Finding $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \left\{\frac{x}{y}\right\} \left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\} dx\space dy\space dz $

Let the considered integral be denoted by $I$. Using the cyclic symmetry of the integrand $f(x,y,z)$ we see that $$I=\underbrace{3\int_{0<x<y<z<1}\frac{x}{y} \frac{y}{z}\left\{\frac{z}{x}\right\}dxdydz}_{A}+\underbrace{3\int_{0<x<z<y<1}\frac{x}{y} \left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\}dxdydz}_{B} $$

• To calculate $A$ we note that $$\eqalign{A&=3\int_{z=0}^1\int_{x=0}^z\frac{(z-x)x}{z}\left\{\frac{z}{x}\right\}dxdz\cr &=3\int_{z=0}^1\int_{t=0}^1(z-tz)t\left\{\frac{1}{t}\right\}zdtdz\cr &=\int_{z=0}^13z^2dz\int_{t=0}^1(1-t)t\left\{\frac{1}{t}\right\}dt=K_1-K_2 }$$ where $K_n=\int_0^1t^n\{1/t\}dt$.
• Similarly, to calculate $B$ we write $$\eqalign{B&=3\int_{y=0}^1\int_{z=0}^y\int_{x=0}^z\frac{x}{y} \left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\}dxdzdy\cr &=3\int_{y=0}^1\int_{z=0}^y\int_{t=0}^1\frac{tz}{y} \left\{\frac{y}{z}\right\}\left\{\frac{1}{t}\right\}zdtdzdy\cr &=3\int_{y=0}^1\int_{s=0}^1\int_{t=0}^1ts \left\{\frac{1}{s}\right\}\left\{\frac{1}{t}\right\}sy^2dtdsdy\cr &=\int_{y=0}^13y^2\int_{s=0}^1s^2 \left\{\frac{1}{s}\right\}ds\int_{t=0}^1t\left\{\frac{1}{t}\right\}dt=K_1K_2. }$$
• It remains to calculate $K_n$ For $n\ge1$. Note that $$\eqalign{K_n&=\int_{1}^\infty\frac{\{x\}}{x^{n+2}}dx=\int_1^\infty\frac{dx}{x^{n+1}}-\int_1^\infty\frac{\lfloor x\rfloor}{x^{n+2}}dx\cr &=\frac1n-\sum_{k=1}^\infty k\int_k^{k+1}\frac{dx}{x^{n+2}}\cr &=\frac1n-\frac{1}{n+1}\sum_{k=1}^{\infty}k\left(\frac{1}{k^{n+1}}-\frac{1}{(k+1)^{n+1}}\right)\cr &=\frac1n-\frac{1}{n+1}\sum_{k=1}^{\infty}\left(\frac{1}{k^{n}}-\frac{1}{(k+1)^{n}}+\frac{1}{(k+1)^{n+1}}\right)\cr &=\frac1n-\frac{1}{n+1}\left(1+\sum_{k=1}^\infty\frac{1}{(k+1)^{n+1}}\right)=\frac1n-\frac{\zeta(n+1)}{n+1}. }$$ Thus $$K_1=1-\frac{\pi^2}{12},\qquad K_2=\frac12-\frac13 \zeta(3).$$ And $$ I= 1-\frac{\pi^2}{8}+\frac{\pi^2}{36}\zeta(3).$$