Evaluate $\int_0^1\frac {\{2x^{-1}\}}{1+x}\,\mathrm d x$ 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?
 A: $$\begin{eqnarray*} \int_{0}^{1}\frac{\{2x^{-1}\}}{1+x}=\int_{1}^{+\infty}\frac{\{2x\}}{x(x+1)}\,dx &=&2\int_{2}^{+\infty}\frac{\{x\}}{x(2+x)}\,dx\\&=&\int_{2}^{+\infty}\left(\frac{\{x\}}{x}-\frac{\{x+2\}}{x+2}\right)\,dx\end{eqnarray*}$$
clearly equals
$$ \int_{2}^{4}\frac{\{x\}}{x}\,dx = \int_{2}^{3}\frac{x-2}{x}\,dx+\int_{3}^{4}\frac{x-3}{x}\,dx = \color{red}{2-4\log 2+\log 3}.$$
A: Splitting up the integral should work. Note that $n\leq 2x^{-1}\leq n+1\iff 2(n+1)^{-1}\leq x\leq 2n^{-1}$. Then splitting into intervals $[2(n+1)^{-1},2n^{-1}]$ for $n\geq 2$, we get 
\begin{align*}
\int_0^1\frac{\{2x^{-1}\}}{1+x}~dx &= \sum_{n=2}^\infty \int_{2(n+1)^{-1}}^{2n^{-1}}\frac{\{2x^{-1}\}}{1+x}~dx = \sum_{n=2}^\infty \int_{2(n+1)^{-1}}^{2n^{-1}}\frac{2x^{-1}-n}{1+x}~dx \\
&= \sum_{n=2}^\infty \int_{2(n+1)^{-1}}^{2n^{-1}}\frac{2}{x(1+x)} - \frac{n}{1+x}~dx.
\end{align*}
A: This is the same as the integral $$\int_1^\infty \frac{\{2x\}}{1+\frac1x} \mathrm dx$$ This is equivalent to $$\sum\limits_{n=2}^\infty \int_{n/2}^{\frac{n+1}2} \frac{2x-n}{1+\frac1x}\mathrm dx$$
This is easily evaluatable.
