# Integral involving up-arrow notation

• I
Saracen Rue
TL;DR Summary
Is it possible to integrate $$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta$$
I was playing around with a graphing program and sketching polar graphs involving tall power towers, when I noticed that ##sin(\theta) \uparrow \uparrow a## has an alternating appearance depending on whether ##a## is odd or even. I also noticed that the area enclosed by these alternating graphs both do in fact seem to be converging to a definite number. I know this is all kind of messy so I'm refining my question to make things clearer. I am wondering if it is possible to evaluate the following intergral:

$$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta$$

Mentor
2022 Award
Can you tell us what this is supposed to mean?

Mentor
Summary: Is it possible to integrate $$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta$$

I was playing around with a graphing program and sketching polar graphs involving tall power towers, when I noticed that ##sin(\theta) \uparrow \uparrow a## has an alternating appearance depending on whether ##a## is odd or even. I also noticed that the area enclosed by these alternating graphs both do in fact seem to be converging to a definite number. I know this is all kind of messy so I'm refining my question to make things clearer. I am wondering if it is possible to evaluate the following intergral:

$$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta$$
I'm 99% certain that this is meaningless. What you have here is an infinitely tall tower of powers. For the benefit of any who don't understand the notation, it is described here:
http://mathworld.wolfram.com/PowerTower.html

BvU
Gold Member
@Mark44’s link provides a simple answer. The infinite power tower ##z^{z^{z\cdots}}## converges iff ##e^{-e}\leq z \leq e^{1/e}##. Since, for example, ##\sin(0) < e^{-e}##, the integral diverges.

sysprog
Gold Member
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Gold Member
What's a GP?
Sorry, Geometric Progression. But my comment here on it is not really meaningful/helpful.

TeethWhitener
Saracen Rue
Sorry for the ambiguity in my original post; let me try to clarify what I meant and better define things in proper mathematical notation. I started by considering the following polar graph:

I then wanted to determine the area enclosed by the curve and the x-axis. To determine the area under a polar curve we use the formula ##\int_{\theta_1}^{\theta_2} \frac {r^2} {2} d\theta##. In this particular case, it would be ##\int_0^{\pi}\frac {1} {2} (sin(\theta)\uparrow \uparrow (2k))^2 d\theta##.

So, I'd like to know how to calculate the value of this integral as ##k \to \infty##

Gold Member
Sorry for the ambiguity in my original post; let me try to clarify what I meant and better define things in proper mathematical notation. I started by considering the following polar graph:

View attachment 252829

I then wanted to determine the area enclosed by the curve and the x-axis. To determine the area under a polar curve we use the formula ##\int_{\theta_1}^{\theta_2} \frac {r^2} {2} d\theta##. In this particular case, it would be ##\int_0^{\pi}\frac {1} {2} (sin(\theta)\uparrow \uparrow (2k))^2 d\theta##.

So, I'd like to know how to calculate the value of this integral as ##k \to \infty##
Please follow the link in post 5 and let us know of any followups. Edit: your expression, if it converges, is of the form ##y=x^y##. This is solved using Lambert's W function, which converges in the interval described in the link.

Gold Member
I plotted the first few iterations using wolframalpha and it's interesting. For each ##k##, the function ##r(\theta)=\sin(\theta)\uparrow\uparrow k## appears to bound a finite area between ##\theta=0## and ##\theta=\pi##, but ##\lim_{\theta\to 0,\pi} r(\theta)## oscillates between 0 and 1 depending on whether ##k## is odd or even (I reckon this is why OP uses ##2k##). I imagine the root cause of this is that ultimately ##0^0## is not well-defined.

This may be a limitation with up arrow notation. I'm not convinced that the limit:
$$\lim_{k\to\infty} \frac {1} {2} \int_0^\pi (\sin(\theta)\uparrow \uparrow (2k))^{2} d\theta$$
is meaningful as written, as ##k## is not meant to be a continuous parameter. So it may not be as simple as looking at the convergence of the power tower. I dunno, I'm out of my depth here.

