We need to evaluate the definite integral from $0$ to $+\infty$ of the function $\frac{[W(x)]^2}{x^2\sqrt{x}}$ with respect to $x$, where $W(x)$ is the Lambert W function. The integral is given by: $\int_{0}^{\infty} \frac{[W(x)]^2}{x^2\sqrt{x}} dx$

1. Problem Description

We need to evaluate the definite integral from $0$ to $+\infty$ of the function $\frac{[W(x)]^2}{x^2\sqrt{x}}$ with respect to $x$, where $W(x)$ is the Lambert W function. The integral is given by:

$\int_{0}^{\infty} \frac{[W(x)]^2}{x^2\sqrt{x}} dx$

2. Solution Steps

To evaluate the integral, we can use the substitution $x = t e^t$, where $t = W(x)$. Then, $dx = (e^t + t e^t) dt = (1+t) e^t dt$.

Also, we need to change the limits of integration. When $x = 0$, $t = W(0) = 0$. As $x \to \infty$, $t = W(x) \to \infty$.

The integral becomes:

$\int_{0}^{\infty} \frac{t^2}{(t e^t)^2 \sqrt{t e^t}} (1+t) e^t dt = \int_{0}^{\infty} \frac{t^2}{t^2 e^{2t} t^{1/2} e^{t/2}} (1+t) e^t dt = \int_{0}^{\infty} \frac{t^2 (1+t) e^t}{t^{5/2} e^{5t/2}} dt$

$= \int_{0}^{\infty} \frac{t^2 (1+t)}{t^{5/2} e^{3t/2}} dt = \int_{0}^{\infty} t^{-1/2} (1+t) e^{-3t/2} dt = \int_{0}^{\infty} t^{-1/2} e^{-3t/2} dt + \int_{0}^{\infty} t^{1/2} e^{-3t/2} dt$

Now, we use the Gamma function formula:

$\Gamma(z) = \int_{0}^{\infty} x^{z-1} e^{-x} dx$

Let $u = \frac{3t}{2}$. Then $t = \frac{2u}{3}$ and $dt = \frac{2}{3} du$.

$\int_{0}^{\infty} t^{-1/2} e^{-3t/2} dt = \int_{0}^{\infty} (\frac{2u}{3})^{-1/2} e^{-u} \frac{2}{3} du = (\frac{2}{3})^{-1/2} \frac{2}{3} \int_{0}^{\infty} u^{-1/2} e^{-u} du = \frac{2}{3} \sqrt{\frac{3}{2}} \Gamma(\frac{1}{2}) = \frac{2}{3} \sqrt{\frac{3}{2}} \sqrt{\pi} = \frac{2 \sqrt{3\pi}}{3\sqrt{2}} = \frac{\sqrt{6\pi}}{3}$.

$\int_{0}^{\infty} t^{1/2} e^{-3t/2} dt = \int_{0}^{\infty} (\frac{2u}{3})^{1/2} e^{-u} \frac{2}{3} du = (\frac{2}{3})^{1/2} \frac{2}{3} \int_{0}^{\infty} u^{1/2} e^{-u} du = \frac{2}{3} \sqrt{\frac{2}{3}} \Gamma(\frac{3}{2}) = \frac{2}{3} \sqrt{\frac{2}{3}} \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{1}{3} \sqrt{\frac{2}{3}} \sqrt{\pi} = \frac{\sqrt{2\pi}}{3\sqrt{3}} = \frac{\sqrt{6\pi}}{9}$.

$\int_{0}^{\infty} t^{-1/2} e^{-3t/2} dt + \int_{0}^{\infty} t^{1/2} e^{-3t/2} dt = \frac{\sqrt{6\pi}}{3} + \frac{\sqrt{6\pi}}{9} = \frac{3\sqrt{6\pi} + \sqrt{6\pi}}{9} = \frac{4\sqrt{6\pi}}{9}$.

3. Final Answer

The final answer is $\frac{4\sqrt{6\pi}}{9}$