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?

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