It is a well known identity that
$$ \int_0^\infty \frac{\log{(x)}}{e^x}\,d{x} = -\gamma $$
This shows how the Euler–Mascheroni constant is directly connected to the exponential function and its inverse. Further, using the exponential integral $\text{Ei}(x)$, the antiderivative of the function can be computed as
$$ \int \frac{\log{(x)}}{e^x}\,d{x} = \text{Ei}(-x) - e^{-x}\log{x} $$
My question is, does there exist any analytic expression for the following definite integral:
$$ ξ=\int_0^\infty \frac{W{(x)}}{xe^x}\,dx $$
WolframAlpha numerically evaluates $\xi\approx0.646503199298577944534$, which, looking up on an inverse symbolic calculator yields no results for it being in terms of other constants.
I've tried some basic methods (integration by parts for example) but nothing has really seemed to simplify the integral. How can I evaluate this integral, and, are there any other definite integrals we can find that also evaluate to this constant (like $\gamma$)?
In the comments, @BobDobbs points out that
$$ \int_0^\infty (x+1)e^{-xe^x}\,dx = \xi $$
And, with the substitution $u=xe^x$ we get
$$ \int_0^\infty \frac{du}{e^{u+W(u)}} = \xi $$
Which is about the most I could simplify it. However, both of these seem equally hard to evaluate.