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Test the convergence of $\int_{0}^{1}\frac{\sin(1/x)}{\sqrt{x}}dx$

What I did

  1. Expanded sin (1/x) as per Maclaurin Series
  2. Divided by $\sqrt{x}$
  3. Integrate
  4. Putting the limits of 1 and h, where h tends to zero

So after step 3, I get something like this:

$S= \frac{-2}{\sqrt{x}}+\frac{2}{5\cdot 3! x^{5/2}}- \frac{2}{9 \cdot 5!x^{9/2}}+\frac{2}{13\cdot 7!x^{13/2}}-...$ Putting Limits: $I=S(1)-S(0)$ But I am stuck at calculating $S(0)$

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    So I take it you meant to write $\int_{0}^{1}\frac{\sin(1/x)}{\sqrt{x}}dx$ in the first line?2012-09-11
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    Change variables $u=1/x$, to get this : $\int_{1}^{+\infty}\frac{\sin(u)}{u}du$. Do you know this integral?2012-09-11
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    @HaraldHanche-Olsen yes indedd I meant $\sqrt{x}$. Apologies2012-09-11
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    I fixed a problem in your edit. Apparently, you did two quick edits in a row, and the second one got lost due to simultaneous editing. I didn't really delete your line about it not being homework.2012-09-11
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    ohkay...not an issue2012-09-11
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    Does it help that $\sin({x})\le x$?2012-09-11
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    yes indeed it does... much thanks.After this observation doesnt the problem become obvious to the point of trivial? interesting username2012-09-11
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    @soham: not really. You have many alternatives here. For example, you may also think of using the Cauchy-Schwarz inequality for integrals. Anyway, I think the answers already offered are some good proofs for your problem.2012-09-11

2 Answers 2

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Change variables $u = \frac{1}{x}$. Then: $$ \int_0^1 \frac{\sin(1/x)}{\sqrt{x}} \mathrm{d}x= \int_1^\infty \sqrt{u} \sin(u) \frac{\mathrm{d}u}{u^2} =\int_1^\infty \frac{\sin(u)}{u^{3/2}}\mathrm{d}u $$ The latter integral is absolutely convergent.

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$$y:=\frac{1}{x}\Longrightarrow dy=-\frac{dx}{x^2}\Longrightarrow \int_0^1\frac{\sin 1/x}{x}\,dx=\int_\infty^1\frac{\sin y}{1/y}\left(-\frac{dy}{y^2}\right)=$$

$$=\int_1^\infty\frac{\sin y}{y}\,dy$$

And since

$$\int_0^\infty\frac{\sin x}{x}\,dx=\frac{\pi}{2}$$

we're done

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    Well, after the editing of the OP my answer makes no more sense, as somebody decided to put $\,\sqrt x\,$ in the denominator instead of the original $\,x\,$...**sigh**2012-09-11
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    I am indeed very sorry for the typo :P Well, what can I say I am a bad editor of my own work, not a very enviable quality :)2012-09-11
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    @DonAntonio you could just delete your answer, you know. You certainly have enough reputation to get by without it.2012-09-11
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    Yes, I know, yet I decide not to for a while, in case either somebody finds my answer usable or else there's a new editing of the OP...it already happened in the past, you know.2012-09-11
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    @DonAntonio comeon Don, cut me a slack!2012-09-11
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    Jeje...don't worry, @Soham.2012-09-11