Improper integral of $\frac{\ln x}x$

Question

Find

$$\int_e^{\infty}\frac{\ln x}{x}\ dx$$

$A.\ \dfrac12$

$B.\ \dfrac{e^2}{2}$

$C.\ \dfrac{\ln(2e)}{2}$

$D.$ DNE (Does not exist)

I tried doing this and this is where I've gone so far:

$$\lim \limits_{b \to \infty}\int_e^b\frac{\ln(x)dx}{x}= \lim \limits_{b \to \infty} \left[\frac{\ln^2(x)}{2}\right]^b_e = \lim \limits_{b \to \infty} \frac{\ln^2(x)}{2} - \frac12 = \infty - \frac12$$

This doesn't match any of my answer choices. Either I'm doing something wrong or I just don't understand, help please.

Answer 1

You can use a comparison to show that a smaller integral diverges:

$$ \left.\int\limits_\epsilon^\infty \frac{1}{x}dx = \ln(x)\right|_\epsilon^\infty \longrightarrow \infty $$

Since $\frac{1}{x} \leq \frac{\ln(x)}{x}$ for all $x \geq e$ (because $\ln(x) \geq 1$ when $x \geq e$), we have:

$$ \int\limits_e^\infty \frac{1}{x}dx \leq \int\limits_e^\infty \frac{\ln(x)}{x}dx $$

Since the smaller integral diverges, the larger one surely does (note that $\int_\epsilon^e \frac{1}{x}dx$ and $\int_\epsilon^e \frac{\ln(x)}{x}dx$ are both finite, so although $\frac{1}{x} \geq \frac{\ln(x)}{x}$ when $0 < \epsilon \leq x \leq e$, it's irrelevant because those two integrals are both finite).

Answer 2

DNE: Does not exist? An improper integral in this case is actually a limit. If it is $\infty$, then it does not exist.