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Given the function $$f(x)=\begin{cases}\frac1{\lfloor1/x\rfloor}&0<x\le1\\0&x=0\end{cases}$$ Knowing it is integrable on $[0,1]$, and knowing the identity: $$\lim_{N\to\infty}\sum_{n=1}^N\frac1{n^2}=\frac{\pi^2}6$$ Find $\int_0^1f(x)\,dx$.

I know it probably has to do with Riemann sums but I can't seem to find a solution.

Clement Yung
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user11555739
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3 Answers3

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$$ \newcommand{\d}[1]{\; \mathrm{d} #1} \newcommand{\f}[1]{\left\lfloor #1 \right\rfloor} \newcommand{\bb}[1]{\left( #1 \right)} $$ You don't require Riemann sum. Observe that, if $\frac{1}{n+1} < x \leq \frac{1}{n}$, then $n \leq \frac{1}{x} < n + 1$, so $\f{\frac{1}{x}} = n$. \begin{align*} \int_0^1 f(x) \d{x} &= \sum_{n=1}^\infty \int_\frac{1}{n+1}^\frac{1}{n} \frac{1}{\f{\frac{1}{x}}} \d{x} \\ &= \sum_{n=1}^\infty \int_\frac{1}{n+1}^\frac{1}{n} \frac{1}{n} \d{x} \\ &= \sum_{n=1}^\infty \frac{1}{n}\bb{\frac{1}{n} - \frac{1}{n+1}} \\ &= \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n(n+1)} \end{align*} The sum on the left is given. Can you evaluate the sum on the right?

Clement Yung
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  • Can you explain the first equality? (The first line of the big chunk of text, idk how to copy it). I don't think we studied that, though it makes sense. Perhaps you see a more basic solution with Riemann/Darboux? – user11555739 Apr 20 '20 at 14:25
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    $\newcommand{\d}[1]{; \mathrm{d} #1}$ Recall the basic property that: $$ \int_a^b f(x) \d{x} = \int_a^c f(x) \d{x} + \int_c^b f(x) \d{x} $$ What I'm doing is basically this, but infinitely many times: $$ \int_0^1 f(x) \d{x} = \int_\frac{1}{2}^1 f(x) \d{x} + \int_\frac{1}{3}^\frac{1}{2} f(x) \d{x} + \int_\frac{1}{4}^\frac{1}{3} f(x) \d{x} + \cdots $$ The solution is as basic as it can get. If you're unsure on how to evaluate the second infinite sum, see this. – Clement Yung Apr 20 '20 at 14:29
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$$\int_0^1f(x)dx=\sum_{n=1}^\infty\int_{\frac1{n+1}}^{\frac1n}f(x)dx=\sum_{n=1}^\infty\int_{\frac1{n+1}}^{\frac1n}\frac1{n}dx$$

Kenta S
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hint

Using the fact that If $$x\in [\frac{1}{k},\frac{1}{k-1}]$$ then $$\frac 1x \in [k-1,k]$$ and $$f(x) = \color{red}{\frac{1}{k-1}}$$ So

$$\int_0^1f=$$ $$\lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f=$$

$$\lim_{n\to+\infty}\sum_{k=2}^n(\frac{1}{k-1}-\frac 1k)\color{red}{\frac{ 1}{k-1}}=$$

$$\frac{\pi^2}{6}-\lim_{n\to+\infty}\sum_{k=2}^n(\frac{1}{k-1}-\frac 1k)$$ $$=\frac{\pi^2}{6}-1$$

hamam_Abdallah
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