Finding if $\int_{1}^{\infty} \frac{\sin(x+2)}{x^2} \, dx $ converges, with two conficting solutions?

Question

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.

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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 $p=2 > 1$

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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

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

Answer 1

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).

Answer 2

You are writing

$$\int f(x)dx\le \int g(x)dx$$

and conclude that if $$ \int g(x)dx$$ diverges, so does $$\int f(x)dx.$$

This is wrong.