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I need to calculate this improper integral. $$\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$$ How do I prove that $$ \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) $$ has an asymptotic equivalence with: $$ \frac{1}{\sqrt{x}} $$ for $x\rightarrow \infty$

And by the p-test that it diverges?

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    $\begingroup$ Please improve your question: Your integral is not indefinite. It seems that you use $x$ with to different meanings. $\endgroup$ May 26, 2014 at 9:17
  • $\begingroup$ There seemed to be a typo in my title and integral, x is replaced by 1 now and by indefinite I meant improper. Sorry, English isn't my native language $\endgroup$ May 26, 2014 at 9:19
  • $\begingroup$ Since the integral does not depend on x, it makes no sense to ask wether it looks like $ \frac{1}{\sqrt{x}}$ for $x\rightarrow \infty.\;$ Do you mean something like this: $$\int_{x}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{t}+1} \right) \right) dt$$ $\endgroup$ May 26, 2014 at 9:25
  • $\begingroup$ I'm trying to ask if the function $$ \sin( \sin (\frac{1}{\sqrt(x) + 1} ) ) $$ has asymptotic equivalence with $ x -> \infty $ $\endgroup$ May 26, 2014 at 9:29

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You don't need an asymptotic equivalence. Since for any $y\in[0,\pi/2]$ $$\sin y\geq\frac{2y}{\pi}$$ holds by convexity, $$\int_{N}^{+\infty}\sin\sin\frac{1}{\sqrt{x}+1}\,dx \geq \frac{4}{\pi^2}\int_{N}^{+\infty}\frac{dx}{\sqrt{x}+1}$$ holds for any $N$ big enough, hence the starting is divergent.

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Yes, as $x\rightarrow \infty\;$ you have the asmptotic relations $$\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) \sim \sin \left( \sin\left( \frac {1}{\sqrt{x}} \right) \right) \sim \sin\left( \frac {1}{\sqrt{x}} \right) \sim \frac {1}{\sqrt{x}} $$ because for small $z\rightarrow 0\;$ you have $\sin(z) \sim z;\;$and therefore the integral $$\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$$ diverges. You can get some better estimate with a CAS e.g.

$$\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) = \sqrt\frac{1}{x}-\frac{1}{x}+\frac{2}{3}\left(\frac{1}{x}\right)^{3/2}+O\left(\frac{1}{x^2}\right)$$

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  • $\begingroup$ How can you prove your first equivalent ? I think the way is to write : $\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) \sim \sin\left( \frac {1}{1+\sqrt{x}} \right) \sim \frac {1}{1+\sqrt{x}} \sim \frac {1}{\sqrt{x}}$ $\endgroup$
    – user146010
    May 26, 2014 at 9:56
  • $\begingroup$ Seems no great difference to your first step (at least for the simple estimate), both use the composition of continous functions. $\endgroup$ May 26, 2014 at 10:07
  • $\begingroup$ You cannot (in general) compose equivalent. $\endgroup$
    – user146010
    May 26, 2014 at 10:10

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