3
$\begingroup$

I'm self-studying analytic number theory from terry tao's blog, there is an exercise (Exercise 33) from the blog that I cannot solve:

Let ${\eta: {\bf R} \rightarrow {\bf C}}$ be a smooth function such that ${\eta(x)}$ vanishes for ${x>C}$ and equals ${1}$ for ${x<-C}$ for some constant ${C>0}$. Let ${s = \sigma+it}$ be in the critical strip. Show that $$ \zeta(s) = \sum_n \frac{1}{n^s} \eta( \log n - \log x) - x^{1-s} \int_{\bf R} e^{(1-s) u} \eta(u)\ du + O_{\eta, A, \sigma}( x^{-A} )\ \ \ \ \ (1)$$ for any ${A>0}$, if one has ${x \geq C' (1+|t|)}$ for some sufficiently large ${C'}$ depending on ${C}$. (Hint: use Lemma 5 of Notes 1, Exercise 31, Lemma 32, and dyadic decomposition.) Conclude in particular that $$\displaystyle \zeta(s) = \sum_n \frac{1}{n^s} \eta( \log n - \log x) + O_{\eta,\sigma,A}( (1+|t|)^{-A} ) \ \ \ \ \ (2)$$ for any ${A>0}$, if ${C'(1+|t|) \leq x \ll (1+|t|)}$.

First note that if we set $y=e^ux$, then by change of variables formula we have $$x^{1-s}\int_{{\bf R}}e^{(1-s)u}\eta(u)\,du=\int_0^\infty \frac{1}{y^s}\eta(\log y-\log x)\,dy$$ Using this and the assumption that $\eta(x)$ vanishes for $x>C$ and equals $1$ for $x<-C$, We can write the main term of the right-hand side of (1) as $$\sum_{n\leq e^{-C}x}\frac{1}{n^s}-\frac{(e^{-C}x)^{1-s}}{1-s}+\sum_{e^{-C}x<n\leq e^C x} \frac{1}{n^s}\eta(\log n-\log x)-\int_{e^{-c}x}^{e^C x}\frac{1}{y^s}\eta(\log y-\log x)\,dy$$ From the integral test $\sum_{x\leq n\leq y}f(n)=\int_x^y f(t)\,dt+O(\int_x^y|f'(t)|\,dt+f(x))$, we see that the above expression is equal to $\zeta(s)+O(\frac{|s|+1}{\sigma}x^{-s})$, thus we have $$\zeta(s) = \sum_n \frac{1}{n^s} \eta( \log n - \log x) - x^{1-s} \int_{\bf R} e^{(1-s) u} \eta(u)\ du + O(\frac{|s|+1}{\sigma}x^{-s})$$ I don't know how to cut down the error to $x^{-A}$ for any $A>0$. I think the hint may be helpful, but I don't how to use it. I don't see how can we deduce (2) from (1)? Thanks for any help

$\endgroup$
  • $\begingroup$ are you supposed to use that for $Re(s) \in (0,1)$ : $\zeta(s) = \lim_{N \to \infty} \sum_{n=1}^N n^{-s} - \int_1^N x^{-s} dx$ ? (or $1- \int_1^N x^{-s} dx = \frac{N^{1-s}}{s-1}$ not sure) $\endgroup$ – reuns Jun 8 '16 at 8:29
  • $\begingroup$ and your first formula diverges, you have to replace $\sum_n$ by $\sum_{n = 1}^N$ $\endgroup$ – reuns Jun 8 '16 at 8:35
  • $\begingroup$ @user1952009 Yes, $s$ is in the critical strip $\{s\in\mathbf{C}:0<\mathrm{Re}(s)<1\}$. Note that $\eta$ vanishes for $x>C$, so the sum is actually a finite sum. $\endgroup$ – Xiang Yu Jun 8 '16 at 13:14
  • $\begingroup$ ok, it makes sense. then as I wrote it is only a matter of proving the analytic continuation of $\zeta(s)$ in the critical strip : $\zeta(s) = s\int_1^\infty \lfloor x \rfloor x^{-s-1} dx$ for $Re(s) > 1$, hence $\zeta(s) = \frac{s}{s-1} + s\int_1^\infty (x-\lfloor x \rfloor) x^{-s-1} dx$ for $Re(s) > 0, s \ne 1$. see en.wikipedia.org/wiki/Abel%27s_summation_formula or directly integrate by parts $\sum_{n=1}^N n^{-s} = \int_{1-\epsilon}^{N+\epsilon} (\sum_n \delta(x-n)) x^{-s} dx$ $\endgroup$ – reuns Jun 8 '16 at 13:22
  • $\begingroup$ you can also prove directly, with $f_N(s) = - \frac{N^{1-s}}{1-s} +\sum_{n=1}^N n^{-s}$ we have $\frac{(s-1)}{(s+1)^2}\zeta(s) =\lim_{N \to \infty} \frac{(s-1)}{(s+1)^2} f_N(s)$ uniformly on $Re(s) > 1$, and since the RHS is analytic and converges uniformly on $Re(s) > \epsilon > 0$, it has to be the analytic continuation of $\zeta(s) (s-1)/(s+1)^2$, so that after all $\zeta(s) = \lim_{N \to \infty} f_N(s)$ in the critical strip $\endgroup$ – reuns Jun 8 '16 at 13:27

Your Answer

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

Browse other questions tagged or ask your own question.