Integrating Floors And Ceilings

Relevant wiki: Integration of Piecewise Functions

We note that for $\dfrac 1k > x \ge \dfrac 1{k+1}$ , where $k = 2, 3, 4, ... 8$ , we have $\left \lceil \dfrac 1x \right \rceil = k+1$ and that $\left \lfloor \ln \left \lceil \dfrac 1x \right \rceil \right \rfloor = \left \lfloor \ln (k+1) \right \rfloor$ , which is a constant. Therefore, we have:

$\begin{aligned} \int_\frac 1{k+1}^\frac 1k \left \lfloor \ln \left \lceil \frac 1x \right \rceil \right \rfloor \ dx & = \left \lfloor \ln (k+1) \right \rfloor \left(\frac 1k - \frac 1{k+1} \right) \\ \implies \int_\frac 18^\frac 12 \left \lfloor \ln \left \lceil \frac 1x \right \rceil \right \rfloor \ dx & = \sum_{k=2}^\color{#D61F06}{7} \left \lfloor \ln (k+1) \right \rfloor \left(\frac 1k - \frac 1{k+1} \right) & \small \color{#3D99F6}{\text{As } \lfloor \ln(3) \rfloor =\lfloor \ln(4) \rfloor =\lfloor \ln(5) \rfloor =\lfloor \ln(6) \rfloor = \lfloor \ln(7) \rfloor = 1, \ \lfloor \ln(8) \rfloor = 2} \\ & = \sum_{k=2}^7 \left(\frac 1k - \frac 1{k+1} \right) + \frac 17 - \frac 18 \\ & = \frac 12 - \frac 18 + \frac 17 - \frac 18 \\ & = \boxed{\dfrac {11}{28}} \end{aligned}$