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

Question

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?

Answer 1

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.

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

Answer 2

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.