Convergence of integral to infinity

Question

Let $p\geq1$. How can I study the convergence of the integral

$$ \int_1^{+\infty} \Bigg( \frac{\log x}{x} \Bigg)^p dx $$

without the substitution $x=\exp(t)$?

Answer 1

For $p \geq 1$, let $f_p$ be the function defined by $$ f_p\left(x\right)=\left(\frac{\ln\left(x\right)}{x}\right)^{p} $$ Then the function $f_p$ is continuous and positive on $\left[1,+\infty\right[$. We search for $x^{\alpha}$ so that $x^{\alpha}f_p\left(x\right) \underset{x \rightarrow +\infty}{\rightarrow}0$ with $\alpha>1$. $$ x^{\alpha}f_p\left(x\right)=x^{\alpha-p}\ln^p\left(x\right) $$

$\bullet$ First, suppose that $p>1$. Then it exists $\alpha \in \left]1,p\right[$. Then $$ x^{\alpha-p}\ln^p\left(x\right)\underset{x \rightarrow +\infty}{\rightarrow}0 $$ meaning that $$ f_p\left(x\right)\underset{(+\infty)}{=}o\left(\frac{1}{x^{\alpha}}\right) $$ With $\alpha>1$ you know that $\displaystyle x \mapsto \frac{1}{x^{\alpha}}$ is integrable on $\left[1,+\infty\right[$ hence $f_p$ is integrable on $\left[1,+\infty\right[$.

$\bullet$ Suppose now that $p=1$. Then for some $N>e$ $$ \int_{1}^{N}f_1\left(x\right)\text{d}x=\int_{1}^{e}f_1\left(x\right)\text{d}x+\int_{e}^{N}f_1\left(x\right)\text{d}x $$ For the second integral, we can use that for $x \in \left[e,+\infty\right[$ $$ \frac{1}{x} \leq f_1\left(x\right) $$ And you know that $x \mapsto \frac{1}{x}$ is not integrable on $\left[1,+\infty\right[$. So $f_1 : x \mapsto \frac{\ln(x)}{x}$ is not integrable on $\left[1,+\infty\right[$.

Answer 2

For $p=1$, we have $\dfrac{\log x}{x}\geq\dfrac{1}{x}$ for all $x\geq 3$, then \begin{align*} \int_{3}^{\infty}\dfrac{\log x}{x}dx\geq\int_{3}^{\infty}\dfrac{1}{x}dx=\infty. \end{align*} For $1<p<\infty$, let $\eta\in(0,1)$ to be determined later. Then $\log x\leq x^{\eta}$ for large $x$, so \begin{align*} \int_{c_{\eta}}^{\infty}\left(\dfrac{\log x}{x}\right)^{p}dx&\leq\int_{c_{\eta}}^{\infty}\dfrac{1}{x^{(1-\eta)p}}dx<\infty, \end{align*} whenever $(1-\eta)p>1$, so $\eta$ can be taken in $(0,1-1/p)$.