I tried to solve it in an intuitive manner, but I am not sure if it's right or wrong. Some feedback would be lovely!
This is how I approached the problem.
Step 1: I used integration by parts.
$ \int_1^\infty \frac{ln(x)}{x^2} dx = \int_1^\infty ln(x) \frac{\mathrm d}{\mathrm d x} \big( -\frac{1}{x} \big) dx = \left.-\frac{ln(x)}{x}\right|_1^\infty + \int_1^\infty \frac{1}{x^2}dx $
Step 2: Verify if $\int_1^\infty \frac{1}{x^2}dx$ converges or not.
Fact:
$\int_1^\infty \frac{1}{x^p}dx$ for p > 1 the area under the graph is finite and the integral converges.
In our case we have: $\int_1^\infty \frac{1}{x^2}dx$ where 2 > 1 $\overset{Fact}{\implies}$ $\int_1^\infty \frac{1}{x^2}dx$ converges.
Step 3: Let's see what happens with $\left.-\frac{ln(x)}{x}\right|_1^\infty$
If we take
$\lim\limits_{b \to \infty} \left.-\frac{ln(x)}{x}\right|_1^b \implies -\lim\limits_{b \to \infty} \frac{ln(b)}{b}-0 \overset{L'Hopital}{\implies} -\lim\limits_{b \to \infty} \frac{1}{b} = 0 $
Therefore I concluded that this part: $\left.-\frac{ln(x)}{x}\right|_1^\infty$ does not affect my convergence since it's zero.
Finally from steps (1), (2) and (3) we can see that $\int_1^\infty \frac{ln(x)}{x^2} dx $ converges.
What do you think guys, did I do something wrong?
Best Answer
More simply we have that as $x \to 1$
$$ \frac{\ln(x)}{x^2}\to 0$$
and as $x \to \infty$
$$\frac{\frac{\ln(x)}{x^2}}{\frac1{x^{3/2}}} \to 0$$
then the integral converges by limit comparison test with $\int \frac1{x^{3/2}}dx$.