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As title says, I need to show that the following integral converges, and I can honestly say I don't really have an idea of where to start. I tried evaluating it using integration by parts, but that only left me with an $I = I$ situation.

$$\int \limits_{0}^{1}{\frac{\sin{x}}{x} \mathrm dx}$$

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Notice that, for all $0 < x < 1$, we have \begin{eqnarray*} \left|\int_0^1 \frac{\sin x}{x} \, \operatorname{d}\!x\right| &\le& \int_0^1 \left|\frac{\sin x}{x} \right| \operatorname{d}\!x \\ \\ &\le& \int_0^1 1 \, \operatorname{d}\!x \\ \\ &\le& 1 \end{eqnarray*}

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    $\begingroup$ I'd replace the last inequality with an equality actually $\endgroup$ – Spine Feast Nov 10 '13 at 20:43
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    $\begingroup$ @DepeHb The notation I'm using means that the first integral on the top left is $\le$ than the last quantity on the bottom right. I could add an equality, but it might cause confusion. My notation does not cause confusion and is still perfectly valid. $\endgroup$ – Fly by Night Nov 10 '13 at 20:44
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    $\begingroup$ The key step is bounding $\frac{sin(x)}{x}$ so I think there should be some justification for this! $\endgroup$ – Deven Ware Nov 10 '13 at 20:58
  • $\begingroup$ @DevenWare: It is well-known that $\sin(x)\leqslant x$ for all $x\geqslant 0$. Indeed, take $f(x)=x-\sin(x)$. Then $f'(x)=1-\cos(x)\geqslant 0$. So $f(x)$ is increasing for $x\geqslant 0$. Since $f(0)=0$, it follows that $f(x)\geqslant 0$ for all $x\geqslant 0$. This is same thing as saying $x\geqslant\sin(x)$, which implies that $|\sin(x)/x|\leqslant 1$. $\endgroup$ – Prism Nov 10 '13 at 21:05
  • $\begingroup$ @Prism of course it is well known. However, this entire question is equally well known. And this is the key fact for the question, so I think it should be elaborated for the questioners sake. Your comment provides adequate justification, however. $\endgroup$ – Deven Ware Nov 10 '13 at 21:27
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Note that $\frac{\sin x}{x} = \sum_{n=1}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!} $.

Then for $|x| \le M$, we have $\left| \frac{\sin x}{x} \right| \le \sum_{n=1}^\infty \frac{x^{2n}}{(2n+1)!} \le \sum_{n=1}^\infty \frac{M^{2n}}{(2n+1)!} \le \frac{1}{M} e^M$.

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Take the function $ f(x)= \left \{ \begin{array}{cc} \frac{\sin x}{x} & x \in (0,1]\\ 1& x=0 \end{array} \right . $

We observe that this function is continuous on $[0,1]$ hence Riemann integrable on $[0,1]$

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