Sophomore's Dream

Theorem
These were discovered in 1697 by.

Proof
The following is a proof of the second identity; the first follows the same lines.

By definition, we can express $x^x$ as:

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.

In this case $m = n$, and they are integers, so:


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

We integrate from $0$ to $1$.

By L'Hôpital's Rule, we have that:
 * $\displaystyle \lim_{x \to 0^+} x^m \left({\ln x}\right)^n = 0$

Because of this, and the fact that $\ln 1 = 0$, all the terms vanish except the last term at $1$.

This yields:


 * $\displaystyle \int_0^1 \frac{x^n \left({\ln x}\right)^n}{n!} \ \mathrm d x = \frac 1 {n!}\frac {1^{n+1}}{n+1} \left({-1}\right)^n \frac{\left({n}\right)_n}{\left({n+1}\right)^n} = \left({-1}\right)^n \left({n+1}\right)^{-\left({n+1}\right)}$

Summing these (and changing indexing so it starts at $n = 1$ instead of $n = 0$ yields the formula.

Also see

 * Freshman's Dream