Integrating The Infinite Power Tower

Calculus Level 5

$\Large f(x)=x^{x^{x^{x^{\dots}}}}$

The function above is what is known as the infinite power tower . At first glance, the function above may seem like it quickly spirals off to infinity, but for select values of $x$ , surprisingly the function converges. If $x\in[e^{-e},e^{\frac{1}{e}}]$ , then $f(x)$ is defined and converges to a finite number. What is the value of the integral below? Give your answer to 3 decimal places.

$\Large \displaystyle\int_{e^{-e}}^{e^{\frac{1}{e}}} f(x)\, dx$

The answer is 1.24413.

3 solutions

Garrett Clarke
Jul 29, 2015

$y=x^{x^{x^{x^{\dots}}}}$

$x^y=x^{x^{x^{x^{\dots}}}}$

$y=x^y$

$x=y^\frac{1}{y}$

$f^{-1}(x)=x^{\frac{1}{x}}$

I present to you the following theorem and its associated proof without words:

$\displaystyle\int_{a}^b f(x)dx+\displaystyle\int_{f(a)}^{f(b)}f^{-1}(x)dx = bf(b)-af(a)$

Letting $a=e^{-e}$ and $b=e^{\frac{1}{e}}$ :

$\left(\frac{1}{e}^{\frac{1}{\frac{1}{e}}}\right) = e^{-e} = a\Longrightarrow f(a)=\frac{1}{e}$

$e^\frac{1}{e} = b \Longrightarrow f(b) = e$

Approximating the integral of the inverse function:

$\displaystyle\int_{\frac{1}{e}}^{e}x^{\frac{1}{x}}dx \approx 2.65861\dots$

Plugging in our values, solving for $\displaystyle\int_{a}^b f(x)dx$ and approximating gives us our desired result:

$\displaystyle\int_{e^{-e}}^{e^{\frac{1}{e}}}f(x)\,dx = \left(e^{\frac{1}{e}}\right)\left(e\right)- \left(e^{-e}\right) \left(\frac{1}{e}\right)-\displaystyle\int_{\frac{1}{e}}^{e}x^{\frac{1}{x}}dx \approx \boxed{1.24413\dots}$

That's exactly how I did it. Terrific shot of that infinite staircase skyscraper there.

Michael Mendrin - 5 years, 10 months ago

lovely solution..

Ananya Aaniya - 5 years, 10 months ago

Even i did the same way. But i used a calculator to solve the approximate integral which u also used. Can u please help me?? @Garrett Clarke

Aditya Kumar - 5 years, 10 months ago

I will admit it's not an easy task! I personally used WolframAlpha to get the final answer, but there certainly are ways to get the correct answer using numerical approximation. Here's one way of doing it:

First off, many methods involve approximation by adding up the areas under the curve in pieces. Basically it works by taking a small portion of the x-axis (known as the step size) and multiplying it by the value of the function at that point. The smaller the step size, the more areas to add up and the better the approximation. When the bounds for our integral are $\frac{1}{e}$ and $e$ , it's not exactly clear what our step size should be; we'd like the step size to be a rational number. This means our first step must be to use U-Substitution to give us rational bounds. Consider the following:

$x=e^u$

$dx = e^u\,du$

$\frac{1}{e} = e^{-1}$

$e = e^1$

$\displaystyle\int_{\frac{1}{e}}^e x^{\frac{1}{x}}\,dx = \displaystyle\int_{-1}^1 e^{u(1+e^{-u})}\,du$

Now I won't say it's a prettier integral, but it is one that we can more easily work with, as our bounds are integers. The approximation method has the follows:

$\displaystyle\int_a^b f(x) = \displaystyle\sum_{i=1}^n hf(x_i)$

Where $h$ is your step size, $n=\frac{b-a}{h}$ , and $x_i=a+ih$ This creates a series of rectangles that can added up to approximate our integral.

Unfortunately, this method does not approximate very quickly. Our step size must be quite small (smaller than 0.001) for us to get an approximation that will yield us an answer close enough to the true answer to be acceptable.

However, I would like to refer you to the comment I made to Aman Raimann below. There I show a different integral that approximates much faster, when $h$ is as small as 0.01. Still, that means you're summing 200 terms, which means you still need a calculator anyway.

In summary, when working with definite integrals of complex functions, sometimes it's best to just let a computer do the work for you. Thanks for the question!

Garrett Clarke - 5 years, 10 months ago

Thanks a lot! That'll help me. I didn't know about the approximation method.

Aditya Kumar - 5 years, 10 months ago

How do you evaluate the integral without approximating? I was stuck on that for a long time and finally gave up.

