Stumbled upon this integral while doing some research.

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

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


$\large{\operatorname{W}(x)}$ = 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 $x=te^t$. $dt = dW(x) = \dfrac{t}{x(1+t)}dx = \dfrac{e^{-t}}{1+t}dx$.

Hence, we have \begin{align} I & = \int_0^{\infty} \dfrac{W(x)}{x\sqrt{x}}dx = \int_0^{\infty} \dfrac{t}{t^{3/2}e^{3t/2}} (1+t)e^tdt = \int_0^{\infty}\left(\sqrt{t}+\dfrac1{\sqrt{t}}\right)e^{-t/2}dt\\ & = \int_0^{\infty} \left(\sqrt{2x}+\dfrac1{\sqrt{2x}}\right)e^{-x} (2dx) = 2\sqrt2 \int_0^{\infty} x^{1/2}e^{-x}dx + \sqrt2 \int_0^{\infty}x^{-1/2}e^{-x}dx\\ & = 2 \sqrt2 \Gamma(3/2) + \sqrt2 \Gamma(1/2) = 2\sqrt2 \dfrac12 \Gamma(1/2) + \sqrt2 \Gamma(1/2) = 2\sqrt2 \Gamma(1/2)\\ & = 2\sqrt2 \sqrt{\pi} = 2\sqrt{2\pi} \end{align}


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