Are there any results known about the asymptotics/bounds for $$\int_0^T\zeta(\tfrac{1}{2}+it)^4\;dt,$$ where we don't have the absolute value on the inside? One could use the triangle inequality to trivially bound it in comparison with the typical fourth moment, but I expect a great deal of cancellation to take place inside the integral so expect this would be a very weak result?
$\begingroup$
$\endgroup$
New contributor
clare31 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
10
These moments are considerably easier than moments with the absolute value. A contour shift argument implies that the moment you write down is $T + O ( T^{4/6+\epsilon})$.
Indeed, we can represent $\int_0^T \zeta(\frac{1}{2}+ it)^4 dt$ as $\int_0^T \zeta(\sigma+ it)^4 dt$ for some large real $\sigma$ plus an integral between $\frac{1}{2}+ iT $ and $\sigma + i T$ an integral between $\frac{1}{2}$ and $\sigma$ (passing over the point $1$ to avoid picking up a residue).
We have $\int_0^T \zeta(\sigma+ it)^4 dt = T + O(T 2^{-\sigma} )$ by expanding the sum
The integral between $\frac{1}{2} $ and $\sigma$ is $O(\sigma)$ as to the right of the critical strip $\zeta$ is $O(1)$ and the integral through the critical strip is independent of $\sigma$ and $T$.
The integral between $\frac{1}{2}+it$ and $\sigma+ it$ is $O( \sigma )+ O(T^{4/6+\epsilon})$ as to the right of the critical strip $\zeta$ is $O(1)$ and in the critical strip $\zeta$ is $O(T^{1/6+\epsilon})$ by subconvexity.
Choosing $\sigma$ between $\log_2 (T)$ and $T^{4/6}$, we get a total error term of $O(T^{4/6+\epsilon})$.
Moments without the absolute value of $L$ functions at the critical point, on the other hand, can be much more difficult.
-
$\begingroup$ Perfect, thank you! may I ask what your comment on moments without absolute values of $L$-functions at the critical point means? $\endgroup$clare31– clare312025-12-04 16:52:20 +00:00Commented Dec 4 at 16:52
-
$\begingroup$ @clare31 I am thinking of problems like estimating $\sum_{ \chi : (\mathbb Z/p)^\times \to \mathbb C^\times} L(1/2,\chi)^4$ for prime $p$, which is more difficult even though for some purposes $L(1/2,\chi)$ and $\zeta(1/2+it)$ behave similarly. $\endgroup$Will Sawin– Will Sawin2025-12-04 17:34:03 +00:00Commented Dec 4 at 17:34
-
$\begingroup$ Nice answer. Do you have any comments about the case when the integrand also has an $x^s$ factor? $\endgroup$tomos– tomos2025-12-04 17:49:32 +00:00Commented Dec 4 at 17:49
-
$\begingroup$ @tomos Assuming $x$ is a positive real number, for $x>1$ this eliminates the main term and we will get a similar bound on the error term, while for $x<1$ this makes the problem more difficult, and one has to use the usual methods for moments of zeta function involving the approximate functional equation. The interesting case is when $x$ is a power of $T$ between $T^{-4}$ and $T^{0}$. Small negative powers should not be too hard, and the most difficult might be somewhere around $T^{-2}$. $\endgroup$Will Sawin– Will Sawin2025-12-04 17:55:28 +00:00Commented Dec 4 at 17:55
-
1$\begingroup$ Perhaps it is worth to emphasize that one should integrate $\zeta(s)^4\,ds$ instead, in the way you indicate. Then it is true that the directed line segment $[1/2,1/2+iT]$ can be replaced by the path you indicate. If $u(t)$ parametrizes the new path, then $\zeta(s)^4\,ds$ gets replaced by $\zeta(u(t))\,u'(t)\,dt$ when expressing the complex line integral as a complex-valued Riemann integral, etc. The point is that $u'(t)$ varies on the path and needs to be taken into account, especially on the semicircle that goes above the residue $s=1$. All this is inconsequential for the final result. $\endgroup$GH from MO– GH from MO2025-12-04 21:29:06 +00:00Commented 2 days ago
