I Integral involving up-arrow notation

Thread starter Saracen Rue
Start date Nov 11, 2019
Tags

Integals Integral Notation Polar

Click For Summary

The discussion revolves around evaluating the integral $$\frac {1} {2} \int_0^\pi (\sin(\theta) \uparrow \uparrow \infty)^{2} d\theta$$, which involves an infinite power tower. Participants note that the convergence of such towers depends on the base value, specifically that convergence occurs when the base is within the interval defined by ##e^{-e}## and ##e^{1/e}##. Since ##\sin(0) < e^{-e}##, the integral is deemed divergent. The conversation also touches on the behavior of the polar graphs generated by these power towers and the implications of oscillation in values as the parameter increases. Ultimately, the consensus suggests that the integral does not yield a meaningful result due to divergence.

Jan 4, 2020

aheight

The graph posted in #15 is misleading at it's endpoints outside of ##(e^{-e},e^{1/e})##: The command used to plot it was:
[CODE title="Mathematica"]Nest[Sin[theta]^# &, 1, 800]
[/CODE]
So that's just iterating the tetration of ##\sin(t)## 800 times. But when ##\sin(t)<e^{1/e}##, the tetration is no longer single valued but rather enters a 2-cycle oscillation and selecting the even number of iterations was just selecting the higher value. Consider the output:
[CODE title="Mathematica"]In[59]:= Nest[Sin[0.02]^# &, 1, 500]
Nest[Sin[0.02]^# &, 1, 501]

Out[59]= 0.884202

Out[60]= 0.0314587[/CODE]

This is what a plot of ##\sin(t)\uparrow\uparrow \infty ## really looks like on ##(0,\pi)##: