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How can I prove that the function $f(x) = \cos (\pi/x)$ when $0 < x \leq 1$, and $f(x)= 0$ when $x=0$ is integrable on $[0,1]$?

I know I can use that the funtion is continous in the interval and using the Riemann's criterion for integrability but I don't know exactly how to start it or if it makes sense.

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  • $\begingroup$ Unfortunately, the function isn't continuous wolframalpha.com/input/?i=plot+cos(pi%2Fx)+for+0+%3C+x+%3C%3D+1 $\endgroup$ – Theo Bendit Oct 5 '17 at 23:02
  • $\begingroup$ try a change of variable y=1/x, so that you integrate from 1 to infinity. $\endgroup$ – j13r Oct 5 '17 at 23:06
  • $\begingroup$ Show that for clever choices of partitions, the upper and lower sums converge to the same limit. $\endgroup$ – Arthur Oct 5 '17 at 23:21
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Here's one thing you could do: fix $\varepsilon > 0$, and partition the domain into $[0, \varepsilon / 4]$ and $[\varepsilon / 4, 1]$. On the latter interval, the function is continuous, and hence integrable.

Therefore, we may find a partition $\mathcal{P}$ of $[\varepsilon/4, 1]$ such that both the upper and lower sums, $U(f, \mathcal{P})$ and $L(f, \mathcal{P})$ respectively, on this partition are within $\varepsilon / 2$ of each other.

Extend $\mathcal{P}$ to a new partition $\mathcal{P}'$ on $[0, 1]$, by taking $\mathcal{P}$ and including $[0, \varepsilon / 4]$. Note that the function is bounded by $1$ and $-1$, and in fact, attains these extreme values infinitely often on $[0, \varepsilon / 4]$. Therefore, the upper sum $U(f, \mathcal{P}')$ will just be $U(f, \mathcal{P}) + \varepsilon/4$. Similarly, $L(f, \mathcal{P}') = L(f, \mathcal{P}) - \varepsilon/4$.

Hence, for any $\varepsilon > 0$, we have constructed a partition $\mathcal{P}'$ on $[0, 1]$ such that $$U(f, \mathcal{P}') - L(f, \mathcal{P}') = U(f, \mathcal{P}) - L(f, \mathcal{P}) + \varepsilon / 2 < \varepsilon.$$ Thus, the function must be integrable.

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  • $\begingroup$ This is the way to go. It shows that a bounded function with a finite number of discontinuities is Riemann integrable. +1 $\endgroup$ – Paramanand Singh Oct 6 '17 at 1:30
  • $\begingroup$ The same argument can also be used to prove that if $D$ is the set of discontinuities of a bounded function and $D$ has a finite number of accumulation points then the function is Riemann integrable. This is perhaps as far as one can go without using measure theory. $\endgroup$ – Paramanand Singh Oct 6 '17 at 1:35
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Any function $f$ that is bounded on $[a,b]$ and is Riemann integrable on $[a+\epsilon,b]$ for all $\epsilon\in (0, b-a)$ is Riemann integrable on $[a,b].$ This is straightforward to prove, and shows the $f$ in this specific case is Riemann integrable on $[0,1].$

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According to Calculus by Michael Spivak (1994), a function is integrable if the lower sum $L$ and the $U$ upper sum converges to same value for any partition $P_n$ when $n \rightarrow\infty$

It means, Let $P_n$ a partition on the interval $[a,b]$, where $n$ is the number of parts of the partition. A function $f$ is integrable, if

$$\lim_{\substack{n\to\infty\\\Vert P_n \Vert \to 0}} U(f,P_n)=\lim_{\substack{n\to\infty\\\Vert P_n \Vert \to 0}} L(f,P_n)$$

Where $\Vert P_n \Vert$ is the norm of the partition.

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