A short answer, there is much more to say about this topic.
$2$: You can bring a minus outside of a floor, but when you do this, the floor becomes a ceiling. In other words,
$$
\lfloor -0.43\rfloor = -1
$$
since the floor function means to round down. On the other hand,
$$
\lfloor -0.43\rfloor = -\lceil 0.43\rceil=-1.
$$
Note that the fractional part of a negative number is sometimes defined in other ways (such as rounding towards zero).
$1$ and $3$: One big application of the mantissa is in how numbers are stored on a computer. There are various ways to do this (and tricks which I am omitting). But, a number is stored as a mantissa $m$ and an exponent $e$. From this pair, the corresponding number is $m2^e$ (on a real computer the representation is slightly different, but I'm just trying to go for the idea). This also shows where logarithms come into play, the exponent $e$ is a (rounded) logarithm of the desired number.
Example (using base $10$, instead of base $2$, which a computer would use): The number $1,578$ could be written as
$$
0.1578\cdot 10^4.
$$
The $0.1578$ would be the mantissa and the $4$ is the exponent. Note that $4\leq\log_{10}1578<5$, a connection to logarithms.
Aside: On a real computer, the representation could be closer to
$$
1.578\cdot 10^3,
$$
where only the mantissa $0.578$ would be stored (on a computer, there is only one leading digit since binary has only numbers, $0$ and $1$.
The upper evaluation limit $\lfloor x\rfloor$ in the OP's original finite sum can be written more simply as $x$ as illustrated in formula (1a) below. The sum can also be written as an infinite sum as defined in formula (1b) below where $\theta(x)$ is the Heaviside step function. There are different conventions for the value of $\theta(0)$ including $\theta(0)=0$, $\theta(0)=\frac{1}{2}$, $\theta(0)=1$, and even leaving $\theta(0)$ undefined, but this answer assumes the definition $\theta(0)=1$. This is only important when evaluating the function $f(x)$, and makes no difference in the evaluation of the integral of $f(x)$.
$$f(x)=\sum\limits_{n=1}^x\frac{\sin(\pi n x)}{x^n}\tag{1a}$$
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\sin(\pi n x)}{x^n}\ \theta(x-n)\right),\quad\theta(x)=\left\{\begin{array}{cc}
0 & x<0 \\
1 & x\geq 0 \\
\end{array}\right.\tag{1b}$$
Formula (3) below is based on term-wise integration where formula (2) below provides the result of the term integral which was evaluated using Mathematica. The $E_n(x)$ terms in formulas (2) and (3) below refer to the generalized exponential integral function.
$$\int\limits_0^\infty\frac{\sin(\pi n x)}{x^n}\ \theta(x-n)\,dx=\int\limits_n^\infty\frac{\sin(\pi n x)}{x^n}\,dx=\frac{i}{2} n^{1-n} \left(E_n\left(i n^2 \pi \right)-E_n\left(-i n^2 \pi \right)\right)\tag{2}$$
$$C=\int\limits_0^\infty f(x)\,dx=\underset{N\to \infty }{\text{lim}}\left(\frac{i}{2}\sum\limits_{n=1}^N n^{1-n} \left(E_n\left(i n^2 \pi \right)-E_n\left(-i n^2 \pi \right)\right)\right)\tag{3}$$
The following table illustrates the value of the integral $C$ defined in formula (3) above for several values of the upper evaluation limit $N$. Note the infinite sum associated with the integral $C$ seems to converge fairly rapidly.
$$\begin{array}{cc}
N & C \\
1 & -0.281141+0. i \\
10 & -0.246292+0. i \\
20 & -0.246292+0. i \\
40 & -0.246292+0. i \\
\end{array}$$
Another way to evaluate the integral of $f(x)$ is illustrated in formulas (4) and (5) below in which case the constant $C$ defined in formula (3) above is evaluated as defined in formula (6) below.
$$\int\limits_n^x\frac{\sin(\pi n t)}{t^n}\,dt=\frac{i}{2}\left(n^{1-n} \left(E_n\left(i n^2 \pi \right)-E_n\left(-i n^2 \pi \right)\right)-x^{1-n} (E_n(i n \pi x)-E_n(-i n \pi x))\right)\tag{4}$$
$$\int\limits_0^x f(t)\,dt=\frac{i}{2}\sum\limits_{n=1}^x\left(n^{1-n}\left(E_n\left(i n^2 \pi \right)-E_n\left(-i n^2 \pi \right)\right)-x^{1-n} (E_n(i n \pi x)-E_n(-i n \pi x))\right)\tag{5}$$
$$C=\underset{x\to\infty}{\text{lim}}\left(\int\limits_0^x f(t)\,dt\right)\tag{6}$$
Figure (1) below illustrates formula (1a) for $f(x)$ in blue and formula (5) for $\int\limits_0^x f(t)\,dt$ in orange. The dashed-gray horizontal reference line is at $y=0.246292$ consistent with the evaluation of formula (3) for $C$ in the table above.
Figure (1): Illustration of formula (1a) for $f(x)$ (blue) and formula (5) for $\int\limits_0^x f(t)\,dt$ (orange)
Best Answer
This function is not integrable in the Lebesgue sense, so you can only evaluate the Cauchy principal value.
That is, what you want to evaluate is the limit $$\lim_{M \rightarrow +\infty} \int_1^M \frac{\{x\}-\frac12}xdx.$$
It is easy to see that it suffices to take the limit for integer values of $M$. We first compute, for every positive integer $k$: $$\int_k^{k + 1}\frac{\{x\}-\frac12}xdx = \int_k^{k + 1}\frac{x- k-\frac12}xdx = 1 - \left(k + \frac 1 2\right) (\ln(k + 1) - \ln k).$$
We then take the sum: $$\int_1^{M + 1} \frac{\{x\}-\frac12}xdx = \sum_{k = 1}^M\left(1 - \left(k + \frac 1 2\right) (\ln(k + 1) - \ln k)\right).$$
This simplifies to: $$M - \left(M + \frac12\right)\ln(M + 1) + \ln M!$$ which, by Stirling's formula, converges to $\ln\frac{\sqrt{2\pi}}e\approx-0.0810614668$.