Question: For fixed $T$, prove the following $$\int_{1/2}^{2} \zeta(\sigma+iT)d\sigma=\mathcal{O}(T^{1/4+\epsilon}) \tag{*}$$ where $\zeta$ denotes the Riemann zeta function.
My Idea: $$\int_{1/2}^{2} \zeta(\sigma+iT)d\sigma=\int_{1/2}^{1}\zeta(\sigma+iT)d\sigma+\int_{1}^{2}\zeta(\sigma+iT)d\sigma\tag{1}$$ We know that for $s=\sigma+it$, $$\zeta(s)=\mathcal{O}(\log(t)), \ \ \ \sigma\geq 1\tag{2}$$
The Phragmen-Lindelof convexity principle implies that,
$$\zeta(s)=\mathcal{O}(t^{(1-\sigma)/2}\log(t))\text{, uniformly in}\ \ 0\leq\sigma\leq 1\tag{3}$$ So using equation $(3)$ and $(2)$ in $(1)$ we get $$\int_{1/2}^{2} \zeta(\sigma+iT)d\sigma=\int_{1/2}^{1}\mathcal{O}(T^{(1-\sigma)/2}\log(T))d\sigma+\int_{1}^{2}\mathcal{O}(\log(T))d\sigma\tag{4}$$ $$\Rightarrow\int_{1/2}^{2} \zeta(\sigma+iT)d\sigma=\int_{1/2}^{1}\mathcal{O}(T^{(1-\sigma)/2}\log(T))d\sigma+\mathcal{O}(\log(T))\tag{5}$$ $$\Rightarrow\int_{1/2}^{2} \zeta(\sigma+iT)d\sigma=2(T^{1/4}-1)+\mathcal{O}(\log(T))\tag{6}$$ How do we prove equation $(*)$. Please guide me.
Since $\log T$ grows slower than $T^\epsilon$ for every $\epsilon > 0$ (use e.g. L'Hôpital's rule to verify this), it follows that $$ 2 (T^{1/4} - 1) + \mathcal{O}(\log T) = \mathcal{O}(T^{1/4 + \epsilon}) $$ because every term in the left-hand side is dominated by $T^{1/4 + \epsilon}$. In particular, the idea is that since for $T$ large enough we have $$ \log T \leq T^\epsilon \leq T^{1/4 + \epsilon}, $$ it follows that anything of size $\mathcal{O}(\log T)$ is dominated by something of size $\mathcal{O}(T^{1/4 + \epsilon})$.