Convergence or divergent of improper integral. [closed]

Determine if this integral converges or diverges: $$\int_1^\infty\dfrac{\ln^3 x}{x}dx$$

I thought of comparison test but had no idea how to do it, so I tried to evaluate and it showed that this integral is diverge.

Is it a correct way to solve by evaluating? If not, how do I do the comparison test?

Solution 1:

The integral $\int_1^e \frac{\ln^3 x}{x}dx$ is convergent and for $x\geqslant e$, $$ \frac{\ln^3 x}{x}\geqslant \ln^2 e\frac{\ln x}{x}=\frac{\ln x}{x}:=g(x) $$ and the divergence of $\int_e^\infty \frac{\ln^3 x}{x}dx$ follows from the observation that the antiderivative of $g$ is $\left(\ln x\right)^2/2$.

Solution 2:

Notice that $$\int_1^{\infty}\frac{\ln(x)^3}{x}\,\mathrm{d}x=\int_1^{\infty}\ln(x)^3\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)\,\mathrm{d}x=\int_0^{\infty}x^3\,\mathrm{d}x$$ by using the change of variables theorem. What this means is that the leftmost integral converges if and only if the rightmost one converges. Now, ask yourself, does the rightmost integral converge? You figure it out, and that will allow you to answer your question. Hopefully, this is a helpful hint.