Infinite tetration giving I*Omega costant=I*0.567143...

[bo198214]

Please could You check analytically or numerically if this derivation is valid?

h(((e^(pi/2))^(1/Omega))*(e^I)) =-W((-pi/(2*Omega)-I))/(pi/(2*Omega)+I)=(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329..

Well, if we dont consider branches we just need to verify that
\( (I\Omega)^{\frac{1}{I\Omega}}=e^{\frac{\pi}{2\Omega}+I} \)

And this is straight forward computing:

\( (I\Omega)^{\frac{1}{I\Omega}}=e^{\frac{\ln(I\Omega)}{I\Omega}} \)

\( \frac{\ln(I\Omega)}{I\Omega}=\frac{\ln(I)}{I\Omega}+\frac{\ln(\Omega)}{I \Omega}=\frac{\pi}{2 I \Omega}-I\frac{\ln(\Omega)}{\Omega} \)

Now you defined \( \Omega \) to satisfy \( \Omega e^\Omega = 1 \) so \( \Omega=e^{-\Omega} \), and \( \ln(\Omega)=-\Omega \). Putting this into the above equation:

\( \frac{\ln(I\Omega)}{I\Omega}=\frac{\pi}{2I\Omega}+I \)

q.e.d.