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?


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

  • $\begingroup$ Beautiful and enlightening. How did you see the last equality in the first set of equations? $\endgroup$ – Tai Aug 16 '18 at 22:52
  • $\begingroup$ @TaisukeYasuda: $\{x\}=\{x+2\}$ and partial fraction decomposition. $\endgroup$ – Jack D'Aurizio Aug 16 '18 at 22:53
  • $\begingroup$ I understand how it works, but I'm asking something more like, is this type of trick for fractional parts well-known in some literature? $\endgroup$ – Tai Aug 16 '18 at 22:56
  • $\begingroup$ @TaisukeYasuda: I honestly do not know, I just started by enforcing a convenient substitution and everything became telescopic soon after. $\endgroup$ – Jack D'Aurizio Aug 16 '18 at 22:56
  • $\begingroup$ The telescopic sum is totally surprising. $\endgroup$ – Szeto Aug 17 '18 at 0:35

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*}


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.


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