Is $f(x)=(1/x)\sin(1/x)$ for $x\not=0$, $f(x)=0$ for $x=0$ Riemann Integrable for $x \in [0,1]$?

Question

Is $f(x)=(1/x)\sin(1/x)$ for $x\not=0$, $f(x)=0$ for $x=0$ Riemann Integrable for $x \in [0,1]$?

I cannot prove this function is integrable because it's not bounded and it doesn't have one sided limit on 0. But I don't know how to prove it's not integrable.

Answer 1

Any Riemann integrable function on a bounded interval $[a, b]$ is bounded.

Suppose that a function $f:[a, b]\to\mathbb{R}$ is unbounded and is integrable. Then by the definition of Riemann integrability, for any $\varepsilon>0$ there is integer $N>0$ such that for $n\ge N$ and any choice of $x_i^\ast\in[x_i, x_{i+1}]$ where $x_ i = a+\frac{b-a}{n}i$, $\Delta x = \frac{b-a}{n}$, we have $$L-\varepsilon < \sum_{k=1}^{n} f(x_i^\ast)\Delta x < L+\varepsilon.$$ But by unboundedness, you can pick an interval $[x_k, x_{k+1}]$ that $f$ is unbounded(why is it?) and making re-choice of $x_k^\ast$, you can make contradiction to the inequality.

Answer 2

We will find a system of partitions $\{P_n\}_n$ for which the limit in the definition of the integral fails to exist. Take $x_k=\frac{2}{\pi+4k\pi}.$ Then, $0<x_{k+1}<x_k<1,\ f(x_k)=\frac{\pi+4k\pi}{2}$ and $x_k-x_{k+1}=\frac{8}{\pi(4k+5)(4k+1)}.$ Then, for the partitions $P_n=\{0,x_{n-2},x_{n-1},\cdots, x_2,1\}$ we have the Riemann sum $\sum_{k=1}^n\left(\frac{\pi+4k\pi}{2}\right)\left(\frac{8}{\pi(4k+5)(4k+1)}\right)=4\sum_{k=1}^n\frac{1+4k}{(4k+5)(4k+1)},$ which consitutes a divergent series.