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The Omega constant $\Omega = \rm{W}_0(1) = 0.56714329\ldots$ is defined as the real root to the equation $x {\rm e}^x = 1$ and corresponds to the value of the Lambert W function for an argument equal to unity. Here ${\rm W}_0(x)$ denotes the principal branch of the Lambert W function.

The following very narrow bound for the Omega constant in terms of the Euler-Mascheroni constant $\gamma$ and its negative exponent, namely $${\rm e}^{-\gamma} < \Omega < \gamma,$$ or approximately, $$0.561<\Omega<0.577$$ has been suggested here.

My question is, is the above bound based on numerical evidence alone or can it be formally proved?

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Let $f(x)=xe^x$ .Then $$f(e^{-\gamma})=e^{e^{-\gamma}-\gamma}$$ $$log(f(e^{-\gamma}))=e^{-\gamma}-\gamma$$ Now, if $e^{-\gamma}<\gamma.$ $f(e^{-\gamma})<1$. $$f(\gamma)=\gamma e^\gamma$$ since ,$1<\gamma e^\gamma$. $1<f(\gamma)$. So we have $f(e^{-\gamma})<1<f(\gamma)$, By IVT ,there exist a root of $f(x)=1$ in the interval $(e^{-\gamma},\gamma)$.

So, it all boils down to proving $\gamma e^\gamma >1.$

Also note that we have not used that $\gamma $ is the Euler-Mascheroni constant anywhere. So, this is true for all $\alpha$ such that $\alpha e^\alpha >1.$

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  • $\begingroup$ From $\alpha {\rm e}^\alpha > 1$, as $f(x) = x {\rm e}^x$ is a monotonic increasing function for $x > -1$, solving for $\alpha$ gives $\alpha > {\rm W}_0(1) = \Omega$. What I want to show is that for the particular case of $\alpha = \gamma$, where $\gamma$ is the Euler-Massceroni constant, that this is true. $\endgroup$ – omegadot Feb 19 '14 at 3:49
  • $\begingroup$ This is also true for $1/ \sqrt{3}$. All am saying is that it might be just be because of its numerical value. $\endgroup$ – viplov_jain Feb 19 '14 at 10:50

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