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?
Best Answer
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.