# Improper integral (comparison)

Normally one would increase/decrease the numerator/denominator and thus deduce whether it is convergent and divergent. This one however is not as simple as that. Here's a proposition of a solution:

We know $x^2$ dominates over $\ln x$, so the integrand can be written

$$\frac 1{x^2 - \ln x} = \frac 1{x^2\left(1 - \frac {\ln x}{x^2} \right)}$$

Since $x^2$ dominates $\ln x$ (i.e. $\frac {\ln x}{x^2} \to 0$ when $x \to \infty$), then surely $\frac {\ln x}{x^2} < \frac 12$ whenever $x$ is greater than some $x_0$. Therefore we have

$$\frac 1{x^2\left(1 - \frac {\ln x}{x^2} \right)} \leq \frac 1{x^2\left(1 - \frac 12 \right)}$$

Now we can integrate from $x_0$ to $\infty$ and show that

$$\int_{x_0}^\infty \frac {\mathrm{d}x}{x^2 - \ln x}$$

converges. But I do not understand how this integral from $x_0$ to $\infty$ implies that the same integral from $1$ to $\infty$ also must converge. Someone care to explain?

$$\forall A\quad\int_{1}^A \frac {\mathrm{d}x}{x^2 - \ln x}= \int_{1}^{x_0} \frac {\mathrm{d}x}{x^2 - \ln x}+ \int_{x_0}^A \frac {\mathrm{d}x}{x^2 - \ln x}$$ so what happens when $A\to\infty$?
• Oh, there we go! The integral far right converges, so we need to ascertain the the first integral, the one from 1 to $x_0$, does not diverge. Jan 11, 2014 at 20:52