For k > 0, \, n \in \mathbb{Z}^+, let
I_n = \int x^k (\log x)^n \, dx.
Prove that
(k+1) I_n = {x^{k+1}(\log x)^n} - {n}I_{n-1}.
Given that
\lim_{x \to 0} x^k (\log x)^n = 0,
Show that
\int_0^1 x^k (\log x)^n \, dx= (-1)^n \frac{n!}{(k+1)^{n+1}},
\frac 1 e \int_0^e (\log x)^n \, dx =...