Gold Member
I am thinking if
I plotted the first few iterations using wolframalpha and it's interesting. For each ##k##, the function ##r(\theta)=\sin(\theta)\uparrow\uparrow k## appears to bound a finite area between ##\theta=0## and ##\theta=\pi##, but ##\lim_{\theta\to 0,\pi} r(\theta)## oscillates between 0 and 1 depending on whether ##k## is odd or even (I reckon this is why OP uses ##2k##). I imagine the root cause of this is that ultimately ##0^0## is not well-defined.

This may be a limitation with up arrow notation. I'm not convinced that the limit:
$$\lim_{k\to\infty} \frac {1} {2} \int_0^\pi (\sin(\theta)\uparrow \uparrow (2k))^{2} d\theta$$
is meaningful as written, as ##k## is not meant to be a continuous parameter. So it may not be as simple as looking at the convergence of the power tower. I dunno, I'm out of my depth here.
Maybe we can use the Monotone Convergence theorem, but it would apply for towers that are monotone. So we need to see what powers give us monotone iterations, i.e., what values of k is the tower monotone. I am not even sure how to do the iteration accurately in Wolfram. Is it Sinx^{Sinx}^{Sinx} ...etc?

mitochan
This may be a limitation with up arrow notation. I'm not convinced that the limit:

limk→∞12∫π0(sin(θ)↑↑(2k))2dθ​

$$\lim_{k\to\infty} \frac {1} {2} \int_0^\pi (\sin(\theta)\uparrow \uparrow (2k))^{2} d\theta$$
is meaningful as written, as kk is not meant to be a continuous parameter. So it may not be as simple as looking at the convergence of the power tower. I dunno, I'm out of my depth here.
From post 4
$$\lim_{k\to\infty} \int_a^{\pi/2} (\sin(\theta)\uparrow \uparrow (2k))^{2} d\theta$$
where
$$e^{-e}<a<\pi/2$$
would converge.

EDIT not ##e^{-e}## but 0. See my next post.

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Gold Member
From post 4
$$\lim_{k\to\infty} \int_a^{\pi/2} (\sin(\theta)\uparrow \uparrow (2k))^{2} d\theta$$
where
$$e^{-e}<a<\pi/2$$
would converge.
Umm...I wrote post 4. See the ensuing discussion.

mitochan
Saracen Rue
Eg:
https://www.wolframalpha.com/input/?i=Plot+r=Sin^sin^sin^sin(x)
for the polar plot. The Cartesian plot is also illuminating:
https://www.wolframalpha.com/input/?i=Sin^sin^sin^sin(x)

Wolfram craps out after about ##\sin(\theta)\uparrow\uparrow 6##. I think what will eventually happen is that the minima in the Cartesian plot will get closer to ##(0,0)## and ##(\pi,0)## as ##k## gets big. I have no idea how to show that, though.

I'm not sure how helpful this is, but I did want to let you know it is possible to use Nest notation to get wolfram to parse uparrow notation (example: https://www.wolframalpha.com/input/?i=Nest[Power[x,+#]&,+1,+8])

This can be used to plot both the polar plot (https://www.wolframalpha.com/input/?i=PolarPlot[(Nest[Power[sin(x),+#]&,+1,+100])]+from+0+to+pi) and the Cartesian plot (https://www.wolframalpha.com/input/?i=Plot[(Nest[Power[sin(theta),+#]&,+1,+800]),+{theta,0,pi}]). Note, for some reason wolfram wasn't playing nice today with the polar plot and wouldn't sketch anything with a k larger than 50. In the past I've definitely gotten it to plot a much larger value of k so I'm not sure what's up with it at the moment.

Regardless, I was still able to get it to generate the Cartesian plot for k=400, and as you can see in the above link the minimums have hardly moved compared to much lower k values, which leads me to believe they are approaching constant values which arent ##(0,0)## and ##(\pi, 0)## respectively.

