# Show $\sin{\frac{1}{x}}$ for $x\not= 0$ and $f(0)=0$ is integrable on $[-1,1]$

Show $\sin{\frac{1}{x}}$ for $x\not= 0$ and $f(0)=0$ is integrable on $[-1,1]$.

I guess my strategy for solving this is to use the following theorem:

• Let $f$ be a function defined on $[a,b]$. If $a<c<b$ and $f$ is integrable on $[a,c]$ and on $[c,b]$, then $f$ is integrable on $[a,b]$ and $\int_a^b{f}=\int_a^c{f}+\int_c^b{f}$.

We split up the interval $[-1,1]$ into three sub-intervals: $[-1,-\epsilon]$, $[-\epsilon, \epsilon]$, $[\epsilon, 1]$. We give the sub-interval $[-1,-\epsilon]$ a partition $P_1$ and $[\epsilon, 1]$ a partition $P_2$.

We can check that $\sin{\frac{1}{x}}$ is continuous for the sub-intervals $[-1,-\epsilon]$ and $[\epsilon, 1]$ so $\sin{\frac{1}{x}}$ is also integrable on those sub-intervals.

Using Darboux's definition of integrability we can say:

• $U(f,P_1) - L(f,P_1) \lt$ some very small number
• $U(f,P_2) - L(,P_2) \lt$ some very small number

I know the goal is to combine partitions $P_1$ and $P_2$ to get a larger partition $P$ for the whole interval $[-1,1]$ but I get stuck at this point.

• Questions:

What "very small number" should I choose and why?

How do I deal with the interval $[-\epsilon, \epsilon]$ when it includes $0$?

Thank you!

• You deal with $[-\epsilon,\epsilon]$ by upper bounding and lower bounding the value of the integral. Use the fact that $-1 \leq \sin(1/x) \leq 1$. – jkabrg Jun 6 '15 at 22:42
• PERFECT! And now, what is the intuition behind this? I kind of get why you would bound the value but I just need a little more clarification to really get it. Thank you! – David South Jun 6 '15 at 22:43
• @user3491648 Forgot to tag you – David South Jun 6 '15 at 22:49
• shrug it's the Darboux definition. Make sure the upper and lower bounds converge to eachother. And the $[-\epsilon,\epsilon]$ is the troublesome bit. – jkabrg Jun 6 '15 at 22:50
• @user3491648 shoot what am I thinking, gotcha! Thanks! – David South Jun 6 '15 at 22:51

## 2 Answers

You can do the following: given $\varepsilon > 0$, choose $\delta=\varepsilon/12$. Use continuity to get partitions of $[-1,-\delta]$ and $[\delta,1]$ such that the upper and lower sums are within $\varepsilon/3$ of each other. (To be concrete, you can take a uniform partition with mesh size $\varepsilon \delta^2/3=\varepsilon^3/432$, since the derivative is bounded by $1/\delta^2$ in magnitude on these intervals.) Now notice the upper and lower sums on $[-\delta,\delta]$ are within $\varepsilon/3$ of each other (here you use the choice of $\delta$ that I suggested). So the upper and lower sums on the entire interval for this overall partition are within $\varepsilon$ of each other.

This is the same approach behind the proof of the Lebesgue criterion for Riemann integrability, in the special case where there is only one point of discontinuity.

• @DavidSouth You don't need to know the Lebesgue criterion to do the proof, that's just a name for this kind of result that you can look up. Anyway, it's not that they're just very close, it's that you can make them arbitrarily close by taking sufficiently fine partitions. That's what I mean when I say "given $\varepsilon > 0$". – Ian Jun 6 '15 at 22:47
• Okay! Thank you for your help thus far, but I have one last question. Why is it that the upper and lower sums on $[-\delta, \delta]$ are within $\frac{\epsilon}{3}$ of each other? You kind of said it as if having the partitions for the first two sub-intervals implies this fact and that confuses me. – David South Jun 6 '15 at 22:48
• The difference between upper and lower sum on $[-\delta, \delta]$ doesn't exceed the area of the rectangle $[-\delta, \delta] \times [-1, 1]$, since the rectangle covers that part of the graph. – Adayah Jun 6 '15 at 23:18

APPROACH 1:

Without loss of generality, we will focus discussion on the integral $I=\int_0^1 \sin(1/x) dx$.

To show directly that $I$ converges, we rely on the following.

First, Riemann's Criterion states that a function $f$ is integrable if for there exists a dissection (aka, a partition) such that the upper and lower sums of that partition differ by at most $\epsilon$.

Second, every continuous function on a closed bounded interval is Riemann integrable. The statement implies that $\int_{\delta}^1\sin(1/x)dx$ exists for every $\delta>0$.

From Riemann's criterion, we can choose a dissection of $[\delta,1]$ such that the upper and lower sums differ by at most $\epsilon/2$.

On $[0,\delta]$, the largest difference between upper and lower sums is $2$. So, if we take $\delta=\epsilon/4$, then the complete dissection has upper and lower sums that differ by at most $\epsilon/2+2\epsilon/4=\epsilon$. And we are done.

APPROACH 2:

Substitute $x=1/u$. Then,

\begin{align} \int_{-1}^{1}\sin(1/x)dx& \color{red}{\equiv} \lim_{\epsilon\to 0^{-}}\int_{-1}^{\epsilon}\sin(1/x)dx+\lim_{\nu\to 0^{+}}\int_{\nu}^{1}\sin(1/x)dx \tag 1\\\\ &=-\lim_{\epsilon\to 0^{-}}\int_{-1}^{1/\epsilon}\frac{\sin u}{u^2}du-\lim_{\nu\to 0^{+}}\int_{1/\nu}^{1}\frac{\sin u}{u^2}du \tag 2\\\\ &=-\lim_{B_1\to \infty}\int_1^{B_1}\frac{\sin u}{u^2}+\lim_{B_2\to \infty}\int_1^{B_2}\frac{\sin u}{u^2}\tag 3 \end{align}

Note that both integrals converge (easily shown by the comparison test). Thus, the integral of interest converges to $0$.

NOTES:

$(1)$ The right-hand side of $(1)$ equation is one way of expressing the left-hand side, provided that the left-hand side exists.

$(2)$ In going from $(1)$ to $(2)$, we enforce the substitution $x=1/u$.

$(3)$ In going from $(2)$ to $(3)$, we made the substitution $u\to -u$ in the first integral, then let $-1/\epsilon=B_1$ in the first integral and $1/\nu =B_2$ in the second integral.

• This is a good method; although I haven't looked too much into it I know what you're doing. However, I was really hoping I could solve this problem using more "pure" definition of integrals such as the Riemann or Darboux definitions – David South Jun 6 '15 at 22:39
• Could you prove the first equality? I feel like we're using what we're trying to prove somewhere underneath. – Adayah Jun 6 '15 at 22:43
• @Adayah The first equality will always work provided the left side makes sense, but what is written there does not really prove that the left side necessarily makes sense. – Ian Jun 6 '15 at 22:45
• Thank you. This transformation seems to facilitate. The "problem" is the discontinuity of $\sin(1/x)$ at $x=0$. The rapid oscillations mean that for any $\epsilon >0$, there exists an $x$ (actually infinitely many of them) in $[-\epsilon,\epsilon]$ such that $|\sin (1/x)|=1$. – Mark Viola Jun 6 '15 at 22:48
• Improper Riemann integral is usually defined to overcome the impossibility of defining an ordinary Riemann integral, i.e. when the integrated function is not bounded. But here the function is bounded, so the red $\color{red}{\equiv}$ is not the definition of the left side. – Adayah Jun 6 '15 at 23:03