Let $f(x)=\sin\left(\frac{1}{x}\right)$ if $0<x\le1$ and $f(x)=0$ if $x=0$. Show that $f$ is Riemann integrable on $[0,1]$ and calculate it's integral on $[0,1]$.

I would like to know if my proof holds, please. And, I would like have a hint on how we can calculate the intergal of this such of functions, please.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.

Let $0<\epsilon<2$ and consider the following intervals $[0,\epsilon/2]$ and $[\epsilon/2,1]$. First of all, as $f(x)$ is continuous on $[\epsilon/2,1]$($\sin(\frac{1}{x})$ is continuous on $[\epsilon/2,1]$), it is integrable on this interval. We would like to show now that $f$ is intergable on $[0,\epsilon/2]$. Consider the partition $\sigma=\{0,\frac{\epsilon}{2}\}$ on $[0,\epsilon/2]$ We have that

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)=\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)$ with $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$ and $m_i=\inf\{f(x):x\in[x_i,x_{i+1}]\}$.

We have that $M_i\le1$ and $m_i\le 0$. Thus,

$\sum_{i=0}^{0}M_i(x_{i+1}-x_i)-\sum_{i=0}^{0}m_i(x_{i+1}-x_i)\le1\cdot \frac{\epsilon}{2}< \epsilon$. Therefore, as $f(x)$ is integrable on $[0,\epsilon/2]$ and on $[\epsilon/2,1]$, we conclude that $f$ is integrable on $[0,1]$.

Now, I would like to calculate it's integral value... If someone could help with it, I would appreciate it. Honestly, I have no idea how to integrate this such of discontinuous functions at finitely many points.


As $f(x)$ is continuous on $[\epsilon/3,1]$, it is integrable on this interval. Therefore, $\forall \epsilon>0 \ \exists$ partition $\tau$ on the interval $[\epsilon/3,1]$: $\overline{S}_{\tau}(f)-\underline{S}_{\tau}(f)<\epsilon/3$

To show that $f$ is integrable on $[0,1]$, we have to show that: $\forall \epsilon>0 \ \exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}<\epsilon$

Suppose $0<\epsilon<3$ and let $\sigma=\tau \ \cup \{0\}$ be a partition on the interval $[0,1]$. We have that (for the following inequality I use the same definition for $M_i$ and $m_i$ as above in the post):

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}=\Big(\sum_{i=0}^{0}\underbrace{M_i}_{\le1}(x_{i+1}-x_i)-\sum_{i=0}^{0}\underbrace{m_i}_{\ge -1}(x_{i+1}-x_i)\Big)+\underbrace{\Big(\sum_{i=1}^{n}M_i(x_{i+1}-x_i)-\sum_{i=1}^{n}m_i(x_{i+1}-x_i)\Big)}_{<\epsilon/3}<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$.

Therefore, $f$ is integrable on $[0,1]$.

  • $\begingroup$ So, could someone give a feedback on the proof, please? $\endgroup$ – Daniil Feb 26 at 20:26
  • 1
    $\begingroup$ You may also look at proof for Riemann integrability for functions with finite set of discontinuities. It should be available somewhere on this site. $\endgroup$ – Paramanand Singh Feb 27 at 2:44
  • 1
    $\begingroup$ For example see this answer of mine. The first part of the answer is exactly what you need here. $\endgroup$ – Paramanand Singh Feb 27 at 2:46

This is about feedback on your proof.

Your goal is to find a partition $\sigma$ such that $\overline{S}_{\sigma} (f) <\underline{S} _{\sigma} (f) +\epsilon $ but instead you are trying to use that $f$ is integrable on $[\epsilon/2,1]$ and prove that $f$ is integrable on $[0,\epsilon/2]$.

This is not what you want. You want a partition $\sigma$ which works as expected.

You can just say that since the function is integrable on $[\epsilon/2,1]$ there is a partition $\sigma_1$ of $[\epsilon/2,1]$ which works for this interval and then take $\sigma=\sigma_1\cup \{0\}$ and using your argument in question show that this particular partition works for $[0,1]$.

You have the correct idea but you need to present it in proper manner.

  • $\begingroup$ Alright, thank you for the feedback! I will try to finish the proof correctly. $\endgroup$ – Daniil Feb 27 at 7:47
  • $\begingroup$ Thank you very much for your encouragement.... $\endgroup$ – Sebastiano Feb 27 at 9:38
  • $\begingroup$ @Paramanand Singh I edited my answer. If you could take a look at it, I would really appreciate it x) Thank you very much for your help and answers! Edit: I think there is a problem in my inequality because of $m_i$ $\endgroup$ – Daniil Feb 27 at 10:11
  • $\begingroup$ @Daniil: looks fine now (but I haven't done proof reading to check for typos). $\endgroup$ – Paramanand Singh Feb 27 at 10:13
  • $\begingroup$ @Paramanand Singh Alright, thank you. But I think m inequality doesn't hold as if i suppose that $m_i\le 0$ it is not true my inequality. Probably, I should have used the interval with $\epsilon/3$ instead of $\epsilon/2$ and use the fact that $m_i\ge-1$ $\endgroup$ – Daniil Feb 27 at 10:22

Integrate by parts $$\int \sin \left(\frac{1}{x}\right) \, dx=x\sin \left(\frac{1}{x}\right) -\int x \left(-\frac{1}{x^2}\right)\cos \left(\frac{1}{x}\right)\,dx=$$ $$=x \sin \left(\frac{1}{x}\right)-\text{Ci}\left(\frac{1}{x}\right)+C$$ Ci is cosine integral $$\text{Ci}(x)=-\int_x^{\infty}\frac{\cos x}{x}$$

$$\int_0^1 \sin \left(\frac{1}{x}\right) \, dx=\sin (1)-\text{Ci}(1)-\lim_{a\to 0^+}\left[a \sin \left(\frac{1}{a}\right)-\text{Ci}\left(\frac{1}{a}\right)\right]$$ set $\frac{1}{a}=t$ $$\lim_{t\to\infty}\left[\frac{1}{t} \sin t-\text{Ci}\left(t\right)\right]=0$$ $$\int_0^1 \sin \left(\frac{1}{x}\right) \, dx=\sin (1)-\text{Ci}(1)\approx 0.504$$

  • $\begingroup$ Thank you for the answer! But, I still didn't see improper integrals. Does it matter in this case? $\endgroup$ – Daniil Feb 26 at 17:28
  • $\begingroup$ Without knowing Cosine Integral function we can't solve the given integral. Or they wanted just an approximeted result? @Daniil $\endgroup$ – Raffaele Feb 26 at 17:32
  • $\begingroup$ No, it was asked to calculate the value. But as it is not a homework (just problem found in the internet), I just would like to know if I'm capable to calculate this such of integrals. My level is Advanced Analysis I (so I didn't see improper integrals). But still I understand the idea of your calculation so thank you very much! But my main problem in the calculation was that I didn't know how to proceed as $ \sin(1/x)$ is undefined at $0$ but I think I see how you calculated it $\endgroup$ – Daniil Feb 26 at 17:38
  • 1
    $\begingroup$ @Daniil I added a few lines to explain how to deal with this improper integral. Hope it is clear $\endgroup$ – Raffaele Feb 26 at 17:49
  • $\begingroup$ It is more clear, thank you! What do you think about my proof on $f$ integrability, please? $\endgroup$ – Daniil Feb 26 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.