TITLE: How to evaluate $ \int_0^\infty {1 \over x^x}dx$ in terms of summation of series?
QUESTION [8 upvotes]: Is there a way to represent this integral in terms of summation of series?
$$ \int_0^\infty {1 \over x^x}dx$$
Like for example:
$$ \int_0^1 {1 \over x^x}dx = \sum_{n=1}^\infty {1 \over n^n}$$
I am not getting an answer from Mathematica.

REPLY [3 votes]: For the integral from $0$ to (finite) $R$ I get a double sum:
$$\int _{0}^{R}\!{x}^{-x}{dx}=\sum _{k=1}^{\infty }  \sum _{j=0}^{
k-1}{\frac { \left( -1 \right) ^{j}{R}^{k} \left( \ln  \left( R
 \right)  \right) ^{j}}{j!\,{k}^{k-j}}}  
$$
The inner sum could be written using an incomplete Gamma function, so:
$$\int_0^R x^{-x}\ dx = \sum _{k=1}^{\infty }{\frac {\Gamma  \left( k,-\ln  \left( R \right) k
 \right) {k}^{-k}}{\Gamma  \left( k \right) }}
$$
I don't see a way to take the limit as $R \to \infty$ on the right, though: each individual term diverges.