Prove that $\int _e^\infty \frac{\ln(x)}{x^p} dx$ is divergent for $p \le1$.
So my textbook divides the problem into first case $p=1$ and integrates and cases $p<1$ in which it uses integration by parts. However, isn't my method also valid (and much faster)? It is:
$$\int _e^\infty \frac{1}{x^p} dx \le\int _e^\infty \frac{\ln(x)}{x^p} dx$$
The integral on the left diverges by the so called $p$-series test for all $p \le 1$ and so does the integral on the right by the comparison test.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int x^{\alpha}\,\dd x & = {x^{\alpha + 1} \over \alpha + 1} \\[3mm] \int x^{\alpha}\ln\pars{x}\,\dd x & = {x^{\alpha + 1}\ln\pars{x} \over \alpha + 1} - {x^{\alpha + 1} \over \pars{\alpha + 1}^{2}} \\[3mm] \int {\ln\pars{x} \over x^{p}}\,\dd x & = {x^{1 - p}\ln\pars{x} \over 1 - p} - {x^{1 - p} \over \pars{1 - p}^{2}} \end{align}
In order to avoid the divergence when $x \to \infty$, it's clear that $p$ must be $> 1$.