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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.

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  • $\begingroup$ It is clear that our function is continuous except at $0$, where it is not defined. (The discontinuity is removable, but we do not need that.) And "except at $0$" is definitely almost everywhere. You are proposing to use a powerful theorem (it is certainly not a standard definition of Riemann integrable). If you already have it, fine. If you do not, you may be expected to use instead the definition of the Riemann integral. $\endgroup$ – André Nicolas Apr 26 '16 at 6:03
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    $\begingroup$ That a function is Riemann integrable on an interval iff it is bounded and continuous almost everywhere. $\endgroup$ – André Nicolas Apr 26 '16 at 6:08
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    $\begingroup$ Your second approach does not look like a good idea. $\endgroup$ – velut luna Apr 26 '16 at 6:11
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    $\begingroup$ So how do I do it? $\endgroup$ – Dark_Knight Apr 26 '16 at 7:45
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    $\begingroup$ It's not Riemann integrable on [-1,1] because it's not defined on all of [-1,1]. $\endgroup$ – zhw. Apr 26 '16 at 18:14
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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 $z\geqslant0$.

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