${\displaystyle \int_0^1 \dfrac{1}{x} \cdot \dfrac{1}{1-x} \cdot \left(1-\dfrac{1}{x} \right) \mathrm dx}$ 
Evaluate
$$\displaystyle \int_0^1 \left\{\dfrac{1}{x}\right\} \cdot \left\{\dfrac{1}{1-x}\right\} \cdot \left\{1-\dfrac{1}{x} \right\} \mathrm dx$$
where $\{x\}$ denotes the fractional part of $x$.

 A: This is one of those problems where it seems like there's only really one way we can start, but it ends up looking intractable. I'll offer what I can; maybe someone can spot how to evaluate this in closed form (or rigorously prove it's not doable), or efficiently do a numerical estimate.
Set $y=1/x,\,z=y-n$ for $n\in\Bbb N$ so the integral is$$\int_1^\infty\{y\}(1-\{y\})\{\tfrac{1}{y-1}\}\tfrac{dy}{y^2}=\sum_{n\ge1}I_n,\,I_n:=\int_0^1z(1-z)\{\tfrac{1}{z+n-1}\}\tfrac{dz}{(z+n)^2}.$$With $u=\tfrac1z$,$$I_1=\int_1^\infty\tfrac1u(1-\tfrac1u)\{u\}\tfrac{du}{(u+1)^2}=\sum_{m\ge1}J_m,\,J_m:=\int_m^{m+1}\tfrac1u(1-\tfrac1u)(u-m)\tfrac{du}{(u+1)^2}.$$In fact$$J_m=\int_m^{m+1}\tfrac{(1-u)(u+m)}{u^2(u+1)^2}du=\tfrac{6}{m+2}-\tfrac{3}{m+1}+(2-6m)\ln(m+1)+(3m-1)\ln(m(m+2)).$$For $n\ge2$,$$I_n=\int_0^1(\tfrac{n-n^2}{z+n-1}+\tfrac{n^2-n-1}{z+n}+\tfrac{n^2+n}{(z+n)^2})dz=(n-n^2)\ln\tfrac{n^2}{n^2-1}-\ln\tfrac{n+1}{n}+1.$$So the original integral is$$\begin{align}I&:=\sum_{m\ge1}(\tfrac{6}{m+2}-\tfrac{3}{m+1}+(2-6m)\ln(m+1)+(3m-1)\ln(m(m+2)))\\&+\sum_{n\ge2}((n-n^2)\ln\tfrac{n^2}{n^2-1}-\ln\tfrac{n+1}{n}+1).\end{align}$$You'll want to double-check I didn't make any mistakes in this, mind.
A: Since $\{1 - \frac{1}{x}\} = \{-\frac{1}{x}\} = 1 - \{\frac{1}{x}\}$, we have
\begin{align}
I &= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} (1 - \{\tfrac{1}{x}\})\mathrm{d}x \\
&= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x
- \int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x.
\end{align}
From 2.10 and 2.12 (page 101) in [1], we have
$$\int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x = 2\gamma - 1$$
and
$$\int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x = \frac{5}{2} - \gamma - \ln (2\pi)$$
where $\gamma $ is Euler-Mascheroni constant.
Thus, we have $I = 3\gamma - \frac{7}{2} + \ln (2\pi) \approx 0.069524062$.
Also, I approximated the integral numerically in Maple which resembles the analytic result above.
Reference: 
[1] Ovidiu Furdui, "Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis".
