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

integrationreal-analysisriemann sum

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!

Best Answer

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.

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