Showing $\frac{\sin(x)}{x}$ is Riemann integrable on $[0,1]$

Question

I am attempting to show that $\frac{\sin(x)}{x}$ is Riemann integrable on $[0,1]$.

As of now, I have managed to show that (1) $\frac{\sin(x)}{x}$ is Riemann integrable on $[a,1]$ for any $a \in (0,1)$. Furthermore, I have shown that (2) $\displaystyle\lim_{x\rightarrow0}{\frac{\sin(x)}{x}}=1$.

I am now unsure how to proceed. Intuitively, I feel like the result should follow form (1) and (2) above. I am struggling however to prove rigorously that $\frac{\sin(x)}{x}$ is Riemann integrable on $[0,1]$.

I would just like to add that my knowledge of Riemann integration is quite basic. Essentially, I've only been given the following definition, for a function $f$ on $[a,b]$: $f$ is Riemann integrable iff $U(f)=L(f)$, where $U(f)=\inf\{\int_{a}^{b}{\phi}:\phi\in S[a,b], \phi\geq f\}$ and $L(f)=\sup\{\int_{a}^{b}{\phi}:\phi\in S[a,b], \phi\leq f\}$. I also know that regulated functions are Riemann integrable.

Answer 1

Let

$f(x)=\frac{\sin(x)}{x}$ for $x\in (0,1] $.

$\int_0^1f$ is an improper integral which is convergent since the function $f$ has a limit at $0^+$.

If we define $g$ by

$$g(x)=f(x) \text{ if } x\ne 0$$ and $$g(0)=\lim_{0^+}f(x)=1$$

$g $ is then Riemann integrable at $[0,1]$ cause it is continuous at $[0,1]$

Answer 2

Hint:

Define $$g(x)=\left\{ \begin{array}{c c} \frac{\sin(x)}{x} & \mbox{ if } x \neq 0 \\ 1 & \mbox{ if } x=0 \\ \end{array} \right.$$

Prove that $g(x)$ is continuous on $[0,1]$ and hence Riemann Integarble.

Compare the Riemann Sums of $f$ and $g$...

Answer 3

A simple and quick way to show that the function is Riemann integerable, is that it is bounded and continuous almost everywhere. Any function on a compact interval is bounded by the EVT, and in this case, the function is clearly continuous everywhere, so its continuous a.e, so it is Riemann integerable.