proving convergence of $\int_1^{\infty} \frac{\ln^5(x)}{x^2}dx$

Question

I wish to prove that the improper integral:

$$\int_1^{\infty} \frac{\ln^5(x)}{x^2}dx$$

converges.

This can be solved using integration by parts multiple times, but I'm sure there is an easier way.

I tried comparing this integral to $\frac{1}{x^n}$ for different $n$ but couldn't get to a solution.

Any suggestions?

Answer 1

Substiute $x=e^u$: $$\int_1^{\infty} \frac{\ln^5 x}{x^2}dx=\int_0^{\infty} u^5 e^{-u}\,du= \Gamma(6)=5!=120<\infty $$

Answer 2

Note that $\ln^5(x)$ grows more slowly then $\sqrt{x}$. Therefore, there exists an $x_0$ such that for all $x>x_0$, $\ln^5(x)< \sqrt{x}$. We can write

$$\int_1^{\infty }\frac{\log^5(x)}{x^2}= \int_1^{x_0}\frac{\log^5(x)}{x^2}+\int_{x_0}^{\infty }\frac{\log^5(x)}{x^2}$$

$$\leq \int_1^{x_0}\frac{\log^5(x)}{x^2} + \int_{x_0}^{\infty }\frac{\sqrt{x}}{x^2}$$

and we see that both terms are finite.

(Edit: if you are curious, $x_0=3.5\times 10^{15}$ suffices).

Answer 3

For any $x\geq 0$ we have $\log(1+x)\leq x$ by concavity, hence $$ \int_{1}^{+\infty}\frac{\log^5(x)}{x^2}\,dx \stackrel{x\mapsto z^6}{=}6\int_{1}^{+\infty}\frac{6^5 \log^5(z)}{z^{7}}\,dz = 6^6 \int_{0}^{+\infty}\frac{\log^5(u+1)}{(u+1)^7}\,du $$ is bounded by $$6^6 \int_{0}^{+\infty}\frac{u^5}{(u+1)^7}\,du = 6^5. $$