How to evaluate $ \int_0^\infty {1 \over x^x}dx$ in terms of summation of series? 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.
 A: A long comment (not an answer)
Split the integration region, and use the Sophomore's dream:
$$
   \int_0^\infty \frac{\mathrm{d} x}{x^x} = \int_0^1 \frac{\mathrm{d} x}{x^x} + \int_1^\infty \frac{\mathrm{d} x}{x^x} = \sum_{n=1}^\infty \frac{1}{n^n} + \int_1^\infty \frac{\mathrm{d} x}{x^x}
$$
Because $x^{-x}$ is strictly decreasing for $x\geqslant 1$, we have
$$
  \sum_{n=1}^\infty \frac{1}{n^n} > \int_1^\infty \frac{\mathrm{d} x}{x^x} > \sum_{n=1}^\infty \frac{1}{(n+1)^{n+1}} = -1 + \sum_{n=1}^\infty \frac{1}{n^n} 
$$
Since the $x^{-x}$ is strictly convex function for $x \geqslant 1$, thus
$$
   \int_1^\infty x^{-x} \mathrm{d} x = \sum_{n=1}^\infty \int_0^{1} (n+y)^{-n-y} \mathrm{d} y < \\ \sum_{n=1}^\infty \int_0^{1} \left(n^{-n}(1-y) + (n+1)^{-n-1} y \right) \mathrm{d} y = -\frac{1}{2} + \sum_{n=1}^\infty \frac{1}{n^n}
$$ 
We thus established that
$$
    -\frac{1}{2} + 2 \int_0^1 x^{-x} \mathrm{d} x > \int_0^\infty x^{-x} \mathrm{d}x > -1 + 2 \int_0^1 x^{-x} \mathrm{d} x 
$$
Numerical confirmation:

A: 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.