TeethWhitener
mitochan
Whether ##a\uparrow\uparrow\infty## converges or not is whether graphs
$$y=a^x$$
and
$$y=log_a x$$ cross or not. They are inverse functions so in other words one of them cross with
$$y=x$$
or not. Graphs show that the lower limit for convergence seems not ##e^{-e}## but 0, i.e.,
$$0<a<e^{1/e}$$ and
$$+0\uparrow\uparrow\infty=0$$
For $$1<a<e^{1/e}$$
we have two cross points. One of them, smaller one, corresponds to the convergent value.

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Gold Member
Whether ##a\uparrow\uparrow\infty## converges or not is whether graphs
$$y=a^x$$
and
$$y=log_a x$$ cross or not. They are inverse functions so in other words one of them cross with
$$y=x$$
or not. Graphs show that the lower limit for convergence seems not ##e^{-e}## but 0, i.e.,
$$0<a<e^{1/e}$$ and
$$+0\uparrow\uparrow\infty=0$$
For $$1<a<e^{1/e}$$
we have two cross points. One of them, smaller one, corresponds to the convergent value.
I'm having a very hard time following your posts. What if ##a=1##? What do we do with ##y=\log_a x##?

mitochan
Hi.
Graph $$y=a^x$$ for a=1 is the line ##y = 1##. So graph of its inverse function $$y=log_a x$$ for a=1 is the line ##x=1##. For
$$y=\frac{ln\ x}{ln\ a}=\frac{0}{0}$$ can be interpreted to show any number.

mitochan
For
$$1<a<e^{1/e}$$
we have two cross points. One of them, smaller one, corresponds to the convergent value.

As graph ##y=e^{x/e}## touches the line ##y=x## at x=e,

0 < the smaller solution < e < the larger solution

Example : a=##\sqrt{2}##

Equation ##\sqrt{2}^x=x## has an obvious solution of x= 2 < e, so

$$\sqrt{2}\uparrow\uparrow +\infty=2$$

mitochan
As we cannot define ##0^0##, we cannot define ##0\uparrow\uparrow+\infty##
However as I mentioned in my previous post for ##\epsilon\rightarrow +0,\ \epsilon\uparrow\uparrow+\infty\rightarrow +0##

Home back to the original post I think we can disregard bad behavior at 0 in the area integral of the original post, so it converges.

As ##sin\theta\uparrow\uparrow+\infty<1## in the region, the integral < ##\frac{\pi}{2}## as a rough estimation.

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Gold Member
But the integral contains set of full measure where the tower diverges : ##[0, e^{-e}]## so it wil diverge, whether Riemann- or Lesbesgue-. Basically, the set ##\{ k: 0< Cos(2k) < e^{-e} \} ## has positive measure. Riemann integrability does not allow unboundedness, Lebesgue only allows them on a set of measure zero.

mitochan
I would like to know from where ##e^{-e}=0.06...## comes as convergent-divergent boundary. I think for an example ##0.06\uparrow\uparrow +\infty## is an ordinary number. Thanks in advance.

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Gold Member
I would like to know from where ##e^{-e}=0.06...## comes as convergent-divergent boundary. I think for an example ##0.06\uparrow\uparrow +\infty## is an ordinary number. Thanks in advance.
This is the way I understand it:

Convergence of ## y =x ^x^{x}...## is equivalented to a solution to ##y =x^y ## which has the solution

##y= W(-ln(x))/ln(x) ## . We can consider a function ##f(y)= x^y ## and use the Banach fixed point theorem

whenever the map is a contraction map to find a fixed point ##y_0: x^{y_0}= y_0## , i.e., a solution. The

interval ##(e^{-e}, e^{1/e})## is the one where ##f(y)## is a contraction map, where a fixed point will exist (where the sequence ##f^{(n)}## of iterations of ##f## will be a contraction map)

according to Banach fixed point theorem. EDIT: But there are issues with the limit being outside of the

integral . Some results allow us to exchange limits, i.e., ##Lim_{x \rightarrow x_n} \int_a^b f(x)dx =

\int_a^b f(Lim_{ x \rightarrow x_n}) dx## but we need to figure out what. Maybe we can use monotone

convergence theorem.

