:05
  • 2
    $\begingroup$ Indeed, this works nicely (and just helped me double-check the formula above). The error term one gets in this way is better than $O(x^{-\beta})$ (for $\beta<\alpha$). Indeed, it is $O_\epsilon\left(x^{-(2\beta/3+\alpha/3)}\right)$ (without using subconvexity; I'm using the bounds on zeta(s) in section 5.1 of Titchmarsh). Now, I need an explicit constant. I'll have to check what explicit bounds on $\zeta(s)$ I can find. My fear is that the constants obtained this way may not be good (unless one would be willing to look for cancellation, but that would most likely be a difficult mess). $\endgroup$ – H A Helfgott Nov 1 '17 at 0:06
  • 1
    $\begingroup$ It seems to me that the simplest way to estimate such an integral is to use Cauchy-Schwarz and then estimate $\frac{1}{2\pi i} \int_{(c)} \frac{|\zeta(s+\alpha)|^2}{|s|^2} ds$ using the fact that the Mellin transform is an isometry. $\endgroup$ – H A Helfgott Nov 1 '17 at 1:05
  • $\begingroup$ That sounds good to me! $\endgroup$ – Lucia Nov 1 '17 at 2:02
  • $\begingroup$ Of course this is not so straightforward, since the fact that the Mellin transform is an isometry is immediately relevant only for $c>1$. $\endgroup$ – H A Helfgott Nov 1 '17 at 7:52
3
$\begingroup$

Let me carry out matters using a complex-analytical approach, as Lucia suggests, and then say where the difficulty lies.

Let $0<\beta<\alpha\leq 1$. First of all, as Lucia says, $$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta} = \frac{1}{2πi} \int_{c-i\infty}^{c+i\infty} \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds$$ for $c>1$. We shift the contour of integration to the left of $\Re(s)=0$, picking up the main terms $$\begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned}$$ plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$ along the way. We are left with the task of estimating an error term $$\frac{1}{2πi} \int_R \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds,$$ where the integral is over a contour $R$ of our choice going from $-r-i\infty$ to $-r+i\infty$, say, and satisfying $\Re s\leq -r$ at all points. The error will be clearly bounded by $O(K x^{-r})$, where $$K = \frac{1}{2πi} \int_R \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds.$$ The problem does reduces to estimating $K$.

Now, there are rigorous-numerics packages that include integration and the possibility to compute the zeta function $\zeta(s)$. (I currently use ARB.) However, (a) computations must obviously be finite (at least assuming mortal mathematicians), and (b) computing $\zeta(s)$ is never a walk in the park, and rigorous integration only adds to the overhead. Integrating an expression such as above from $-1/2 - i T$ to $1/2 + i T$ takes 15 minutes for $T = 10000$ (says a better programmer than I), but we should not expect to go much further than $T = 100000$ programming casually on our laptops.

The problem that remains, then, is how to bound a tail $$\frac{1}{2\pi i} \left(\int_{-r-i\infty}^{-r-i T} + \int_{-r+i T}^{-r + i \infty} \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds\right).$$

The most obvious approach is to use Backlund's explicit bounds (1918) on $\zeta(\sigma + it)$ (see http://iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html#Size). They are of the quality $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{(1-\sigma)/2} \log t$$ for $0\leq \sigma\leq 1$ and $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{1/2-\sigma} \log t$$ for $-1/2\leq \sigma\leq 0$. The problem here is that convergence is painfully slow. If, say, $\alpha = 1$, $\beta=1/2$ and $r =-1/4$ (reasonable values all around), the tails will be bounded by a constant times $(\log T)^2/\sqrt{T}$. For $T=10000$, $(\log T)^2/\sqrt{T} > 0.848\dotsc$ - not exactly small; for $T=100000$, the same equals $0.419\dotsc$ - barely an improvement.

Notice, however, that why Backlund's bounds are essentially tight for $\Re s<0$, that is not the case for $0<\Re(s)<1$. Of course, they are convexity bounds, so improving on them explicitly would amount to translating into explicit terms rather non-trivial material. However, as long as we are satisfied with $r>-\beta$, what we can do instead is give $L_2$ bounds for the tails, that is, bound $$\int_{r-i \infty}^{r-i T} \frac{|\zeta(s)|^2}{|s|^2} ds + \int_{r+iT}^{r+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$ (The integral $\int_{r-i \infty}^{r-i T}$ is obviously the same.) Then we use Cauchy-Schwarz to bound the tail of the integral we were discussing.

This is non-trivial, and takes us further afield, so I will make it into a separate question: $L_2$ bounds for tails of $\zeta(s)$ on a vertical line .

$\endgroup$
1

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.