= Solution
For $m>-1$, integrate the base <power> to obtain $J(m)=\int_0^1x^m\,dx=1/(m+1)$. Since $\partial_m^n x^m=x^m(\log x)^n$, differentiating $n$ times gives
$$
\boxed{\int_0^1x^m(\log x)^n\,dx
=\frac{d^n}{dm^n}\frac1{m+1}
=\frac{(-1)^nn!}{(m+1)^{n+1}}}.
$$
Here $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x|^n$, whose <integral> becomes $\int_0^\infty e^{-(m_0+1)t}t^n\,dt$ under $x=e^{-t}$ and is finite. The <dominated convergence theorem> establishes the needed <differentiation under the integral sign> and the <logarithmic moments of a power on the unit interval>.
Back to article page
