Integral with floor function

Evaluate the integral:

$$I=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )\, {\rm d}x$$

Solution



Successively we have:

$$\begin{aligned} I&=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )dx\\ &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\frac{1}{x}} \\ {dx\,=\,-\frac{1}{t^2}\,dt} \end{subarray}}\,\int_{1}^{\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\\ &=\sum_{n=1}^{\infty }\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\\ &=\sum_{n=1}^{\infty }\frac {1}{(n+1)(2n+1)}\\
&=\sum_{n=0}^{\infty}\frac{1}{(n+1)\left ( 2n+1 \right )}- 1\\&=\sum_{n=0}^{\infty}\left [ \frac{2}{2n+1}- \frac{1}{n+1} \right ]-1\\&= \psi(1)- \psi \left ( \frac{1}{2} \right )-1 =\cancel{-\gamma} +2\log 2 +\cancel{\gamma} -1\\ &=2\log{2}-1\end{aligned}$$