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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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    @soham: not really. You have many alternatives here. For example, you may also think of using the $C$auchy-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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    Jeje...don't worry, @Soham.2012-09-11