I need to show that the function $\frac{\sin x}{x}$ is Riemann integrable on $[-1 , 1]$.
A function is called Riemann integrable if and only if it is bounded and continuous almost everywhere on its domain. The boundedness condition is satisfied. To see the continuity part, the only point which seems to cause problem is $x=0$.Now, I haven't showed that the function is continuous in the entire domain except at $0$ .
Also, I had a second approach in mind. I can write $$\int_{-1}^{1} \frac{\sin x}{x} = \lim_{\epsilon \rightarrow 0} \Big[\int_{-1}^{\epsilon}\frac{\sin x}{x} + \int_{\epsilon}^{1} \frac{\sin x}{x} \Big]$$
But I don't know how to proceed from here.
Best Answer
Let $f:[-1,1]\to\mathbb R$ be defined by $$f(x)=\begin{cases}\dfrac{\sin x}x,& x\ne 0\\ 0,&x=0\end{cases}$$ and $f_n:[-1,1]\to\mathbb R$ by $$f\chi_{\left[-1,-\frac1{n+1}\right]\cup\left[\frac1{n+1},1\right]}. $$ Then $f_n\to f$ a.e. and $|f_n|\leqslant 1$, so by dominated convergence $$\int f\ \mathsf d m = \lim_{n\to\infty} \int \ f_n\ \mathsf d m = \lim_{n\to\infty}\left[ 2 \mathsf{Si}(1) - 2 \mathsf{Si}\left(\frac1{n+1} \right)\right]= 2\mathsf{Si}(1) $$ where $\mathsf{Si}$ denotes the sine integral defined by $$z\mapsto \int_0^z \frac{\sin t}t\ \mathsf dt $$ for $z\geqslant0$.