Does $\int_1^\infty \frac{\ln(x)}{x^2} dx $ converge or diverge?

Question

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?

Answer 1

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 $\int \frac1{x^{3/2}}dx$.

Answer 2

I like to also post an answer here to give a bit more intuition.

Let me first start off with saying that your proof is perfectly fine (although the typesetting might not be entirely optimal).

So, the first thing you should try to remember when you are solving this exercise is that your intuition should tell you that $\log(x)$ diverges rather slowly to infinity. In fact, $x^a$, where $a>0$, diverges quicker to infinity, as you can see from evaluating the limit $\log(x)/x^a$ for $x$ tending to infinity. To be more specific, we could conclude from this that for any $a>0$, we can find an $X_a>0$ such that for all $x>X_a$, we have that $$ \frac{\log x}{x^2} \leq \frac{x^a}{x^2}. $$ In particular, we could choose $a=1/2$ but any will do.

The second thing you could do is trying to use this fact in this exercise. For example, we know that $$\int_1^{\infty} \frac{x^a}{x^2}\,\mathrm{d} x$$ is finite if $a<1$. So, this limits our choice for $a$ from $(0,\infty)$ to $(0,1)$ but we still have enough choice here.

Now, combine these two facts to solve this exercise. So we pick an $0<a<1$ and we split up the integral in two parts: the first finite part and its tail. Note that its tail usually determines whether or not the integral will be finite. We get $$\int_1^{\infty} \frac{\log x}{x^2}\,\mathrm{d} x=\int_1^{X_a} \frac{\log x}{x^2}\,\mathrm{d} x+\int_{X_a}^{\infty} \frac{\log x}{x^2}\,\mathrm{d} x\leq\int_1^{X_a} \frac{\log x}{x^2}\,\mathrm{d} x+\int_{X_a}^{\infty} \frac{x^a}{x^2}\,\mathrm{d} x.$$ Finally we say that both integrals are finite. The first one because we are integrating a continuous function on an interval of finite length. The second one because $a < 1$ as we concluded earlier.