Integration of fractional part and floor function

According to the given identity we can rewrite the internation in this form $I =\displaystyle\int_{0}^{1} \bigg\{ (-1)^{\left\lfloor \frac{1}{x} \right\rfloor} \frac{1}{x} \bigg\} dx = \displaystyle\int_{0}^{1} (-1)^{\left\lfloor \frac{1}{x} \right\rfloor} \frac{1}{x} dx - \displaystyle \int \limits^{1}_{0}\left\lfloor \left( -1\right) ^{\left\lfloor \frac{1}{x} \right\rfloor }\frac{1}{x} \right\rfloor dx$ Now let $\frac{1}{x} = y$ . We get the internation above as $I =\displaystyle\int \limits^{\infty }_{1}\left( -1\right) ^{\left\lfloor y\right\rfloor }\frac{1}{y} dy - \displaystyle\int \limits^{\infty }_{1}\left\lfloor \left( -1\right) ^{\left\lfloor y\right\rfloor }y\right\rfloor \frac{1}{y^{2}} dy$ Now $I_{1} = \displaystyle\int \limits^{\infty }_{1}\left( -1\right) ^{\left\lfloor y\right\rfloor }\frac{1}{y} dy = \large\sum \limits^{\infty }_{m=1}\left( -1\right) ^{m}\displaystyle\int \limits^{m+1}_{m}\frac{dy}{y} = \large\sum \limits^{\infty }_{m=1}\left( -1\right) ^{m} ln\bigg( \frac{m+1}{m} \bigg)$ so $I_{1} = -ln\bigg[\large\prod \limits^{\infty }_{m =1}\left( \frac{2m}{2m -1} \times \frac{2m }{2m +1} \right) \bigg] = -ln\bigg(\frac{\pi}{2} \bigg)$ $I_{2} =\displaystyle\int \limits^{\infty }_{1}\left\lfloor \left( -1\right) ^{\left\lfloor y\right\rfloor }y\right\rfloor \frac{1}{y^{2}} dy =\large\sum \limits^{\infty }_{m=1}\displaystyle \int \limits^{m+1}_{m}\left\lfloor \left( -1\right) ^{m}y\right\rfloor \frac{1}{y^{2}} dy$ then $I_{2} = \large\sum \limits^{\infty }_{m =1}2m \left( \int \limits^{2m +1}_{2m}\frac{1}{y^{2}} dy - \int \limits^{2m}_{2m -1}\frac{1}{y^{2}} dy\right)$ so $I_{2}= \large\sum \limits^{\infty }_{m=1}\frac{1}{2m +1} - \large\sum \limits^{\infty }_{m=1}\frac{1}{2m -1} = -1$ $\text{And} : I = I_{1}- I_{2} = 1- ln\bigg(\frac{\pi}{2}\bigg) = \boxed {0.548}$