How to calculate it?

$\displaystyle \int_0^1 \left\{(-1)^{\left\lfloor \frac{1}{x}\right\rfloor}\cdot\dfrac{1}{x}\right\}dx = \int_{1}^{\infty}\frac{(-1)^{\lfloor x\rfloor}x - \lfloor (-1)^{\lfloor x\rfloor}x\rfloor}{x^2}dx\quad\text{ by }\frac{1}{x}\rightarrow x$

Split into regions of floor(odd.xyz) = odd and floor(even.xyz) = even

$\displaystyle\sum_{n=1}^{\infty}\int_{2n-1}^{2n}\frac{-1}{x}dx+\int_{2n}^{2n+1}\frac{1}{x}dx-\sum_{n=1}^{\infty}\int_{2n-1}^{2n}\frac{-2n}{x^2}dx+\int_{2n}^{2n+1}\frac{2n}{x^2}dx =$

$\displaystyle\sum_{n=1}^{\infty}-\ln{(2n)}+\ln{(2n-1)}+\ln{(2n+1)}-\ln{(2n)} + \sum_{n=1}^{\infty}\frac{2n}{2n-1}+\frac{2n}{2n+1}-2 =$

$\ln{\big( \prod_{n=1}^{\infty}1-\frac1{4n^2}\big)} + \sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}$

$\displaystyle\prod_{n=1}^{\infty}1-\frac1{4n^2} = \boxed{\frac{2}{\pi}}$ from the Wallis Product

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1} = \sum_{n=1}^{\infty}\int_{0}^{1}(y^2)^{n-1}dy-(y^2)^ndy =\int_{0}^{1}\sum_{n=1}^{\infty}(y^2)^{n-1}dy-(y^2)^ndy =$

$\displaystyle\int_{0}^{1}\frac{1}{1-y^2}-( \frac{1}{1-y^2}-1) = \boxed{1}\quad$ by noticing the infinite geometric sequence and reversing the sum and integral.

$\displaystyle\Huge\ln{\frac{2}{\pi}}+1 \approx \boxed{.548}$

p.s. Great problem!