We know that $f$ is Lebesgue Integrable iff $|f|$ is Lebesgue Integrable. We have shown by contour integration that $\int^{\infty}_0 \frac{\sin(x)}{x}$ is $\frac{\pi}{2}$ - yet we can also show using a summation trick on the integral and considering limits $n\pi$ and $0$ and letting n tend to $\infty $ -which I'm sure many are aware of- that $\int^{\infty}_0 \frac{\left|\sin(x)\right|}{x}$ tends to $\infty$ we means by our iff that $\frac{\sin(x)}{x}$ is not Lebesgue Integrable on $(0, \infty)$. This confuses me greatly- we've assigned a value to it but it's not Lebesgue Integrable? How?
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1Great question. We assign a value to it by defining its value to be $\lim_{t\rightarrow \infty}\int_0^t\sin(x)/xdx$. This limit converges despite the fact that the function is not Lebesque integrable. – Tim kinsella May 27 '16 at 10:36
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But I thought the defn of being Lebesgue Integrable is that it's Lebesgue integral is finite...? – Arcane1729 May 27 '16 at 21:07
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$f$ is lebesgue integrability iff the lebesgue integral of $|f|$ is finite, yes. – Tim kinsella May 27 '16 at 21:29
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1Basically this integral corresponds to a series which is convergent but not absolutely convergent. Lebesgue theory says "let's not think about those integrals because they have pathological properties, and assigning values to them depends on the order inwhich you sum. The integral shouldn't be sensitive to an ordering imposed on your measure space, and fact that $\mathbb{R}$ has a natural ordering isn't something you want your integration theory to see. The ordering of $\mathbb{R}$ is something peculiar to $\mathbb{R}$ and it doesnt effect it's measure-theoretic properties. – Tim kinsella May 27 '16 at 21:49
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@Timkinsella I understand now thank you - if you'd posted as an answer I'd have accepted :) but if rep points don't bother you that's fine :p – Arcane1729 May 28 '16 at 08:21
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no problem. Glad I could help. – Tim kinsella May 28 '16 at 09:17