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Is it possible to get an upper bound better than $\ll_\sigma T^{3/2-\sigma}$ for $$\int_{0}^{T}|\zeta (\sigma +it)|\,dt,\qquad 0<\sigma<1/2\,?$$

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    $\begingroup$ No. By Stirling's formula and the functional equation $|\zeta(\sigma+it)|$ is about size $|t|^{1/2 -\sigma} |\zeta(1-\sigma+it)|$. So what you write is of size $T{1/2-\sigma} \times T$. $\endgroup$
    – Lucia
    Commented Jan 13, 2022 at 16:36

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This answer is based on Lucia's remark, and is included for completeness.

By (8.111) in Ivić's book "The theory of the Riemann zeta function with applications", we have $$\int_T^{2T}|\zeta(\sigma+it)|\,dt\asymp_\sigma T,\qquad T\geq 1,\quad 1/2<\sigma<1.$$ Hence, by the functional equation for $\zeta(s)$ and Stirling's approximation, we also have $$\int_T^{2T}|\zeta(\sigma+it)|\,dt\asymp_\sigma T^{3/2-\sigma},\qquad T\geq 1,\quad 0<\sigma<1/2.$$ In particular, the answer to the original question is negative.

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