For $s \notin (-\infty, 0]$, Stirling's formula is: $$ \log \Gamma(s) = \left(s - \frac12 \right) \log s - \frac12 \log 2\pi - \int_0^\infty \frac{P_1(x)}{s + x} dx $$ where $P_1 (x) = x - \lfloor x \rfloor - 1/2$ is the sawtooth function. In Davenport's Multiplicative Number Theory, it says that $$ \int_0^\infty \frac{P_1(x)}{s + x} dx = O(\vert s \vert^{-1}) $$ as $\vert s \vert \to \infty$ on $-\pi + \delta < \arg s < \pi - \delta$ for each $\delta > 0$. However, I am wondering if it holds on whole $\mathbb{C} \setminus (-\infty, 0]$; here is my attempt for proof.
Write $s = \sigma + it$ with $\sigma, t \in \mathbb{R}$, and let $P_2(x) = (x^2 - x)/2$ for $x \in [0, 1]$ and $P_2(x) = P_2(x+1)$ for all $x$. Then $P_2' = P_1$ so by integration by parts, $$ \int_0^\infty \frac{P_1(x)}{s + x} dx = \int_0^\infty \frac{P_2(x)}{(s+x)^2} dx. $$ Also $P_2$ is bounded by some $M > 0$, $$ \left\vert \int_0^\infty \frac{P_2(x)}{(s+x)^2}dx \right\vert \le \int_0^\infty \frac{Mdx}{\vert s+ x \vert^2} = M \int_0^\infty \frac{dx}{(x+\sigma)^2 + t^2}. $$ By symmetry, let us consider $t \ge 0$. Then $$ \int_0^\infty \frac{dx}{(x+\sigma)^2 + t^2} = \frac{1}{t} \left(\frac{\pi}{2} - \tan^{-1} \frac{\sigma}{t}\right) = \frac{\tan^{-1} (t/\sigma)}{t}. $$ where the value $\tan^{-1} (t/\sigma) / t$ is interpreted as $\pi / 2t$ for $\sigma = 0$ and $1/\sigma$ for $t = 0$.
Write $s = R e^{i\theta}$ with $0 \le \theta < \pi$. Let us consider the three cases: $0 \le \theta < \pi / 2$, $\theta = \pi/2$, and $\pi/2 < \theta < \pi$. For $0 \le \theta < \pi / 2$, $$ \frac{\tan^{-1} (t/\sigma)}{t} = \frac{\theta}{R \sin \theta} \le \frac{\pi/2}{R}. $$ For $\theta = \pi / 2$, $\tan^{-1} (t/\sigma) / t = \pi / 2R$. For $\pi/2 < \theta < \pi$, $$ \left\vert\frac{\tan^{-1} (t/\sigma)}{t}\right\vert = \frac{\pi - \theta}{R \sin \theta} \le \frac{\pi/2}{R}. $$ Thus $\tan^{-1} (t/\sigma) / t = O(R^{-1})$ on $\mathbb{C} \setminus (-\infty, 0]$. Is there anything wrong in my proof?