Prove that $\int _e^\infty \frac{\ln(x)}{x^p} dx$ is divergent for $p \le1$. 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 $p&lt;1$ 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.
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\begin{align}
\int x^{\alpha}\,\dd x &amp; = {x^{\alpha + 1} \over \alpha + 1}
\\[3mm]
\int x^{\alpha}\ln\pars{x}\,\dd x &amp; =
{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 &amp; =
{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 $&gt; 1$.
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\begin{align}
\int x^{\alpha}\,\dd x &amp; = {x^{\alpha + 1} \over \alpha + 1}
\\[3mm]
\int x^{\alpha}\ln\pars{x}\,\dd x &amp; =
{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 &amp; =
{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 $&gt; 1$.