Solution
= Solution
Since $\log x\leq0$ on $(0,1]$, substitute $x=e^{-s}$ to obtain
$$
\int_0^1|x^{t-1}\log x|\,dx=\int_0^\infty se^{-ts}\,ds=\frac1{t^2}<\infty\qquad(t>0).
$$
The <functions> are measurable, so this proves <Lebesgue integrability>. Also $0<e^{-x}\leq1$ gives $|h_t(x)|\leq|g_t(x)|$. Hence \b[$h_t$ is integrable on $(0,1]$ as well].