3
$\begingroup$

Denote by $\zeta$ the Riemann zeta function.

It is known that

$$\int_{0}^{T} \zeta(1/2 + it) \mathrm{d}t = T + O(T^{1/2}).$$

But is a similar result for $\int_{0}^{T} \zeta(\sigma + it) \mathrm{d}t$, where $0<\sigma<1$ also known ?

Improve this question
$\endgroup$
3
  • $\begingroup$ Should the integrand be the absolute value of zeta rather than zeta itself? $\endgroup$
    – Yemon Choi
    Commented Dec 1, 2018 at 21:03
  • $\begingroup$ @Yemon Choi, the integrand is indeed zeta itself. $\endgroup$
    – sigma
    Commented Dec 1, 2018 at 21:07
  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Dec 2, 2018 at 15:30

3 Answers 3

Reset to default
7
$\begingroup$

We have, for any $0<\sigma<1$, $$\int_0^T\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma}).$$ To see this, it suffices to show that $$\int_T^{2T}\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma}).$$ By Theorem 4.11 in Titchmarsh: The theory of the Riemann zeta-function, we have $$\zeta(\sigma+it)=\sum_{n\leq T}n^{-\sigma-it} + O_\sigma(T^{-\sigma}),\qquad T\leq t\leq 2T,$$ therefore by explicit integration $$\int_T^{2T}\zeta(\sigma+it)\,dt = T + \sum_{2\leq n\leq T}O(n^{-\sigma}) +O_\sigma(T^{1-\sigma}) = T+O_\sigma(T^{1-\sigma}).$$ We are done.

Improve this answer
$\endgroup$
1
$\begingroup$

For $1/2<\sigma<1$, the same result should be known, and indeed should follow from the same proof: $\zeta(s)$ can be approximated by a suitably chosen truncation of its Dirichlet series (despite that series not converging inside the critical strip), and integrating term by term yields $T$ from the $1^{-s}$ term and $o(T)$ from all the $n^{-s}$ terms with $n\ge2$. (I don't know whether this method yields the strong error term cited, but the idea is worth sharing anyway: the asymptotic comes from the constant term in the Diriclet series.)

For $0<\sigma<1/2$, the answer will probably be different: for example, on the Lindelöf hypothesis, $|\zeta(\sigma+it)| \asymp t^{1/2-\sigma}$. So the integral will presumably have size around $T^{3/2-\sigma}$. I'm sure this is known (but not by me at this moment...).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ If we have some aproximation $\zeta(\sigma + it) = \sum_n V(n) n^{-s}$ for smooth weight function $V$, the non-constant terms contribute $\sum_n V(n) i (n^{iT} - 1) /( n^\sigma \log n)$. If I remember correctly, $V$ should have height about one and width about $T$, which would make the nonconstant terms contribute at most $O ( T^{1 - \sigma})$. The main term is still $T$. I don't think Lindelöf is relevant. $\endgroup$
    – Will Sawin
    Commented Dec 2, 2018 at 1:08
-2
$\begingroup$

What you say is incorrect. It is known that $$ \int_{0}^{T} \zeta(\tfrac 12 + it) dt = o(T). $$ Can you please correct the question, and ask again?

Improve this answer
$\endgroup$
4
  • $\begingroup$ Im so curious now. Can you give a reference for this claim? $\endgroup$
    – BigM
    Commented Dec 1, 2018 at 19:43
  • 2
    $\begingroup$ Well, i hadn't put a refernce because i assumed the result is fairly famous. Indeed, what i wrote in my question is known, see Montgomery-Vaughan, p. 458 $\endgroup$
    – sigma
    Commented Dec 1, 2018 at 20:01
  • 1
    $\begingroup$ @wqeqwe, it is not true that $\int_{0}^{T} \zeta(1/2 + it) \mathrm{d}t =o(T)$. See Montgomery-Vaughan's ''Multiplicative Number Theory: The Classical Theory'', p.458. $\endgroup$
    – sigma
    Commented Dec 1, 2018 at 21:11
  • $\begingroup$ wqeqwe: Lemma 14.10 in the book @sigma refers to is not consistent with your claim. (The main step in the proof establishes that $\int_1^T \zeta(1/2+it)\,dt = \int_1^T \zeta(2+it)\,dt + O(T^{1/2})$ and we can estimate the second integral easily using the Dirichlet series) $\endgroup$
    – Yemon Choi
    Commented Dec 1, 2018 at 23:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .