how to show $\int_{1}^{\infty}\frac{(\log x)^a}{x^p}\,dx$ is improper integrable

Question

how to show $\int_{1}^{\infty}\frac{(\log x)^a}{x^p}dx$, if $p \gt 1$ and $a \gt 0$???. I can show this is not improper integrable in case $p \lt 1 $, but I was stuck in showing the other case. Could you let me know how to handle it??

Answer 1

Do some asymptotic analysis:

Choose $\varepsilon>0$ such that $1<q=p-\varepsilon$, and rewrite the integrand as $$\frac{\log^ax}{x^p}=\frac{\log^ax}{x^\varepsilon}\frac 1{x^{q}}=o\Bigl(\frac1{x^q}\Bigr),$$ and the latter function has a convergent integral on $[1,\infty)$ since $q>1$.