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

## Theorem

 $\ds \int_0^1 x^{-x} \rd x$ $=$ $\ds \sum_{n \mathop = 1}^\infty n^{-n}$ $\ds$ $=$ $\ds 1.29128 \ 5997 \ldots$

## Proof

We can write:

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

So:

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

Numerical computation of partial sums gives the decimal approximation.

$\blacksquare$