Determine if the function $1/\lfloor 1/x\rfloor$ is integrable on $[0,2]$

Question

Is this function integrable on $[0,2]$? $$\cfrac{1}{\left\lfloor\cfrac{1}{x}\right\rfloor}$$

I have suspicion that it is, but I'm unsure of how I could determine if that's true.

Answer 1

$\displaystyle\int_0^2\frac1{\bigg\lfloor\dfrac1x\bigg\rfloor}~dx~=~\int_\tfrac12^\infty\frac1{\lfloor u\rfloor}~\dfrac{du}{u^2}~,~$ which clearly diverges, since $\lfloor u\rfloor=0$ for $u\in(0,1)$.

Answer 2

Note that for $x>1$ we have $\newcommand{\dcc}[1]{\left\lfloor#1\right\rfloor}\newcommand{\dd}{\,\mathrm{d}}\dcc{\frac1x}=0$, so this function is undefined for $x>1$. If we work with the interval $[0,1]$ instead we get

$$\int_0^1 \frac1{\dcc{\frac1x}} \dd x = \sum_{n=1}^\infty \int_{\frac1{n+1}}^{\frac1n} \frac1{\dcc{\frac1x}} \dd x \overset{(*)}= \sum_{n=1}^\infty \frac1n \left(\frac1n -\frac1{n+1}\right)= \sum_{n=1}^\infty \frac1n \cdot \frac 1{n(n+1)} = \sum_{n=1}^\infty \frac1{n^2(n+1)} < +\infty.$$

$(*)$: We used that for $\frac1{n+1}<x\le\frac1n$ we have $n\le\frac1x<n+1$ and thus $\dcc{\frac1x}=n$

So the integral from $0$ to $1$ is finite.

Drawing graph of the function might help with better understanding of the problem, so I'll add some plots from WolframAlpha. (But you should be able to draw something like this also by hand, without using any software.)

Graph of $\dcc{\frac1x}$: plot 1/floor(1/x) from 0 to 2

Graph of $\frac1{\dcc{\frac1x}}$: plot 1/floor(1/x) from 0 to 2

Area under the graph of $\frac1{\dcc{\frac1x}}$: integrate 1/floor(1/x) from 0 to 2