How can i evaluate $\displaystyle\int _0^{\infty }W\left(\frac{1}{x^3}\right)\:\mathrm{d}x$ in an easy manner i managed to end up with this $$3\int _0^{\infty }\frac{W\left(\frac{1}{x^3}\right)}{W\left(\frac{1}{x^3}\right)+1}\:\mathrm{d}x$$ But how can i continue?.
$W(x)$ is the lambert w function
You can generalize this integral $$\underbrace{\int _0^{\infty }W\left(\frac{1}{x^n}\right)\:dx}_{t=W\left(\frac{1}{x^n}\right)}$$ $$=\frac{1}{n}\int _0^{\infty }t^{1-\frac{1}{n}}e^{-\frac{t}{n}}\:dt+\frac{1}{n}\int _0^{\infty }t^{-\frac{1}{n}}e^{-\frac{t}{n}}\:dt=n^{1-\frac{1}{n}}\Gamma \left(2-\frac{1}{n}\right)+n^{-\frac{1}{n}}\Gamma \left(1-\frac{1}{n}\right)$$ $$=-n^{-\frac{1}{n}}\Gamma \left(-\frac{1}{n}\right)$$ So $$\boxed{\int _0^{\infty }W\left(\frac{1}{x^n}\right)\:dx=-n^{-\frac{1}{n}}\Gamma \left(-\frac{1}{n}\right)}$$ So for your integral, letting $n=3$ gets $$\int _0^{\infty }W\left(\frac{1}{x^3}\right)\:dx=-3^{-\frac{1}{3}}\Gamma \left(-\frac{1}{3}\right)=3^{\frac{2}{3}}\Gamma \left(\frac{2}{3}\right)\approx2.816678$$