If $n \in \mathbb{N}$, then, on $]n;n+1[$, the fractional part function is differentiable and its derivative equals $1$ ?
Then, is it possible to integrate by parts this :$\displaystyle \int \limits_{n}^{n+1} \operatorname{frac}(x)~ dx$? where $\operatorname{frac}(x) = \{x\}$. Like...
$$\int \limits_{n}^{n+1} \operatorname{frac}(x)~ dx = \int \limits_{n}^{n+1} x \operatorname{frac}(x) ~dx \\ = x \operatorname{frac}(x)\bigg\vert_{n}^{n+1} - \int \limits_{n}^{n+1} x ~\frac{d \operatorname{frac}(x)}{dx} ~dx \\ = 0 - \int \limits_{n}^{n+1} x ~dx = -n - \frac{1}{2} \quad ?$$
but $\displaystyle \int \limits_{n}^{n+1}\{x\} dx = \frac{1}{2}$, the goal of the question is not to calculate the integral but to integrate it by parts in the most simple way, if it is possible. Sorry for eventual obvious mistakes...
The limits of the first $xfrac(x)$ are not $n$ and $n+1$, but $n^+$ and $n+1^-$ They are the limits as you approach $n$ and $n+1$. The left-hand limit of $xfrac(x)$ at $n+1$ is $n+1$, not $0$