An integral resulting in harmonic number

[Tolaso J Kos]

Let $m \in \mathbb{N}$. Prove that:

$$\int_0^1 x^m \log(1-x)\, {\rm d}x=-\frac{\mathcal{H}_{m+1}}{m+1}$$

Generalization

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Can you find an explicit formula for the integral

$$\int_0^1 x^m \log^n (1-x)\, {\rm d}x$$

if $m,n \in \mathbb{N}$. Stirling numbers will come in handy.

[Riemann]

Let $m \in \mathbb{N}$. Prove that:

$$\int_0^1 x^m \log(1-x)\, {\rm d}x=-\frac{\mathcal{H}_{m+1}}{m+1}$$

Here is a solution. Recalling the fact that:

\begin{equation} \mathcal{H}_m = \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ] \end{equation}

Proposition: It holds that:

$$\mathcal{H}_m = \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ]$$

Proof:

\begin{align*}
\mathcal{H}_m &= \int_{0}^{1}\frac{1-x^m}{1-x}\, {\rm d}t\\
&= \int_{0}^{1}\left ( 1-x^m \right ) \sum_{k=0}^{\infty} x^k \, {\rm d}x\\
&= \sum_{k=0}^{\infty} \int_{0}^{1}\left ( x^k- x^{m+k} \right )\, {\rm d}x\\
&= \sum_{k=0}^{\infty} \left [ \frac{1}{k+1} - \frac{1}{m+k+1} \right ] \\
&= \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ]
\end{align*}

Now for our integral we have successively:

\begin{align*}
\mathcal{J} &=\int_{0}^{1} x^m \log(1-x) \, {\rm d}x \\
&= - \int_{0}^{1}x^m \sum_{n=1}^{\infty} \frac{x^n}{n} \, {\rm d}x \\
&= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1}x^{m+n} \, {\rm d}x\\
&= - \sum_{n=1}^{\infty} \frac{1}{n \left ( n+m+1 \right )}\\
&= - \sum_{n=1}^{\infty}\left[\frac{1}{(m+1)n} - \frac{1}{(m+1) \left ( m+n+1 \right )} \right]\\
&= - \frac{1}{m+1} \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \frac{1}{n+m+1} \right ]\\
&\overset{(1)}{=} - \frac{\mathcal{H}_{m+1}}{m+1}
\end{align*}

For the generalization I seem to recall that $\log^n (1-x)$ has a Taylor series of the form involving stirling numbers and that the sum begins from $n$, that is $k=n$ to $\infty$ but I cannot seem to remember the exact formula .. Anyone wishes to try or at least present the formula ?