I want to find the closed form of $\begin{align}\int_{0}^{\infty}x^{1-x^{2-x^{3-...}}}dx\end{align}$.

Quick disclaimer: I have no reason to believe one actually exists

Using Desmos, the closest I have gotten is $1.2421832267$.

I have noticed that when it is repeated an odd amount of times, for example $x^{1-x^{2-x^3}}$ the integral does not converge and I would have to change it to $x^{1-x^{|2-x^3|}}$. Is there any way to avoid having to do this (perhaps with a slightly altered equation)? My question is still for the closed form.

I first tried to find the closed form by putting it in Desmos and then plugging the decimal it gave me into WolframAlpha in hopes of getting a closed form but it didn't give me anything, even after I wrote a Python script so I could copy and paste 200 repetitions into Desmos in an attempt to get a more accurate decimal.

Here is my other approach:

The closed form of $\begin{align}\int_{0}^{\infty}x^{1-x^{2-x^{3-...}}}dx\end{align}$ is equal to $\begin{align}\lim_{a\to\infty}\int_{0}^{a}x^{1-x^{2-x^{3-...}}}dx\end{align}$.

This is where I am stuck. Thanks in advance for the help!

  • 1
    $\begingroup$ No, I mean $x^{1-(x^{2-(x^{...})})}$ $\endgroup$ Mar 16, 2023 at 23:19
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    $\begingroup$ Your integrand is this Wolfram Alpha command if it helps $\endgroup$ Mar 17, 2023 at 1:28
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    $\begingroup$ Is this the Somophore's nightmare ? (kidding). Interesting problem. If you want 50 exact decimal places, it is $$1.2421832266975400643278225490476835939896793761402$$ which is not recognized by inverse symbolic calculators. $\endgroup$ Mar 17, 2023 at 4:39
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    $\begingroup$ Thanks! What did you do to get 50 decimals? I wanted to try the inverse symbolic calculator but I could only find as many characters as I had given before. $\endgroup$ Mar 17, 2023 at 11:57
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    $\begingroup$ This is some kind of recursive definition, but I fail to see the precise definition here. How is that sequence of functions really defined? How do we know it is convergent? $\endgroup$
    – freakish
    Mar 27, 2023 at 11:30

1 Answer 1


I was hoping to provide something for the $\huge \pi$ day.

For an error of $1.33\times 10^{-51}$, the number

$${\large\Xi}=1.2421832266975400643278225490476835939896793761402$$ (obtained using $88$ levels) is such that it could write $$8058940\,{\large\Xi}=-5866165 \binom{\pi }{\pi !}+4785910 \binom{\pi !}{\pi }-14882994 \binom{\pi !}{\log (\pi )}+$$ $$12531710 \binom{\log (\pi )}{\pi !}-14673958 \binom{\pi }{\log (\pi )}-761197 \binom{\log (\pi )}{\pi }$$ which $\cdots\cdots$ does not mean anything.


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