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?
 A: Let $F(a,m)$ the given integral. By integration by parts we have
$$F(a,m)=-\frac{m}{a+1}\int_0^1x^{a}\ln(x)^{m-1}dx=-\frac{m}{a+1}F(a,m-1)$$
and by simple induction we get
$$F(a,m)=\frac{(-1)^mm!}{(a+1)^m}F(a,0)=\frac{(-1)^mm!}{(a+1)^{m+1}}$$
A: I get
$$\begin{align}
\int_0^1 x^a (\ln x)^m\,dx &= \int_0^\infty e^{-at} (-t)^m e^{-t}\,dt\tag{$x = e^{-t}$}\\
&= (-1)^m \int_0^\infty t^m e^{-(a+1)t}\,dt\\
&= \frac{(-1)^m}{(a+1)^{m+1}} \int_0^\infty u^m e^{-u}\,du \tag{$u=(a+1)t$}\\
&= \frac{(-1)^m m!}{(a+1)^{m+1}}.
\end{align}$$
Does that agree with Mathematica's result?
A: $\newcommand{\+}{^{\dagger}}%
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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}
