Integral $\int_{0}^{\infty}x^{-x}dx$ I'm trying to find a closed form for this integral:$$\int_{0}^{\infty}x^{-x}dx$$
Here's the integrand graph:

Clearly it is convergent. My attempt is to obtain a closed form for the area under the curve. Is this possible?

$\int\limits_0^\infty x^{-x}dx=\int\limits_0^1 x^{-x}dx+\int\limits_0^1 x^{-2+1/x}dx$ 
The second integral is unpleasant to develope into a series.  
But if we can use the solution $\,z_0\,$ for $\,\int\limits_0^1 (x^{xz} - x^{-2+1/x})dx=0$  
with $\,z=z_0\approx 1.45354007846425$ , then we get:  
$$\displaystyle \int\limits_0^\infty x^{-x}dx=\sum\limits_{n=1}^\infty \frac{1+(-z_0)^{n-1}}{n^n}$$
