# Evaluation of an integral involving the Lambert W function

Wikipedia claims that

$$\int_0^\infty W\left(\frac{1}{x^2}\right) \,\text dx=\sqrt{2\pi}$$

and a numerical computation seems to confirm this.

How can this result be proven?

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Writing $w=W(1/x^2)$, we have $we^w=1/x^2$ and hence
$$(w+1)e^wdw=-2\frac{dx}{x^3}\implies-\frac{(w+1)e^w}{2(we^w)^{3/2}}dw=dx.$$
$$\int_0^\infty wdx=\int_\infty^0-w\frac{(w+1)e^w}{2(we^w)^{3/2}}dw=\int_0^\infty\frac{w^{1/2}+w^{-1/2}}{2}e^{-w/2}dw$$
$$=\int_0^\infty (\sqrt{2}u^{1/2}+u^{-1/2}/\sqrt{2})e^{-u}du=\sqrt{2}\Gamma(3/2)+\frac{\Gamma(1/2)}{\sqrt{2}}=\sqrt{2\pi}.$$