TITLE: Is there a closed form for the integral $\int_0^1 x^n \log^m (1-x) \, {\rm d}x$?
QUESTION [5 upvotes]: Let $n \in \mathbb{N}$. We know that:
$$\int_0^1 x^n \log(1-x) \, {\rm d}x = - \frac{\mathcal{H}_{n+1}}{n+1}$$
Now, let $m , n \in \mathbb{N}$. What can we say about the integral
$$\int_0^1 x^n \log^m (1-x) \, {\rm d}x$$
For starters we know that $\displaystyle \log^m (1-x)=m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k$ where $s(k, m)$ are the Stirling numbers of first kind. 
Thus
\begin{align*}
\int_{0}^{1} x^n \log^m (1-x) \, {\rm d}x &=m! \int_{0}^{1}x^n \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k  \\ 
 &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \int_{0}^{1}x^{n+m} \, {\rm d}x\\ 
 &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \frac{1}{m+n+1}
\end{align*}
Can we simplify? I know that Striling numbers are related to the Harmonic number but I don't remember all identities.

REPLY [3 votes]: Another closed form follows by differentiating the beta function multiple times and applying the Faà di Bruno's formula.

Claim. For positive integers $m$ and $n$,
  $$ \mathcal{J}_{n,m} := \int_0^1 x^{n-1}\log^m (1-x) \, \mathrm{d}x = (-1)^m \frac{m!}{n} \sum_{\alpha\in I_m} \prod_{k=1}^m \frac{1}{\alpha_k!} \bigg(\frac{H_n^{(k)}}{k}\bigg)^{\alpha_k} \tag{1} $$
  where $\alpha$ runs over the set of indices
  $$I_m = \{(\alpha_1,\cdots,\alpha_m)\in\Bbb{N}_0^m : 1\cdot\alpha_1+\cdots+m\cdot\alpha_m=m\}.$$

This formula gives an almost explicit formula for $\mathcal{J}_{n,m}$ in terms of polynomial of $H_n^{(1)}, \cdots, H_n^{(n)}$ at the expense of introducing certain combinatorial object, namely $I_m$.
Proof. Notice that
$$ \int_0^1 x^{n-1}(1-x)^s \, \mathrm{d}x = \frac{(n-1)!}{(s+1)\cdots(s+n)} = (n-1)!\exp\left(-\sum_{j=1}^n \log(s+j) \right). $$
Letting $f(s) = -\sum_{j=1}^n \log(s+j) $ and applying the Faà di Bruno's formula, we have
$$ \mathcal{J}_{n,m} = (n-1)!e^{f(0)} \sum_{\alpha \in I_m} m! \prod_{k=1}^{m} \frac{1}{\alpha_k !} \bigg( \frac{f^{(k)}(0)}{k!} \bigg)^{\alpha_k}. \tag{2}$$
Plugging $f(0) = -\log n!$ and
$$ f^{(k)}(0) = \sum_{j=1}^n (-1)^k (k-1)! (s+j)^{-k} \bigg|_{s=0} = (-1)^k (k-1)! H_n^{(k)} $$
into $\text{(2)}$ and simplifying the resulting expression yields $\text{(1)}$.