Answer
Convergent for $p\gt 1$.
Work Step by Step
$\int_{1}^{\infty}\frac{lnx}{x^{p}}dx=\int_{0}^{\infty}ue^{-u(p-1)}du=[\frac{ue^{-u(p-1)}}{p-1}-\frac{e^{-u(p-1)}}{(p-1)^{2}}]_{0}^{\infty}$
$=\frac{\infty e^{-\infty(p-1)}}{p-1}-\frac{e^{-\infty (p-1)}}{(p-1)^{2}}$
which is convergent for $p\gt 1$.
