# Definite Integral from 0 to 1 of x to the x

## Theorem

 $\ds \int_0^1 x^x \rd x$ $=$ $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^n}$ $\ds$ $=$ $\ds -\sum_{n \mathop = 1}^\infty \paren {-n}^{-n}$ $\ds$ $=$ $\ds 0.78343 \ 05107 \ 12 \ldots$

## Proof

We can write:

 $\ds x^x$ $=$ $\ds \map \exp {x \ln x}$ Definition of Power to Real Number $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {x^n \paren {\ln x}^n} {n!}$ Definition of Exponential Function

So:

 $\ds \int_0^1 x^x \rd x$ $=$ $\ds \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \frac {x^n \paren {\ln x}^n} {n!} }\rd x$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\int_0^1 x^n \paren {\ln x}^n \rd x}$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {n + 1}^{n + 1} } }$ Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {n + 1}^{n + 1} }$ Gamma Function Extends Factorial $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^n}$ shifting the index and using $\paren {-1}^{n + 1} = \paren {-1}^2 \paren {-1}^{n - 1} = \paren {-1}^{n - 1}$ $\ds$ $=$ $\ds -\sum_{n \mathop = 1}^\infty \paren {-\frac 1 n}^n$ $\ds$ $=$ $\ds -\sum_{n \mathop = 1}^\infty \paren {-n}^{-n}$

Numerical computation of partial sums gives the decimal approximation.

$\blacksquare$