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## Homework Statement

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.

## Homework Equations

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).

## The Attempt at a Solution

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.

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