# Show that $f(x)=\sin(1/x)$ is integrable on any interval (you can define $f(0)$ to be anything).

Q. a) Show that $f(x)=\sin(1/x)$ is integrable on any interval (you can define $f(0)$ to be anything). b) Compute $\int_{-1}^1\sin(1/x)dx.$ (Mind the discontinuity)

Let $r , \delta\in R^+$ for $r>\delta$. Then, $f$ is continuous at $[-r,-\delta] \cup [\delta,r].$ Therefore, it is Riemann integrable on this interval. The problem here is to show $f$ is Riemann integrable on $[-\delta, \delta].$ I found the accepted answer solving this problem. But, although I read several times, I am still struggling to understand the answer. Especially, why the difference between upper and lower sum is within $\varepsilon/3?$ Could you elaborate on this?

• Since f is continuous in those suggested intervals it is also integrable. A function is integrable on an interval if for every $\epsilon >0$ there exists a partition of this interval such that the difference between the upper and lower sums is less than $\epsilon$. In the proof it's chosen $\frac{\epsilon}{3}$ so you can then add it up together and form a whole $\epsilon$. It's a standard technique that's being used in most epsilon-delta proofs. – DreaDk Jun 7 '18 at 9:44
• I understand what you said. But, my problem to understand this answer is that I don't understand how to deal with $[-\delta, \delta]$. – Sihyun Kim Jun 7 '18 at 10:44
• You could just try to prove that it's integrable on [-1,1] since it's the only troublesome part of any wider interval (use the same method from the link you posted). Then use the theorem that if it's integrable on [$-\delta,-1$] and on [$-1,\delta$] then it's integrable on every [$-\delta,\delta$]. Another approach is since $\sin(\frac{1}{x})$ is continuous everywhere except at $0$ and is bounded on any closed interval we can use the following theorem: math.stackexchange.com/questions/643042/… – DreaDk Jun 7 '18 at 11:58