We are asked to evaluate the 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 are asked to evaluate the 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 $I$ be the given integral. We have:

$$$$

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

$$$$

Let $x = te^t$. Then $W(x) = t$ and $dx = (e^t + te^t) dt = e^t(1+t)dt$. Therefore,

$$$$

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

$$$$

$$$$

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

$$$$

$$$$

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

$$$$

We can write this as

$$$$

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

$$$$

Recall the Gamma function:

$$$$

\Gamma(z) = \int_{0}^{+\infty} t^{z-1} e^{-t} dt

$$$$

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

So,

$$$$

\int_{0}^{+\infty} t^{z-1} e^{-at} dt = \int_{0}^{+\infty} (\frac{2}{3} u)^{z-1} e^{-u} \frac{2}{3} du = (\frac{2}{3})^{z-1} \frac{2}{3} \int_{0}^{+\infty} u^{z-1} e^{-u} du

$$$$

$$$$

\int_{0}^{+\infty} t^{z-1} e^{-at} dt = (\frac{2}{3})^z \int_{0}^{+\infty} u^{z-1} e^{-u} du = (\frac{2}{3})^z \Gamma(z)

$$$$

For the first term, $z-1 = -1/2$, so $z = 1/2$.

$$$$

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

$$$$

For the second term, $z-1 = 1/2$, so $z = 3/2$.

$$$$

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

$$$$

Therefore,

$$$$

I = \sqrt{\frac{2\pi}{3}} + \frac{1}{3}\sqrt{\frac{2\pi}{3}} = \frac{4}{3} \sqrt{\frac{2\pi}{3}} = \frac{4}{3\sqrt{3}} \sqrt{2\pi} = \frac{4\sqrt{3}}{9} \sqrt{2\pi} = \frac{4\sqrt{6\pi}}{9}

$$$$

Using the result that $\int_0^\infty [W(x)]^n x^{-m} dx = \frac{n}{m-1}(\frac{m}{m-1})^{m-1} \Gamma(m-1)$ with $n=2$, $m = 5/2$, then we have

$$$$

\int_{0}^{+\infty} \frac{[W(x)]^2}{x^{5/2}} dx = \frac{2}{5/2-1} (\frac{5/2}{5/2-1})^{5/2-1} \Gamma(5/2-1) = \frac{2}{3/2} (\frac{5/2}{3/2})^{3/2} \Gamma(3/2)

$$$$

$$$$

= \frac{4}{3} (\frac{5}{3})^{3/2} \frac{1}{2} \Gamma(1/2) = \frac{2}{3} \frac{5\sqrt{5}}{3\sqrt{3}} \sqrt{\pi} = \frac{10\sqrt{15}}{27}\sqrt{\pi} = \frac{10}{27} \sqrt{15\pi}

$$$$

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

3. Final Answer

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