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I don't really know how I would actually show this. The only thing I can think of is to look at the graph of the function to see that it is convergent. However, how would I do it algebraically?

$$ \int_{0}^{1} \frac{\sin(x)dx}{x} $$

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Integrate by parts and see what you get. You should be able to easily prove convergence from there. – TenaliRaman Nov 20 '12 at 17:01
@TenaliRaman: Thanks, will try that. – Curtain Nov 20 '12 at 17:12
up vote 1 down vote accepted

The function $\,\dfrac{\sin x}{x}\,$ is continuous and bounded on $\,(0,1]\,$ , and the discontinuity point at zero is removable, so the integral exists.

In fact, this wouldn't usually be considered an improper integral but in fact a proper, definite one.

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How is it removable? The limit when x tends to zero is 1, but that is not the function value, or am I wrong? – Curtain Nov 20 '12 at 17:17
Since $$\lim_{x\to 0}\frac{\sin x}{x}\,\,\text{exists finitely}$$ is exactly the definition of "removable discontinuity point" when the function isn't defined at $\,x=0\,$ (or when it has a value there different from the limit's), then we have a removable point here. – DonAntonio Nov 20 '12 at 17:21
Yes, of course. Thanks! – Curtain Nov 20 '12 at 17:29

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