Question

Integrate:
\[\int_0^1 x^n(\ln x)^k\,dx\]
where \(n,k\in\mathbb{N}\), where \(\mathbb{N}=\{1,2,3,...\}\).

Answer 1

You're welcome to use any method (power series, gamma/beta function, etc) where applicable. I have a relatively short solution at the moment, but I'm looking for more interesting approaches.

Answer 2

By "more interesting approaches" I mean both "more and interesting" and "more interesting than mine." :)

Answer 3

i smell gamma function but struggling a bit with gettting into the required form @Marki

Answer 4

\[\begin{align}
\int\limits_0^1 x^n (\ln x)^k dx~~ &\stackrel{x = e^{\frac{-u}{n+1}}}{\leadsto
} ~~ \int\limits_{\infty}^0 e^{\frac{-un}{n+1}} \cdot \left(\frac{-u}{n+1}\right)^k \cdot \frac{-e^{\frac{-u}{n+1}}}{n+1}du\\~\\
&= \frac{(-1)^k}{(n+1)^{k+1}}\int\limits_0^{\infty} u^k\cdot e^{-u}~ du\\~\\
&= \frac{(-1)^k}{(n+1)^{k+1}}\Gamma(k+1)\\~\\

\end{align}\]

since \(k \in \mathbb{N}\), \(\Gamma(k+1) = k!\)

Answer 5

That's right! Here was my approach.
\[\begin{align*}
\int_0^1x^n(\ln x)^k\,dx&=\int_0^1\frac{\partial^k}{\partial n^k}x^n\,dx\\\\
&=\frac{\partial^k}{\partial n^k}\int_0^1x^n\,dx\\\\
&=\frac{\partial^k}{\partial n^k}\left[\frac{1}{n+1}\right]\\\\
&=\frac{(-1)^kk!}{(n+1)^{k+1}}
\end{align*}\]

Answer 6

Amazing! thats a cute feynman way of evaluating it xD