It is those puzzeling improper integrals that I can't get my head around....
Does the (improper) integral $\frac 1{x^2}$ from 1 to $\infty$ coverges because it is converging "fast" or because it has converging anti-derivative? in case of the former, what do you mean by "fast"?
Likewise the integral $\frac 1x$ from $1$ to $\infty$ is divergent, it that because it does not appraoch the x-axis "fast enought"? Eventaully it does approach the x-axis as we take the limit as x $\to \infty$. For me it is more logical to say that is has an anti-derivative (namely, $\ln(x)$) that blows up as $x$ $\to \infty$
Which reasoning is correct and why?
The comparison tests (either direct or via a limit) are concrete, rigorous statements expressing the intuition behind the statement that $$\int_a^{\infty} f(x) \, dx$$ converges if $f(x)\rightarrow 0$ as $x\rightarrow\infty$ "fast enough".
The direct comparison test, for example, states that if $0\leq f(x)\leq g(x)$ and if $$\int_a^{\infty} g(x) \, dx$$ converges then $$\int_a^{\infty} f(x) \, dx$$ must also converge. Now, if $0\leq f(x) \leq g(x)$ then I think it's reasonable to say that $f(x)\rightarrow 0$ at least as fast as $g(x)$ so, again, this is one expression of your intuition.
Of course, one thing that's nice about the comparison test, is that we don't need to find an anti-derivative to apply it. For example, $$\int_1^{\infty} \frac{\sin^{8/3}(x)}{x^2}\,dx$$ converges by comparison with $1/x^2$, even though we'll probably never find a nice anti-derivative for the integrand.
It might be worth mentioning that we often compare to functions of the form $1/x^p$ and the $p$-test provides another expression of your statement. That is,
$$\int_1^{\infty} \frac{1}{x^p} \, dx$$ converges if and only if $p>1$. Furthermore, the larger $p$ is, the faster $1/x^p \rightarrow 0$ in a quantitative sense that can be made precise with inequalities.