= Logarithmic moments of a power on the unit interval
{title2=$\int_0^1x^m(\log x)^n\,dx=(-1)^nn!/(m+1)^{n+1}$}
For real $m>-1$ and integer $n\geq0$, start from $\int_0^1x^m\,dx=1/(m+1)$ and differentiate $n$ times with respect to $m$. Each derivative adds one factor $\log x$, giving
$$
\int_0^1x^m(\log x)^n\,dx=\frac{(-1)^nn!}{(m+1)^{n+1}}.
$$
To justify <differentiation under the integral sign>, restrict $m$ to a compact interval with lower endpoint $m_0>-1$. The derivatives are dominated by $x^{m_0}|\log x|^n$, which is integrable: $x=e^{-t}$ turns its integral into $\int_0^\infty e^{-(m_0+1)t}t^n\,dt$. A <power> of $t$ is integrable against this decaying exponential. This also explains why $m=-1$ is the convergence threshold.
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