Find $\displaystyle \int_0^1x^a \ln(x)^m \mathrm{d}x$ Find $$\int_0^1x^a \ln(x)^m\ \mathrm{d}x$$ where $a>-1$ and $m$ is a nonnegative integer. I did a subsitiution and changed this into a multiple of the gamma function. I get $(-1)^m m! e^a$ as the solution but Mathematica does not agree with me. Can someone confirm my answer or provide a solution?

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\begin{align}
&\color{#0000ff}{\large\int_{0}^{1}x^{a}\ln^{m}\pars{x}\,\dd x}
= \lim_{\mu \to 0^{+}}\totald[m]{}{\mu}\int_{0}^{1}x^{a}x^{\mu}\,\dd x
=\lim_{\mu \to 0^{+}}\totald[m]{}{\mu}\bracks{1 \over \mu + a + 1}
\\[3mm]&=\lim_{\mu \to 0^{+}}\bracks{\pars{-1}^{m}\,m! \over \pars{\mu + a + 1}^{m + 1}}
=\color{#0000ff}{\large%
{\pars{-1}^{m}\,m! \over \pars{a + 1}^{m + 1}}}
\end{align}
