Fractional part of the floor function Integral

Question

Let $\lfloor\rfloor\ $ and $\{\}$ denote the floor function and the fractional part funtion, respectively. Then calculate in closed-form the following integral

$$\int_{0}^{1}\bigg\{\frac{1}{x}\bigg\lfloor\frac{1}{x}\bigg\rfloor\bigg\}dx$$

Answer 1

Hint: Can you compute the value of $$I_n\equiv\int\limits_{1/n}^{1/(n+1)}\left\{\frac1x \left\lfloor \frac 1x\right\rfloor\right\}\; dx$$ and consider $$\sum_{n=1}^{\infty}I_n$$?

Answer 2

$$\int_{0}^{1}\bigg\{\frac{1}{x}\bigg\lfloor\frac{1}{x}\bigg\rfloor\bigg\}dx=\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}}\bigg\{\frac{1}{x}\bigg\lfloor\frac{1}{x}\bigg\rfloor\bigg\}dx=\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}}\bigg\{\frac{k}{x}\bigg\}dx$$also$$\int_{\frac{1}{k+1}}^{\frac{1}{k}}\bigg\{\frac{k}{x}\bigg\}dx=\sum_{l=0}^{k-1}\int_{\frac{k}{k^2+l+1}}^{\frac{k}{k^2+l}}\dfrac{k}{x}-k^2-ldx=\sum_{l=0}^{k-1}k\ln\dfrac{k^2+l+1}{k^2+l}-(k^2+l)\dfrac{k}{(k^2+l)(k^2+l+1)}=k\ln(1+\dfrac{1}{k})-kH_{k^2+k}+kH_{k^2}$$so the integral would be$$\Large I=\sum_{k=1}^{\infty}k\ln(1+\dfrac{1}{k})-kH_{k^2+k}+kH_{k^2}\approx0.4350$$here is a figure that how the integral converges: