Finding if $\int_{1}^{\infty} \frac{\sin(x+2)}{x^2} \, dx $ converges, with two conficting solutions? Consider the problem where the following integral converges or not:
$$\int_{1}^{\infty} \frac{\sin(x+2)}{x^2} \,dx $$
I tried to solve it in two different ways but the results conflict. I am not sure why.

First Solution:
Using comparison criterion we can prove that it converges because
$$ \frac {\sin(x+2)}{x^2} \leq \frac {1}{x^2} $$
and  $$\int_{1}^{\infty} \frac{1}{x^2} dx < + \infty$$ converges as a p-intergral with $p=2 > 1$

Second Solution
$$ \frac {\sin(x+2)}{x^2} \leq \frac {x+2}{x^2} = \frac {1}{x} + \frac {2}{x^2} $$
Where this converges 
$$\int_{1}^{\infty} \frac{2}{x^2} dx$$
but this diverges
$$ \int_{1}^{\infty} \frac{1}{x} dx$$
Thus the initial integral also diverges because one part of its sum diverges


The solutions conflict and I know that something is wrong with the
  second solution. But I cannot spot what went wrong. Any ideas?


The problem is integrabilty near $0$. Since $\frac {\sin x} x \to 1$ and $ \int_0^{1} \frac  1 x dx$ does not converge, the given integral does not converge. 
It is also not true that $\int_0^{\infty} \frac 1 {x^{2}} d x<\infty$.
Answer for the edited version: The integral is convergent because $-\frac 1 {x^{2}} \leq \frac {\sin x} {x^{2}} \leq \frac 1 {x^{2}}$.
[I do not understand why you are considering $\sin (x+2)$]. 
(The question was edited again after  I posted this answer). 
