Fixed point representation

The defining identity can be expressed, for example, as

${\displaystyle \ln({\tfrac {1}{\Omega }})=\Omega .}$

or

${\displaystyle -\ln(\Omega )=\Omega }$

as well as

${\displaystyle e^{-\Omega }=\Omega .}$

Computation

One can calculate Ω iteratively, by starting with an initial guess Ω₀, and considering the sequence

${\displaystyle \Omega _{n+1}=e^{-\Omega _{n}}.}$

This sequence will converge to Ω as n approaches infinity. This is because Ω is an attractive fixed point of the function e^−x.

It is much more efficient to use the iteration

${\displaystyle \Omega _{n+1}={\frac {1+\Omega _{n}}{1+e^{\Omega _{n}}}},}$

because the function

${\displaystyle f(x)={\frac {1+x}{1+e^{x}}},}$

in addition to having the same fixed point, also has a derivative that vanishes there. This guarantees quadratic convergence; that is, the number of correct digits is roughly doubled with each iteration.

Using Halley's method, Ω can be approximated with cubic convergence (the number of correct digits is roughly tripled with each iteration): (see also Lambert W function § Numerical evaluation).

${\displaystyle \Omega _{j+1}=\Omega _{j}-{\frac {\Omega _{j}e^{\Omega _{j}}-1}{e^{\Omega _{j}}(\Omega _{j}+1)-{\frac {(\Omega _{j}+2)(\Omega _{j}e^{\Omega _{j}}-1)}{2\Omega _{j}+2}}}}.}$

Integral representations

An identity due to ^{[citation needed]}Victor Adamchik^{[citation needed]} is given by the relationship

${\displaystyle \int _{-\infty }^{\infty }{\frac {dt}{(e^{t}-t)^{2}+\pi ^{2}}}={\frac {1}{1+\Omega }}.}$

Other relations due to Mező^[1][2] and Kalugin-Jeffrey-Corless^[3] are:

${\displaystyle \Omega ={\frac {1}{\pi }}\operatorname {Re} \int _{0}^{\pi }\log \left({\frac {e^{e^{it}}-e^{-it}}{e^{e^{it}}-e^{it}}}\right)dt,}$

${\displaystyle \Omega ={\frac {1}{\pi }}\int _{0}^{\pi }\log \left(1+{\frac {\sin t}{t}}e^{t\cot t}\right)dt.}$

The latter two identities can be extended to other values of the W function (see also Lambert W function § Representations).

Transcendence

The constant Ω is transcendental. This can be seen as a direct consequence of the Lindemann–Weierstrass theorem. For a contradiction, suppose that Ω is algebraic. By the theorem, e^−Ω is transcendental, but Ω = e^−Ω, which is a contradiction. Therefore, it must be transcendental.^[4]