# Decide whether $\int_{0}^{1}\left[\frac{1}{x}\right]dx$ and $\int_0^{1}\frac{1}{\left[\frac{1}{x}\right]}dx$ converge or diverge [closed]

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.

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

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• @EricTowers edited – Amit Zach May 3 at 16:30
• So, you've told us about a tool you do not have. What tools do you have to show Riemann integrability? – Eric Towers May 3 at 16:58

## 4 Answers

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 $$(0,1] = \bigcup_{k \geq 1} \left[\frac{1}{k+1}, \frac{1}{k} \right].$$

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

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

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 $$1/g(x) \geq (1/x) - 1$$.

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