Evaluate the following integral $ \int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx$ $$\int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx$$
Here $\lbrace\cdot\rbrace$ denotes the fractional part.
I found this challenging integral, and I'm curious about the solution, so I decided to do some efforts to solve it, but sadly I didn't, any hints?
Attempts:
\begin{align}
\int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx&=\int_1^{\infty} \frac{1\lbrace x\rbrace-1}{2x} dx\\
&=\int_1^{\infty}\frac{\lbrace x\rbrace}{x}-\frac{1}{2x}dx\\
&=\int_1^\infty \frac{x-\lfloor x\rfloor-1}{x}-\frac{1}{2x}dx\\
&=\int_1^\infty \frac{x-\lfloor x\rfloor-1}{x} dx -\int_1^\infty \frac{dx}{2x}
\end{align}
I thought about this property:
$$\int_0^\infty \varphi (x) dx=\lim_{a\to \infty} \int_0^a \varphi(x) dx$$
So I applied it only for the second fraction because its antiderivative was easy enough, and here's what I've got:
\begin{align}
\int_1^\infty \frac{dx}{2x}&=\lim_{a\to \infty} \int_1^a \frac{dx}{2x}\\
&=\lim_{a\to \infty}\frac{\ln (x)}{2}\bigg\vert_0^a\\
&=\lim_{a\to \infty}\frac{\ln (a)}2 -\frac{\ln (0)}{2}
\end{align}
And here I felt that I'm wrong I can't get $\infty -\infty$, So any thoughts or hints, I'll be thankfull!

Euler-Maclaurin formula applied to $\log N!$ gives
$$
\begin{aligned}
\log N!
&=\int_1^N\log x\mathrm dx+{\log N\over2}+\int_1^N{\overline B_1(x)\over x}\mathrm dx \\
&=\left(N+\frac12\right)\log N-N+1+\int_1^N{\overline B_1(x)\over x}\mathrm dx
\end{aligned}
$$
However, according to Stirling's approximation, we have
$$
\log N!=\left(N+\frac12\right)\log N-N+\frac12\log2\pi+\mathcal O\left(\frac1N\right)
$$
Consequently as $N\to\infty$ we get
$$
\int_1^\infty{\overline B_1(x)\over x}\mathrm dx=\int_1^\infty{\{x\}-1/2\over x}\mathrm dx=-1+\frac12\log2\pi
$$
