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.
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I don't have the tools provided by the Fundamental Theorem of Calculus; just a note. – Senya Jan 22 '15 at 05:27
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So, are you to argue using partitions? The function is bounded. The discontinuities can be enumerated. Does this connect with any results you have about integrals? – Jan 22 '15 at 05:31
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I thnk it is not integrable. – soumajit das Jan 22 '15 at 05:34
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The boundedness, yes; the discontinuities, no. – Senya Jan 22 '15 at 05:34
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I think it is integrable, I'm just unsure of how to show that it is. – Senya Jan 22 '15 at 05:36
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So it's bounded, but would that mean I have to show that $U(f,P) - L(f,P) < \epsilon$? – Senya Jan 22 '15 at 05:47
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3Wait a moment. When $x\in (1,2]$, the denominator is $0$. So.. does not look very good for integrability. Unless there's a typo in the statement. – Jan 22 '15 at 05:50
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2Hint: Divide $[0,2]$ in smaller intervals on which the function is constant. – Janko Bracic Jan 22 '15 at 05:50
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1@Fundamental So it can't be integrated from $(1,2]?$ What about $(0, 1]?$ – Senya Jan 22 '15 at 05:53
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1Try this: http://www.math.utah.edu/~yael/3210_public/exams/Integral.pdf – user1537366 Jan 22 '15 at 06:09
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$\displaystyle\int_0^2\frac1{\bigg\lfloor\dfrac1x\bigg\rfloor}~dx~=~\int_\tfrac12^\infty\frac1{\lfloor u\rfloor}~\dfrac{du}{u^2}~,~$ which clearly diverges, since $\lfloor u\rfloor=0$ for $u\in(0,1)$.
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The variable change was $x\mapsto\dfrac1x$. If, on the other hand, the upper limit would have been $1$ instead of $2$, then it would have converged to $\zeta(2)-1=\dfrac{\pi^2}6-1$. – Lucian Jan 22 '15 at 06:12
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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
Martin Sleziak
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The sum we got is calculated, for example, in these posts: Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ or Sum of series with addition: $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ – Martin Sleziak Sep 10 '20 at 04:11