 Logarithmic moments of a power on the unit interval

For real $m>-1$ and integer $n\ge 0$, start from ${\int }_0^1{x}^{m}dx=1/(m+1)$ and differentiate $n$ times with respect to $m$. Each derivative adds one factor $\log x$, givingTo 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.