TITLE: Interesting integral with fractional part function.
QUESTION [7 upvotes]: Question: Find
$$\int_1^\infty \frac{\{x\}}{x(x+1)}dx,$$
where $\{x\}$ means $x - \lfloor x \rfloor$.
I have attempted to split this into two integrals, namely
$$\int_1^\infty \frac{x}{x(x+1)}dx - \int_1^\infty \frac{\lfloor x \rfloor}{x(x+1)}dx,$$
however did not get anywhere significant. I have also considered that the average value of $\{x\}$ is $\frac{1}{2}$ (I think). With mostly just instincts and jotting down a few numbers, I obtained an answer of $\ln{\sqrt{2}}$, which seems reasonable since Desmos approximated the answer (when the upper limit is large) as 0.30…
I am only 17 so try to keep notation at my level, though this shouldn't really involve anything too complicated.

REPLY [11 votes]: \begin{align}
\int_1^\infty \frac{\{x\}}{x(x+1)}dx
=& \sum_{k=1}^\infty \int_k^{k+1}\frac{x-k}{x(x+1)} \overset{y=x-k}{dx}\\
=& \sum_{k=1}^\infty \int_0^{1}\frac{y}{(y+k)(y+k+1)}dy\\
=& \sum_{k=1}^\infty (k+1)\ln\frac {k+2}{k+1}-k\ln\frac{k+1}k\\
=& \ \lim_{k\to \infty} (k+1)\ln\frac {k+2}{k+1}- \ln2
= 1-\ln2
\end{align}