# Closed form of \begin{align}\int_{0}^{\infty}x^{1-x^{2-x^{3-...}}}dx\end{align}.

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!

• No, I mean $x^{1-(x^{2-(x^{...})})}$ Mar 16, 2023 at 23:19
• Your integrand is this Wolfram Alpha command if it helps Mar 17, 2023 at 1:28
• 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. Mar 17, 2023 at 4:39
• 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. Mar 17, 2023 at 11:57
• 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? Mar 27, 2023 at 11:30

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.