In a paper I am reading, I've come across the estimate
$$ 2\pi\sum_{\substack{\sigma<\beta<\sigma_0\\ T<\gamma\leq 2T}}(\beta-\sigma)=\underbrace{\int_T^{2T}\log|\zeta(\sigma+it)|\,dt}_{I_\sigma}-\underbrace{\int_T^{2T}\log|\zeta(\sigma_0+it)|\,dt}_{I_{\sigma_0}}+O(\log T), $$ where $\frac{1}{2}\leq\sigma\leq1$ is fixed, $\sigma_0=4$, and $\rho=\beta+i\gamma$ is a nontrivial zero of Riemann's zeta function.
I'm fine up to this point. Shortly after this, the author makes the statement that the above estimate holds for all complex numbers $a$, so in particular, when $a=0$, it becomes
$$ I_\sigma=O(\log T). $$
I don't understand how to prove this estimate.
Assuming Riemann's hypothesis (RH), the sum on the left-hand side becomes $0$. Thus, we need only to show that $I_{\sigma_0}=O(\log T)$.
A trivial estimate doesn't work because that will only give us $O(T)$ which isn't good enough for my purposes. I've attempted to search for estimation on the logarithm of the zeta function but have failed at successfully finding anything helpful, so I'm asking here.
We can show something stronger. If $\sigma>1 $ we know that holds, by Euler product, $$\log\left(\zeta\left(s\right)\right)=\sum_{p}\sum_{k\geq1}\frac{1}{kp^{ks}} $$ then $$\int_{T}^{2T}\log\left|\zeta\left(4+it\right)\right|dt=\textrm{Re}\left(\sum_{p}\sum_{k\geq1}\frac{1}{kp^{4k}}\int_{T}^{2T}\exp\left(-itk\log\left(p\right)\right)dt\right)\ll\sum_{n\geq1}\frac{1}{n^{4}}\ll1. $$