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mitochan
Thanks! Very exciting. I experienced even-odd alternative convergent behavior in that region of a for first 100 or so stories of tower by myself.　It does not diverge to infinity but is oscillating between the two finite values like the series ##(-1)^n##.
So then how do I get this oscillation or alternative convergence along with the obvious one crossing point mentioned in the attached graph ?

Should the statement
If there exist a value of ##a\uparrow\uparrow\infty##, it corresponds to one of crossing points.
be right, its inverse
If there exist crossing points, one of them corresponds to ##a\uparrow\uparrow\infty##
would not stand ?

Wn(e)/e is not ##e^{-e}\uparrow\uparrow\infty## anymore ?

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aheight
Summary:: Is it possible to integrate $$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta$$

I would use it's analytic-continuation into the Lambert function:

We have:
$$\large ^{\infty}\sin(z)=-\frac{\text{W}(-\log(\sin(z)))}{\log(\sin(z))}$$
And therefore:
$$\int_0^{\pi} \large^{\infty}\sin(z)dz=\int_0^{\pi}-\frac{\text{W}(-\log(\sin(z)))}{\log(\sin(z))}dz\approx -2.32$$

Gold Member
##\sin(0)=0##, therefore ##\ln(\sin(0))## is undefined. Also, take a look at the plots from post #15: the integral is clearly positive.

aheight
##\sin(0)=0##, therefore ##\ln(\sin(0))## is undefined. Also, take a look at the plots from post #15: the integral is clearly positive.

Ok sorry, missed a sign. However, I believe the (improper) integral
$$\int_{0}^{\pi} \frac{\text{W}(-\log(\sin(z)))}{\log(\sin(z))}dz$$
nevertheless converges in the same sense as ##\displaystyle\int_0^1 \log(x)dx##. Would make an interesting problem to prove.

Gold Member
Ok sorry, missed a sign. However, I believe the (improper) integral
$$\int_{0}^{\pi} \frac{\text{W}(-\log(\sin(z)))}{\log(\sin(z))}dz$$
nevertheless converges in the same sense as ##\displaystyle\int_0^1 \log(x)dx##. Would make an interesting problem to prove.
The problem (as stated in post #21) is still that the integrand diverges over the interval ##[0,e^{-e}]##, so the integral doesn't exist.

aheight
The problem (as stated in post #21) is still that the integrand diverges over the interval ##[0,e^{-e}]##, so the integral doesn't exist.

Post #21 refers to the integral of the "power tower" which has a region of convergence on the real line equal to ##(e^{-e},e^{1/e})## which is exceeded by the limits placed on the ##\sin## argument in that integral so yes I agree the integral of the power tower beyond it's region of convergence, does not exist. However, I'm simply suggesting we could evaluate the integral using it's analytic-continuation:

$$\frac {1} {2} \int_0^\pi (sin(\theta)\uparrow \uparrow \infty)^{2} d\theta \equiv 1/2 \int_0^{\pi}\frac{-\text{W}(-\log(\sin(z))}{\log(\sin(z))}dz$$

And I argue the integral of the analytic-continuation is well-defined and converges.

Also, just for the reference so that we're thorough: The region of convergence of the power tower extends into the Complex Plane beyond the region normally given as ##(e^{-e},e^{1/e})##. That region is plainly visible in a picture of the tetration fractal.

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TeethWhitener
Gold Member
You might be onto something. At least intuitively, 2.32 is reasonable (looking at the graph, it has to be between 0 and ##\pi##). Very interesting.

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:
Mathematica:
Nest[Sin[theta]^# &, 1, 800]
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:
Mathematica:
In[59]:= Nest[Sin[0.02]^# &, 1, 500]
Nest[Sin[0.02]^# &, 1, 501]

Out[59]= 0.884202

Out[60]= 0.0314587

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

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TeethWhitener