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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? – Harald Hanche-Olsen Sep 11 '12 at 15:35
Change variables $u=1/x$, to get this : $\int_{1}^{+\infty}\frac{\sin(u)}{u}du$. Do you know this integral? – Lucien Sep 11 '12 at 15:37
@HaraldHanche-Olsen yes indedd I meant $\sqrt{x}$. Apologies – Soham Sep 11 '12 at 15:39
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. – Harald Hanche-Olsen Sep 11 '12 at 15:43
ohkay...not an issue – Soham Sep 11 '12 at 15:52
up vote 3 down vote accepted

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** – DonAntonio Sep 11 '12 at 15:41
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 :) – Soham Sep 11 '12 at 15:43
@DonAntonio you could just delete your answer, you know. You certainly have enough reputation to get by without it. – Harald Hanche-Olsen Sep 11 '12 at 15:44
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 already happened in the past, you know. – DonAntonio Sep 11 '12 at 15:47
@DonAntonio comeon Don, cut me a slack! – Soham Sep 11 '12 at 16:01

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