Is it possible to get a good upper bound for the integral $$\int_{0}^{T}\zeta ^{3}(\frac{1}{3}+it)dt$$ (unconditionally)?
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4$\begingroup$ $=T+O(1)-i\int_{2+iT}^{1/3+iT} \zeta^3(s)ds$, there is the trivial bound $\zeta^3(s)=O(s^3)$, which can be improved a lot with the partial results toward the Lindelof hypothesis extended to the whole critical strip with Hadamard 3 lines theorem. $\endgroup$– reunsCommented Jan 11, 2022 at 4:50
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2$\begingroup$ You can improve it further by writing $\left|\int \zeta^3\right|\leq |\zeta|\cdot \int |\zeta|^2$ and using $L^2$ bounds on $\zeta$ on vertical lines together with subconvexity bounds on $\zeta$. $\endgroup$– H A HelfgottCommented Jan 11, 2022 at 6:37
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2$\begingroup$ Reuns' contour shifting means one is now integrating on horizontal lines rather than vertical lines, so the L^2 mean value theorem is no longer of much use (and applying it before contour shifting would destroy too much cancellation to give a strong bound). $\endgroup$– Terry TaoCommented Jan 11, 2022 at 16:28
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$\begingroup$ @reuns I think you meant to write $O(T)$ instead of $T$. Is it the term corresponding to $\int_0^T \zeta^3 (2+it) dt$? $\endgroup$– Sungjin KimCommented Jan 11, 2022 at 18:30
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2$\begingroup$ @SungjinKim Direct calculation gives $\int_0^T \zeta^3(2+it)\ dt = T + \sum_{n=2}^\infty \frac{d_3(n) (1-n^{-iT})}{i n^2 \log n} = T + O(1)$. $\endgroup$– Terry TaoCommented Jan 11, 2022 at 19:27
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