$\def\LR#1{\left(#1\right)}$ $\def\LRsq#1{\left[#1\right]}$ $\def\LRbar#1{\left|#1\right|}$ $\def\Rbar#1{\left.#1\right|}$ $\def\BigRbar#1{\left.#1\right\rvert}$ $\def\bigO{\mathcal{O}}$ $\def\bigOLR#1{\bigO\mathopen{}\left(#1\right)}$ $\def\InCurly#1{\{#1\}}$ $\def\intdx{\, dx}$ $\def\intdu{\, du}$
Background. Let $s = \sigma + it$, let $\zeta(s)$ be the zeta function and fix any $\epsilon > 0$. I am studying the proof that all but an infinitesimal proportion of the non-trivial zeros of $\zeta(s)$ lie in the mini critical strip $1/2 - \epsilon < Re(s) < 1/2 + \epsilon$. My primary sources are Riemann's Zeta Function, Edwards (pp. 193-199) and The Theory of the Riemann Zeta Function, Titchmarsh (2nd Ed)(p. 230 and many references back).
One of the steps in the proof is to develop the following approximation for $\zeta(s)$ for $1/2 < \sigma \leq 1$ \begin{equation*} \zeta(s) = \sum_{n\leq |t|} \frac{1}{n^s} + \bigOLR{|t|^{-\sigma}}. \end{equation*} The starting point is the following formula, valid for $\sigma > 0$ and $s\neq 1$ \begin{equation*} \zeta(s) = \sum_{n=1}^{N} \frac{1}{n^s} - \frac{N^{1-s}}{1-s} - s\int_N^{\infty} \InCurly{x} x^{-s-1}\intdx, \end{equation*} where $N\in\mathbb{N}$ and $[x]$ and $\InCurly{x}$ are the integral and fractional parts of the real number $x$.
Using the equation above and letting $N = [|t| + 1]$, our approximation formula is established if we can show that \begin{equation*} \frac{[|t| + 1]^{1-\sigma}}{|1-s|} + |s| \LRbar{\int_{[|t| + 1]}^{\infty} \InCurly{x} x^{-s-1}\intdx} = \bigOLR{|t|^{-\sigma}}. \end{equation*} A simple estimate gives the needed result for the first term, so it only remains to show the second term is $\bigOLR{|t|^{-\sigma}}$.
The proofs in Edwards and Titchmarsh are quite involved and will not be discussed here. Instead, my question relates to an alternate proof suggested (not shown) in the lecture notes of A.J. Hildebrand (Introduction to Analytic Number Theory, Math 531 Lecture Notes, Fall 2005), where he says on page 166:
Using the trivial bound $|\InCurly{u} u^{-s-1}| \leq u^{-\sigma - 1}$, the integral would give for the second term the bound $|s||t|^{-\sigma}/\sigma \ll t^{1-\sigma}$. This is too weak, but a more careful estimate of the integral, using integration by parts and the estimate $\int_0^x \InCurly{u} du = (1/2)x + \bigOLR{1}$, shows that this term is also of order $\ll t^{-\sigma}$.
Question. Using Hildebrand's hint, I developed the proof below. It includes a step where I substitute an $\bigO$-estimate in an integral (and then continue to use the equal sign in following steps). So, the proof might be a little loose. I'm also not sure if my approach is what Hildebrand was suggesting. Comments, suggestion, corrections and/or improvements are requested.
I will focus on the integral. The $|s|$ factor outside the integral will be included at the end. Let $f(x) = x^{-s-1}$ and $g(x) = \int_0^x \InCurly{u}\intdu$, so that $f'(x) = (-s-1)x^{-s-2}$ and $g'(x) = \InCurly{u}$. Using integration by parts \begin{align*} \int_{\LRsq{|t| + 1}}^{\infty} u^{-s-1}\InCurly{u}\intdu &= \int_{\LRsq{|t| + 1}}^{\infty} f(x) g'(x)\intdx = \BigRbar{f(x) g(x) }_{\LRsq{|t| + 1}}^{\infty} - \int_{\LRsq{|t| + 1}}^{\infty} f'(x) g(x)\intdx \\ &= \BigRbar{x^{-s-1} g(x) }_{\LRsq{|t| + 1}}^{\infty} - \int_{\LRsq{|t| + 1}}^{\infty} (-s-1)x^{-s-2} g(x)\intdx \\ \text{We now use the estimate } &g(x) = x/2 + \bigOLR{1} = x/2 + K,\text{ giving} \\ &= \BigRbar{x^{-s-1} \LR{x/2 + K} }_{\LRsq{|t| + 1}}^{\infty} - \int_{\LRsq{|t| + 1}}^{\infty} (-s-1)x^{-s-2} \LR{x/2 + K}\intdx \\ &= \BigRbar{ \LR{ \LRsq{ \frac{x^{-s}}{2} + Kx^{-s-1} } - \LRsq{ \frac{(s+1)}{2} \frac{x^{-s}}{s} + K x^{-s-1} } } }_{\LRsq{|t| + 1}}^{\infty} \\ &= \BigRbar{ \LR{ \LRsq{ \frac{x^{-s}}{2} } - \LRsq{ \frac{(s+1)}{2} \frac{x^{-s}}{s} } } }_{\LRsq{|t| + 1}}^{\infty} \\ &= \BigRbar{ \LR{ \frac{1}{2} x^{-s} \LRsq{1 - \frac{s+1}{s} } } }_{\LRsq{|t| + 1}}^{\infty} \\ &= \BigRbar{ \LR{ \frac{1}{2} x^{-s} \LRsq{\frac{-1}{s} } } }_{\LRsq{|t| + 1}}^{\infty} \\ &= \BigRbar{ \LR{ \frac{-x^{-s}}{2s} } }_{\LRsq{|t| + 1}}^{\infty} \\ &= \frac{\LRsq{|t| + 1}^{-s}}{2s} \\ \text{Now factoring in the } &|s|, \text{ we have for the full second term} \\ |s| \cdot \LRbar{\frac{\LRsq{|t| + 1}^{-s}}{2s}} &= \bigOLR{|t|^{-\sigma}}. \end{align*}