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Calculus and Beyond Homework Help
Comparison of asymptotic behavior of two Lebesgue integrals.
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[QUOTE="Jezuz, post: 3125304, member: 43744"] [h2]Homework Statement [/h2] The problem is to show that [tex] \int_1^x \frac{| \sin(t) |}{t} dt \underset{x \rightarrow \infty}{\sim} \frac{2}{\pi} \log(x). [/tex] The integral is (in case it is important) a Lebesgue integral. [h2]Homework Equations[/h2] A theorem is stated which says (I do not currently have the book in front of me so I am unsure about the details): Given an two measurable functions [tex]f , g : [a,b) \rightarrow \mathbb R[/tex], we have that if both functions are integrable on any interval [a,c] where c<b and where g is not integrable (on the full interval [a,b) ) we have that [tex] f \underset{b}{\sim} g [/tex] gives [tex] \int_{a}^{x} f(t) dt \underset{x \rightarrow b}{\sim} \int_{a}^{x} g(t) dt . [/tex] The definition of the equivalence relation is that [tex] f \underset{b}{\sim} g [/tex] iff [tex] \lim_{x \rightarrow b} f(x)/g(x) = 1 [/tex]. (Alternatively that f - g = o(g) at b). [h2]The Attempt at a Solution[/h2] Since I would like to apply this theorem I have rewritten the right hand side as an integral. [tex] \log(x) = \int_1^\infty \frac{1}{t} dt [/tex]. Then I would like to show that this integrand is asymptotically equivalent to the integrand on the left hand side. However, this does not seem to be the case since [tex] \frac{f(x)}{g(x)} = \frac{ \pi | \sin x|}{2} [/tex] which does not tend to 1! Have I missed something? Perhaps it is not this theorem I should use. Or I might have misinterpreted the definition of the equivalence relation of functions (it is not very clearly stated in the book as it is only given for sequences). I started working on this problem and now I can not stop! I need to find the solution. Any help is greatly appreciated. [/QUOTE]
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Calculus and Beyond Homework Help
Comparison of asymptotic behavior of two Lebesgue integrals.
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