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

Question

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!

Answer 1

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.

Answer 2

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.

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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 $0$.

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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.