# Improper Integral $\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$

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?

• Please improve your question: Your integral is not indefinite. It seems that you use $x$ with to different meanings. May 26, 2014 at 9:17
• 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 May 26, 2014 at 9:19
• 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: $$\int_{x}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{t}+1} \right) \right) dt$$ May 26, 2014 at 9:25
• I'm trying to ask if the function $$\sin( \sin (\frac{1}{\sqrt(x) + 1} ) )$$ has asymptotic equivalence with $x -> \infty$ May 26, 2014 at 9:29

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.

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.

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

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