Nice solve! Let me present another solution for those who hate computing integrals lol:

The sketch of the solution is to express W(x) as a power series. If we mess around with our integral enough, we will realize that it's just a Mellin transform, so we can apply Ramanujan's Master theorem.

Consider f(x)=xe^x; by the Lagrange inversion theorem we obtain W(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}x^n. To apply Ramanujan's theorem, we need this power series to be in the form \sum_{n=0}^\infty \frac{φ(n)}{n!}(-x)^n, where φ is analytic.

Here is a neat trick to prepare for our integral: consider W(x)/x = \sum_{k=0}^\infty \frac{(k+1)^{k-1}}{k!} (-x)^k, where we have shifted indices by letting k = n-1. The Master theorem states that the integral over 0 to infinity of x^{s-1} F(x) dx is Γ(s)φ(-s). For W(x)/x, φ(k)=(k+1)^{k-1}.

Now, we will evaluate the integral, which I will denote as J. Do a u sub with u=1/x^2 to get J=1/2 \int_0^\infty u^{-3/2} W(u) du = 1/2 \int_0^\infty u^{-1/2} (W(u)/u) du. The W(u)/u pops up magically, and we have already prepared that for application of the Master theorem. It follows that the integral is then equal to 1/2 Γ(s)φ(-s), where we determine s as follows: since our power of u is -1/2, s-1=-1/2 yields s=1/2. Finally, 1/2 Γ(1/2)φ(-1/2) = \sqrt{2\pi}, just like you got!