Asymptotics for $\int_{0}^{T} \zeta(\sigma+ it) \mathrm{d}t$ 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 ?
 A: 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.
A: 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...).
A: 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?
