Definite Integral from 0 to 1 of x to the x

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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.

$\blacksquare$