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

1. Problem Description

We need to evaluate the definite integral:

$\int_{0}^{+\infty} \frac{[W(x)]^2}{x^2\sqrt{x}} dx$, where $W(x)$ is the Lambert W function.

2. Solution Steps

Let the integral be $I$.

$I = \int_{0}^{+\infty} \frac{[W(x)]^2}{x^2\sqrt{x}} dx = \int_{0}^{+\infty} \frac{[W(x)]^2}{x^{5/2}} dx$

Let $x = te^t$, so $W(x) = t$. Also, $dx = (e^t + te^t)dt = (1+t)e^t dt$.

The limits of integration need to be updated.

When $x = 0$, $te^t = 0$, so $t = 0$.

When $x \to \infty$, $te^t \to \infty$, so $t \to \infty$.

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

$I = \int_{0}^{+\infty} t^{-1/2} e^{-3t/2} dt + \int_{0}^{+\infty} t^{-3/2} e^{-3t/2} dt$

Recall the gamma function:

$\Gamma(z) = \int_0^{\infty} x^{z-1} e^{-x} dx$

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

Then,

$\int_0^{\infty} t^{-1/2} e^{-3t/2} dt = \int_0^{\infty} (\frac{2u}{3})^{-1/2} e^{-u} \frac{2}{3} du = (\frac{2}{3})^{1/2} \frac{2}{3} \int_0^{\infty} u^{-1/2} e^{-u} du = (\frac{2}{3})^{1/2} \frac{2}{3} \Gamma(\frac{1}{2}) = \frac{2\sqrt{2}}{3\sqrt{3}} \sqrt{\pi} = \frac{2\sqrt{6\pi}}{9}$

Similarly,

$\int_0^{\infty} t^{-3/2} e^{-3t/2} dt = \int_0^{\infty} (\frac{2u}{3})^{-3/2} e^{-u} \frac{2}{3} du = (\frac{2}{3})^{-3/2} \frac{2}{3} \int_0^{\infty} u^{-3/2} e^{-u} du = (\frac{3}{2})^{3/2} \frac{2}{3} \Gamma(-\frac{1}{2})$. This diverges since $\Gamma(-1/2)$ is undefined.

However, we can integrate by parts:

$\int_0^{\infty} t^{-3/2} e^{-3t/2} dt = \left[ \frac{t^{-1/2}}{-1/2} e^{-3t/2} \right]_0^{\infty} - \int_0^{\infty} \frac{t^{-1/2}}{-1/2} (-\frac{3}{2})e^{-3t/2} dt = [-2 t^{-1/2} e^{-3t/2}]_0^{\infty} - 3 \int_0^{\infty} t^{-1/2} e^{-3t/2} dt$

The term $[-2 t^{-1/2} e^{-3t/2}]_0^{\infty}$ is undefined at $0$.

However, let's assume we can evaluate this integral using the gamma function. For the integral to converge, $z-1 > -1$, $z>0$.

We need $z-1 = -3/2$ so $z = -1/2$. $\Gamma(-1/2)$ is undefined. There must be an error.

The integral $\int_0^{\infty} (t^{-1/2} + t^{-3/2}) e^{-3t/2} dt$ diverges at $t=0$ since $\int_0^{\epsilon} t^{-3/2} dt$ diverges for any $\epsilon>0$.

3. Final Answer

The integral diverges.