Use the Comparison Test Use the Comparison Test to determine for what values of $p$ the integral: $\int_{8}^{\infty} \frac{1}{x^p} \ ln(x)  \ dx$ converges. (Use interval notation.) $$\\$$
This is what I have so far:
I'm comparing to $\frac{1}{x}$, which diverges.
I'm unsure of what to do next. I know it starts to converge from 2.
I would really appreciate to your help.
 A: For any $p > 1$ choose $a$ such that $0 < a < p-1$.
We have
$$\frac{\ln x}{x^p} = \frac{\ln x^a}{ax^p} < \frac{x^a}{ax^p} = \frac{1}{ax^{p-a}}$$
Since $p-a > 1$ we have convergence for any $p > 1$.
The integral diverges if $p \leqslant 1$ by an easy comparison: $\ln x / x^p > 1/x^p$
A: Note that for $x>1,\ln { x } <x$ so $$\\ \int _{ 8 }^{ \infty  }{ \frac { \ln { x }  }{ { x }^{ p } } dx<\int _{ 8 }^{ \infty  }{ \frac { dx }{ { x }^{ p-1 } } dx }  } \\ \\ \\ \\ \\ \\ $$ and RHS converges when $p-1>1$ or $p>2$
A: Take an arbitrary $\eta >8.$ Integration by parts yields$$\int\limits_{8}^{\eta} \frac{1}{x^p}\ln{x} \,dx = -\dfrac{1}{p-1}\int\limits_{8}^{\eta}\ln{x} \, d\left(x^{-(p-1)} \right) = \\ = -\dfrac{1}{p-1}\left(\dfrac{\ln{x} }{x^{p-1}} \Bigg\vert_{8}^{\eta} - \int\limits_{8}^{\eta} \dfrac{dx}{x^p}\right).$$
Then using  L'Hospital's rule,
$$\lim\limits_{\eta\to+\infty} {\dfrac{\ln{x} }{x^{p-1}} \Bigg\vert_{8}^{\eta}} = \begin{cases} 0 - \dfrac{\ln{\eta} }{\eta^{p-1}}  \  \text{(exists and is finite) for } \  p > 1, \\ +\infty, \   \text{if} \  p \leqslant 1  .
\end{cases}$$
The limit $$\lim\limits_{\eta\to+\infty}\int\limits_{8}^{\eta} \dfrac{dx}{x^p}$$ exists for $p>1$ and is infinite, if $ p \leqslant 1  .$
