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Show that $f(x)$ defined by $f(x)=\frac{1}{n}$, if $\frac{1}{n+1}<x<\frac{1}{n}, n=1,2,3,...$ and $f(0)=0$ is Riemann integrable on $[0,1]$. Also show that $$\int_0^1f(x) dx=\frac{\pi ^2}{6}-1$$

I know that I can use $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6} $ to solve this. But I can't get the required form.

Ravi Upadhyay
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  • By taking the partition $P = {0, \frac{1}{n},..., \frac{1}{2},1}$, I can show that $U(p,f)-L(P,f)=1/n^2$. Thereby it's easy to show that f is Riemann integrable. But how do I go about the value of integration. If I use the same partition P given above $sup L(p,f)= \displaystyle \sum_{i=1}^{\infty}\frac{1}{i^2.(i+1)}$. How do I find it's sum. – Ravi Upadhyay Nov 24 '12 at 07:31
  • Why $1/n^2$? you partitioned $[0,1]$ as $P$ above,so we can take $\xi$ such that $\frac{1}{n}<\xi<\frac{1}{n+1}$. This leads us to evaluate just $\lim_{n\to\infty}\sum_1^n\frac{1}{i}$. – Mikasa Nov 24 '12 at 07:43
  • First the partition should depend on epsilon. How is your partition related to a given epsilon? Second I wouldn't try to use the partition to get the integral because you have to take the limit as epsilon goes to zero. Partition is good for proving integrability but not for actually finding the integral. Just draw the graph and then the answer will be obvious. Go back to the basics just pick lots of points in [0,1] and make a table and plot the graph. – Fixed Point Nov 24 '12 at 08:39
  • @PrinceAli for any $\epsilon \gt 0$, I can choose $n$ sufficiently large (or exactly $n=[\sqrt{\frac{1}{\epsilon}}]$) so that $U(P,f)-L(P,f)<\epsilon$.

    $\frac{1}{i^2(i+1)}=\frac{1}{i^2}-\frac{1}{i(i+1)}$. You can easily show that sum of these terms equals the $\pi^2/6-1$.

     – Ravi Upadhyay Nov 24 '12 at 15:16
  • Ahhhhh the missing link. In your comment above you didn't say what $n$ is or how it is related to $\epsilon$. – Fixed Point Nov 24 '12 at 22:20

3 Answers3

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$$ \begin{align} \int_0^1f(x)\,\mathrm{d}x &=\sum_{k=1}^\infty\int_{\frac1{n+1}}^{\frac1n}f(x)\,\mathrm{d}x\\ &=\sum_{n=1}^\infty\int_{\frac1{n+1}}^{\frac1n}\frac1n\,\mathrm{d}x\\ &=\sum_{n=1}^\infty\frac1n\left(\frac1n-\frac1{n+1}\right)\\ &=\sum_{n=1}^\infty\frac1{n^2}-\sum_{n=1}^\infty\frac1{n(n+1)}\\ &=\frac{\pi^2}{6}-\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)\\ &=\frac{\pi^2}{6}-1 \end{align} $$ Note that any sub-partition of $P=\left\{\left[\frac1{n+1},\frac1n\right]\right\}$ yields the same answer (since $f$ is constant on each interval). To reduce this to a finite partition, we only need to notice that on $\left[0,\frac1n\right]$, $|f(0)|\le\frac1n$. Therefore, any finite sub-partition of $P$ contained in $\left[\frac1n,1\right]$ yields the same answer and is within $\frac1{n^2}$ of any sub-partition of $P$ on a larger interval. This shows that $f$ is Riemann integrable.

robjohn
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  • This doesn't show that $f$ is Riemann integrable. – WimC Dec 25 '12 at 11:32
  • I was relying on the fact that any sub-partition of $\left{\left[\frac1{n+1},\frac1n\right]\right}$ yields the same answer to handle that. – robjohn Dec 25 '12 at 12:29
  • Riemann integrability is about finite partitions of the interval. – WimC Dec 25 '12 at 12:39
  • Since the function and interval are bounded, including all the sub-intervals longer than some $\epsilon>0$ should get us close enough. Of course, not knowing what the OP is allowed to use, the fact that $f$ is bounded and has a countable number of discontinuities also shows that $f$ is Riemann integrable. However, to be honest, I didn't get the impression that the integrability was the author's difficulty, but I could be wrong. – robjohn Dec 25 '12 at 13:02
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You are asking about $$\int_0^1 \frac{1}{Floor(\frac{1}{x})}dx$$ I love this problem. I saw this first when I was taking Calc I and now I always give this to my students. Here is a snippets from a detailed writeup I have.

Fact I: The function $f(x)=g(x)$ on the interval [0,1] where \begin{eqnarray*} g(x)=\left\{ \begin{array}{ll} \frac{1}{n} & \textrm{ for }x \in \left(\frac{1}{n+1},\frac{1}{n}\right] \textrm{ where } n \in \mathbb{N}\\ 0 & \textrm{ if } x=0 \end{array}\right. \end{eqnarray*}

Proof: At $x=0,\,f(x)=g(x)$ obviously.

Pick any $x\in(0,1]$ and let $n\in \mathbb{N}$ be the number such that $x\in\left(\frac{1}{n+1},\frac{1}{n}\right]$

$\Rightarrow \frac{1}{n+1}<x\leq\frac{1}{n}$

$\Rightarrow n+1>\frac{1}{x}\geq n$

$\Rightarrow Floor(1/x)=n$ by the definition of the floor function

$\Rightarrow \frac{1}{Floor(\frac{1}{x})}=1/n$

$\Rightarrow f(x)=g(x).$

This tells us that at every point $x$ where $x$ is a unit fraction, our function $f(x)$ has a jump discontinuity and everywhere else, $f(x)$ is a constant.

Now for the first part of your problem, the number of jump discontinuities is countable therefore the function is Riemann integrable. But for a formal proof of the level you probably want, you have to use the definition of a Riemann integrable function. Given an arbitrary $\epsilon > 0$ you have to find a partition of $[0,1]$ such that the difference between the upper sum and the lower sum over that partition of $f(x)$ is less than $\epsilon$.

And then for the last part of your question, just graph the function very carefully and then finding the area under the curve should become very obvious. I suggest you do this part by hand. Using a calculator won't help you much. After getting you started I'll leave both of these parts up to you.

Fixed Point
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  • Also a quick comment that my $f(x)$ is slightly different from your $f(x)$ at the jump discontinuities. Your function (the way you have it written) is undefined at the unit fractions so you have a bunch of holes. My function is defined everywhere on [0,1]. But these differences are only countable and a set of measure zero so the Riemann integral is exactly the same. – Fixed Point Nov 24 '12 at 07:22
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To prove the function is Riemann Integrable, you need to check the two conditions

i) boundedness,

ii) the set of discontinuity of the function is countable or more generally, by Lebesgue criteria for Riemann integrability, it has measure zero.

Mhenni Benghorbal
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