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:


 * $\displaystyle \int_0^1 x^x dx = \sum_{n \mathop = 0}^\infty \int_0^1 \frac{x^n \left({\ln x}\right)^n}{n!} \ \mathrm d x$

We can evaluate this by Integration by Parts.

Integrate:
 * $\displaystyle \int x^m \left({\ln x}\right)^n \ \mathrm d x$

by taking $u = \left({\ln x}\right)^n$ and $\mathrm d v = x^m \mathrm d x$, which gives us:


 * $\displaystyle \int x^m \left({\ln x}\right)^n \ \mathrm d x = \frac{x^{m+1}\left({\ln x}\right)^n}{m+1} - \frac n {m+1} \int x^{m+1} \frac{\left({\ln x}\right)^{n-1}} x \mathrm d x \qquad \text{ for } m \ne -1$

for $m \ne -1$.

Thus, by induction:


 * $\displaystyle \int x^m \left({\ln x}\right)^n \ \mathrm d x = \frac {x^{m+1}} {m+1} \sum_{i \mathop = 0}^n \left({-1}\right)^i \frac{\left({n}\right)_i}{\left({m+1}\right)^i} \left({\ln x}\right)^{n-i}$

where $\left({n}\right)_i$ denotes the falling factorial.