I have shown that $f(x)=\sin(1/x)$ is Riemann integrable on $(0,1]$, but I am wondering if $f(x)=(\sin (1/x))^4$ is Riemann integrable on $(0,1]$?
It isn't hard to show that $\sin(1/x)$ is Riemann integrable on $(0,1]$.
Proof.
Given $\epsilon > 0$, consider the partition $[0, \epsilon]$ and $[\epsilon, 1]$. Since $f$ is bounded and continuous on $[\epsilon, 1]$ it is integrable, so there is a partition $P_1$ such that $U – L < \epsilon$. On $[0, \epsilon]$, the most $U – L$ could be (for any partition, $P_\epsilon$) is $2 \epsilon$ (since $\max_{x \in \mathbb{R}} \sin x = 1$, $\min_{x \in \mathbb{R}} \sin x = – 1$, and length of the interval is $\epsilon$). Now consider the partition of $[0, 1]$ given by $P_1$ union $P_e$ and look at $U – L$. We have $|U – L|$ on $[0, 1]$ is less than or equal to the sum of $U – L$ for the two partitions $P_1$ and $P_e$, which is less than $\epsilon + 2 \epsilon = 3 \epsilon$. So $U – L < 3\epsilon$, which means it can be made arbitrarily small. Thus $f$ is integrable on $[0, 1]$.
Is there any way I can adapt this proof to address $f(x)=(\sin (1/x))^4$?
Best Answer
The product of two Riemann integrable functions is Riemann integrable function. The product of two Riemann integrable functions is integrable. $f(x)^4$ can be written as a product of Riemann integrable functions.