# Find $\int_1^\infty \frac{\{x\}}{x(x+1)}dx,$ where $\{x\}$ means $x - \lfloor x \rfloor$.

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. 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.

## 2 Answers

Telescopic approach: note that for $$R>2$$, \begin{align} \int_1^R \frac{\{x\}}{x(x+1)}\,dx &=\int_1^R \left(\frac{\{x\}}{x}-\frac{\{x\}}{x+1}\right)\,dx\\ &=\int_1^R \frac{\{x\}}{x}\,dx-\int_1^R \frac{\{x+1\}}{x+1}\,dx \qquad(\{x+1\}=\{x\})\\ &=\int_1^R \frac{\{x\}}{x}\,dx-\int_2^{R+1} \frac{\{x\}}{x}\,dx\\ &=\int_1^2 \frac{\{x\}}{x}\,dx-\int_R^{R+1} \frac{\{x\}}{x}\,dx\\ &=\int_1^2 \frac{x-1}{x}\,dx-\int_R^{R+1} \frac{\{x\}}{x}\,dx. \end{align} What happens as $$R\to +\infty$$?

Show that, in the last line, the first integral is equal to $$1-\ln(2)\approx 0.30685$$, whereas the second one goes to zero.

\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}