How can you prove this equality? I am trying to figure out the following equality, but cannot seem to get anywhere. I tried integrating by parts, but that blew up when I set u = (log x)^n and tried to take log (0). I also tried differentiating the right side but got stuck when I did not know the derivative of $n!$
How can we prove that the following is true:
$$
\int_0^1 x^a(\log x)^n \,dx= \frac{(-1)^n(n!)}{(a+1)^{n+1}}
$$
Any suggestions would be greatly appreciated.
 A: Applying integration by parts:
$$
\int_0^1 x^a(log x)^ndx = \lim_{b\to 0^+}\left[(\log(x))^{n}\frac{1}{a+1}x^{a+1}\Big|_{b}^1\right]-\frac{n}{a+1}\int_0^1 x^{a}(\log x)^{n-1}dx
$$
Now if $a>-1$, you can show with L'Hopital's rule that
$$\lim_{b\to 0^+}\left[(\log(b))^{n}b^{a+1}\right] = 0$$
Hence
$$\int_0^1 x^a(log x)^ndx = -\frac{n}{a+1}\int_0^1 x^{a}(\log x)^{n-1}dx$$
See the inductive pattern now? Why you will have $n$ factors of $ (-1)$ when you're finished?
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{0}^{1}x^{a}\ln^{n}\pars{x}\,\dd x
     = {\pars{-1}^{n}\,n! \over \pars{a + 1}^{n+1}}:\ {\large ?}}$

With $\ds{a > -1}$:
  \begin{align}
\color{#00f}{\large\int_{0}^{1}x^{a}\ln^{n}\pars{x}\,\dd x}&=
\lim_{\mu \to 0}\partiald[n]{}{\mu}\int_{0}^{1}x^{a + \mu}\,\dd x
=
\lim_{\mu \to 0}\partiald[n]{}{\mu}\pars{1 \over a + \mu + 1}
=
\partiald[n]{}{a}\pars{1 \over a + 1}
\\[3mm]&=
\partiald[n - 1]{}{a}\pars{-1 \over \pars{a + 1}^{2}}
=
\partiald[n - 2]{}{a}\pars{2 \over \pars{a + 1}^{3}}
=
\partiald[n - 3]{}{a}\pars{-3\times 2 \over \pars{a + 1}^{4}}
\\[3mm]&=\cdots=
\partiald[n - k]{}{a}\pars{\pars{-1}^{k}k! \over \pars{a + 1}^{k + 1}}
=\cdots=\color{#00f}{\large{\pars{-1}^{n}\,n! \over \pars{a + 1}^{n + 1}}}
\end{align}

A: Note that $x^a \log^n(x) = \dfrac{\partial^n}{\partial a^n} x^a$, so you just need to start with $\int_0^1 x^a\ dx$ and take some derivatives.   Of course you do need $a > -1$ for this to converge.
