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?
 A: It is highly unlikely that you will be able to find a closed-form expression for your integral. Two special variants of your integral, however, are rather infamous as the "Sophomore's Dream" integrals. They are the integrals
$$\int_0^1 x^{x} dx$$
and
$$\int_0^1 x^{-x} dx$$
And, as of yet, no closed-form has been obtained for either of them. However, your integral
$$\int_0^\infty x^{-x} dx$$
Converges incredibly quickly, even more so than an exponential function, and so a very good approximation can be obtained rather quickly. Wolfram Alpha yields the approximation
$$\int_0^\infty x^{-x} dx\approx 1.99545595750014...$$
And so $2$ should be a good enough approximation. Even the inverse symbolic calculator doesn't yield anything for the approximation given by WA, so I doubt that it has any kind of closed form using the elementary functions or any other known constants.
A: We have the following approximation:
Let $S$ be one of the Sophomore's dreams:
$$S=\int_0^1x^{-x}~\mathrm dx=\sum_{n=1}^\infty n^{-n}$$
Then we have the integral $I$ in question,
$$I=\int_0^\infty x^{-x}~\mathrm dx=S+\int_1^\infty x^{-x}~\mathrm dx$$
The second integral has a quick trapezoidal sum approximation:
$$\int_1^\infty x^{-x}~\mathrm dx\approx-\frac12+\sum_{n=1}^\infty n^{-n}$$
Thus, we have
$$I\approx2S-\frac12=2.08257$$
The error is approximately $0.08711$
A: $\int\limits_0^\infty x^{-x}dx=\int\limits_0^1 x^{-x}dx+\int\limits_0^1 x^{-2+1/x}dx$ 
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:  
$$\displaystyle \int\limits_0^\infty x^{-x}dx=\sum\limits_{n=1}^\infty \frac{1+(-z_0)^{n-1}}{n^n}$$
