Solution
= Solution
Substitute $t=\log x$, with $dx=e^t\,dt$. The nonnegative integral becomes
$$
\int_0^\infty t e^{-(\alpha-1)t}\,dt=\frac1{(\alpha-1)^2},
$$
by scaling and <integration by parts>. The exponential decays because $\alpha>1$, with zero boundary term at infinity, proving integrability and
$$
\boxed{\int_1^\infty x^{-\alpha}\log x\,dx=(\alpha-1)^{-2}}.
$$