User:Caliburn/sandbox

just so this stuff doesn't get lost

to adapt for Definite Integral from 0 to 1 of Power of x by Power of Logarithm of x/Proof 2

Thus the exercise devolves into the following sum of integrals:

$\ds \int_0^1 x^x dx = \sum_{n \mathop = 0}^\infty \int_0^1 \frac {x^n \paren {\ln x}^n} {n!} \rd x$

We can evaluate this by Integration by Parts.

Integrate:

$\ds \int x^m \paren {\ln x}^n \rd x$

by taking $u = \paren {\ln x}^n$ and $\d v = x^m \rd x$, which gives us:

$\ds \int x^m \paren {\ln x}^n \rd x = \frac {x^{m + 1} \paren {\ln x}^n} {m + 1} - \frac n {m + 1} \int x^{m + 1} \frac {\paren {\ln x}^{n - 1} } x \rd x$

for $m \ne -1$.

Thus, by induction:

$\ds \int x^m \paren {\ln x}^n \rd x = \frac {x^{m + 1} } {m + 1} \sum_{i \mathop = 0}^n \paren {-1}^i \frac {\paren n_i} {\paren {m + 1}^i} \paren {\ln x}^{n - i}$

where $\paren n_i$ denotes the falling factorial.