TITLE: Evaluate $\int_{0}^1 x^{p}(\log x)^q dx$
QUESTION [6 upvotes]: Evaluate 
$$\int_{0}^1 x^{p}(\log x)^q dx$$
for $p \in \mathbb{N}$ and $q \in \mathbb{N}$.

REPLY [3 votes]: For $\mathbb{R} \ni p>-1$ and $q \in \mathbb{R}$ by setting $x = e^{-t/(p+1)}$ we get
$$\int_0^1 x^{p}(\log x)^q\,dx=\frac{(-1)^q}{(p+1)^{q+1}} \cdot \int_0^\infty e^{-t}t^q \, dt.$$
By definition of the upper incomplete gamma function we get for
$$\int_0^\infty e^{-t}t^q \, dt = \Gamma(1+q).$$
Using this we get
$$\int_0^1 x^{p}(\log x)^q\,dx = \frac{(-1)^q}{(p+1)^{q+1}} \cdot \Gamma(q+1).$$
Assuming that $p,q \in \mathbb{N}$ we have
$$\int_0^1 x^{p}(\log x)^q\,dx = \frac{(-1)^q}{(p+1)^{q+1}} \cdot q!.$$