TITLE: Evaluate $\int_0^1\frac {\{2x^{-1}\}}{1+x}\,\mathrm d x$
QUESTION [10 upvotes]: I am trying to evaluate $$\int_0^1\frac {\{2x^{-1}\}}{1+x}\,\mathrm d x$$
where $\{x\}$ is the fractional part of $x$. 
I have tried splitting up the integral but it gets quite complicated and confusing. Is there an easy method for such integrals?

REPLY [8 votes]: $$\begin{eqnarray*} \int_{0}^{1}\frac{\{2x^{-1}\}}{1+x}=\int_{1}^{+\infty}\frac{\{2x\}}{x(x+1)}\,dx &=&2\int_{2}^{+\infty}\frac{\{x\}}{x(2+x)}\,dx\\&=&\int_{2}^{+\infty}\left(\frac{\{x\}}{x}-\frac{\{x+2\}}{x+2}\right)\,dx\end{eqnarray*}$$
clearly equals
$$ \int_{2}^{4}\frac{\{x\}}{x}\,dx = \int_{2}^{3}\frac{x-2}{x}\,dx+\int_{3}^{4}\frac{x-3}{x}\,dx = \color{red}{2-4\log 2+\log 3}.$$