Deeparaj Bhat - 5 years, 2 months ago

OhMyGod This is so beautiful!

Pi Han Goh - 5 years, 10 months ago

I agree 100%, this is probably my favorite problem I've made so far.

Garrett Clarke - 5 years, 10 months ago

i got the answer. yes

Mugundhan Nallathambi - 5 years, 10 months ago

Aditya Kumar
Aug 9, 2015

$Now,\quad I'll\quad simplify\quad the\quad approximation\quad of\quad \int _{ \frac { 1 }{ e } }^{ e }{ { x }^{ \frac { 1 }{ x } } } dx\\ \int _{ \frac { 1 }{ e } }^{ e }{ { x }^{ \frac { 1 }{ x } } } dx\quad =\quad \int _{ \frac { 1 }{ e } }^{ e }{ { e }^{ \frac { logx }{ x } } } dx\quad =\quad I\\ Since\quad we\quad want\quad to\quad approximate\quad the\quad value\quad of\quad I,\\ we\quad can\quad use\quad taylor\quad series\quad of\quad expansion\quad of\quad { e }^{ x }\\ I\quad =\quad \int _{ \frac { 1 }{ e } }^{ e }{ \left( 1+\frac { \frac { logx }{ x } }{ 1! } +\frac { { \left( \frac { logx }{ x } \right) }^{ 2 } }{ 2! } +\frac { { \left( \frac { logx }{ x } \right) }^{ 3 } }{ 3! } +... \right) } dx\\ on\quad splitting\quad the\quad integrals\quad and\quad solving\quad them,\quad we\quad get\\ I\quad \approx \quad 2.65861$

@Garrett Clarke @Pi Han Goh @Michael Mendrin @NILAY PANDE can u verify it as i learnt approximate integrals recently.

Aditya Kumar - 5 years, 10 months ago

Yeah this is right too. But how do you know the convergence rate? You need to plug in one number at a time to increase the accuracy. That is, how many terms do we need to have the sufficient number of significant figures?

Furthermore, it's very tedious to integrate all these terms by hand.

Damn. I didn't thought of this. Your solutions are an eye opener.

Pi Han Goh - 5 years, 10 months ago

U r right I should've added more terms to it as it would've have brought in more accuracy. It took me 2 pages to just approximate this integral.

Thanks for the compliment!

Aditya Kumar - 5 years, 10 months ago

Aman Rajput
Jul 31, 2015

I used : $x^{x^{x^{\dots}}} = \frac{W(-\log x)}{-\log x}$

where $W$ is the Lambert's W function , also called omega function.

Then using numerical method or wolframalpha, we get

$\displaystyle \int\limits_{1/e^e}^{e^{1/e}} \frac{W(-\log x)}{-\log x} dx = \boxed{1.244}$

You know I tried putting that function into WolframAlpha and it wouldn't accept my input, as it often does for computations involving obscure functions like the Lambert W function. There is a way we can rewrite the function above, however, using U-Substitution that allows Wolfram to understand the input much more easily. Observe the following:

First, let $u=W(-ln(x))$ :

$-ln(x)=ue^u\Longrightarrow x=e^{-ue^u}$

$dx = -e^{u-e^u u} (1+u)\, du$

$W(-ln(\frac{1}{e^e}))=W(e)=1,\, W(-ln(e^{\frac{1}{e}}))=W(-\frac{1}{e})=-1$

Subbing in for $x$ , $dx$ and our bounds:

$\displaystyle\int_{1}^{-1} \frac{u}{ue^u}(-e^{u-e^u u} (1+u))\, du$

$\displaystyle\int_{-1}^{1} e^{-e^u u} (1+u)\, du$

This integral is a bit easier to wrap our head around and avoids the use of non-elementary fuctions. Good insight to use this approach though!

Garrett Clarke - 5 years, 10 months ago

you should know how to write that function as an input...

type productlog[-logx]

Aman Rajput - 5 years, 10 months ago

Ahhh yes that worked! My mistake was just typing in "integrate W(-ln(x))/(-ln(x)) from 1/e^e to e^(1/e)" rather than using "productlog[-ln(x)]". I'll be sure to remember that, thanks! And don't worry I've got plenty more calculus questions where that came from!

Garrett Clarke - 5 years, 10 months ago

@Garrett Clarke – i'll be waiting for your questions my friend. ;)

Aman Rajput - 5 years, 10 months ago

and @Garrett Clarke please keep posting good calculus question ... loved this one.

Aman Rajput - 5 years, 10 months ago

Integrating The Infinite Power Tower

The answer is 1.24413.

3 solutions

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