# Lambert-W Function Integral

Stumbled upon this integral while doing some research.

I'd love to see the different methods used to prove this subtle integral!

$$\Large{\int_{0}^{\infty}{\dfrac{ \operatorname{W}(x)}{x\sqrt{x}} }\, dx =2\sqrt{2\pi}}$$

where

$\large{\operatorname{W}(x)}$ = Lambert W-Function

Thank you to everyone who participates :)

We have $W(x)$ to be the function $W(x) e^{W(x)} = x$, i.e., $$\log(W(x)) + W(x) = \log(x) \implies \dfrac{dW(x)}{W(x)} + dW(x) = \dfrac{dx}x \implies dW(x) = \dfrac{W(x)}{x(1+W(x))}dx$$
Set $t=W(x)$, we then have $x=te^t$. $dt = dW(x) = \dfrac{t}{x(1+t)}dx = \dfrac{e^{-t}}{1+t}dx$.