Evaluate \(\int_{0}^{1} x^{a} \mathrm{ln}^{k}(x) dx\)

This problem was inspired by a simpler problem posed at Mathematics Stack Exchange.

Consider
\begin{equation}
\int\limits_{0}^{1} x^{a} dx = (a+1)^{-1}
\end{equation}

\begin{equation}
\frac{\partial}{\partial a}\int\limits_{0}^{1} x^{a} \mathrm{ln}(x) dx = \frac{\partial}{\partial a} (a+1)^{-1}
= -(a+1)^{-2}
\end{equation}

Repeat until the pattern emerges to find the solution to our problem
\begin{equation}
\int\limits_{0}^{1} x^{a} \mathrm{ln}^{k}(x) dx = \frac{(-1)^{k} k!}{(a+1)^{k+1}}
\end{equation}

We also obtain the solution of a simpler problem
\begin{align}
\int\limits_{0}^{1} \mathrm{ln}^{k}(x) dx &= \lim_{a \to 0} \int\limits_{0}^{1} x^{a} \mathrm{ln}^{k}(x) dx \\
&= \lim_{a \to 0} \frac{(-1)^{k} k!}{(a+1)^{k+1}} \\
&= (-1)^{k} k!
\end{align}

Here is a more efficient method of evaluating the original integral inspired by a solution at the MSE link referenced above.
\begin{align}
\tag{a}
\int\limits_{0}^{1} x^{a} \mathrm{ln}^{k}(x) dx
&= (-1)^{k} \int\limits_{0}^{\infty} z^{k} \mathrm{e}^{-(a+1)z} dz \\
\tag{b}
&= \frac{(-1)^{k}}{(a+1)^{k+1}} \int\limits_{0}^{\infty} y^{k} \mathrm{e}^{-y} dy \\
&= \frac{(-1)^{k} \Gamma(k+1)}{(a+1)^{k+1}}
\end{align}

a. \(z=-\mathrm{ln}(x)\)

b. \(y=(a+1)z\)

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