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Let $g:[0,1] \rightarrow \mathbb R$ be the function: $$g(x)=\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}\quad\forall x\in (0,1]$$

with $g(0)=0$, whereas $\lfloor{}\cdot{}\rfloor$ is the floor function.

Decide whether:

$$(a) \int_{0}^{1}\left\lfloor\frac{1}{x}\right\rfloor dx\qquad (b)\int_0^{1}\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

converge of diverge. Thanks!

P.S.: I don't know how to handle series yet.

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closed as off-topic by RRL, José Carlos Santos, NCh, Lee David Chung Lin, Cesareo May 4 at 8:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @EricTowers edited $\endgroup$ – Amit Zach May 3 at 16:30
  • $\begingroup$ So, you've told us about a tool you do not have. What tools do you have to show Riemann integrability? $\endgroup$ – Eric Towers May 3 at 16:58
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Write $$\int_0^1 \left[ \frac{1}{x} \right] dx = \int_0^1 \frac{1}{x} dx - \int_0^1 \left\{ \frac{1}{x} \right\} dx,$$ where the first is $\infty$ and the second is bounded by 1. The second is handled best by writing the integral over $$(0,1] = \bigcup_{k \geq 1} \left[\frac{1}{k+1}, \frac{1}{k} \right].$$

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Hint: $[\frac{1}{x}] = k$ $\Longleftrightarrow$ $k\le \frac{1}{x}<k+1$ $\Longleftrightarrow $ $\frac{1}{k+1}<x\le\frac{1}{k}$

$$\int_0^1 \left[\frac{1}{x}\right]dx=\sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}} \left[\frac{1}{x}\right]dx$$

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Hint: $$ \int_0^1 g(x) \, \mathrm{d}x = \sum_{n=1}^\infty \frac{1}{n} \left( \frac{1}{n} - \frac{1}{n+1} \right) \text{.} $$

This comes from the fact that $\int 1/g(x) \,\mathrm{d}x$ is an improper integral with discontinuities at $\{1 / n : n \in \mathbb{Z}_{\geq 0} \}$, so we break the interval of integration and take limits approaching these discontinuities from both sides.

Also, $ x - 1 < \lfloor x \rfloor \leq x$, so $1/g(x) \geq (1/x) - 1$.

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Hint: $$\int_{0}^{1}{\lfloor{\frac1x}\rfloor}dx=\sum_{n=1}^{\infty}{\int_{\frac1{n+1}}^{\frac{1}n}{\lfloor{\frac1x}\rfloor}}dx$$ And: $$x\in(\frac1{n+1},\frac1{n})\iff \frac1x\in (n,n+1)\Longrightarrow \lfloor{\frac1x}\rfloor=n$$

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