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On MSE I asked whether each of $\int_0^\infty\frac{dx}{x^x},\int_0^\infty\frac{dx}{x^{x^{x^x}}},\int_0^\infty\frac{dx}{x^{x^{x^{x^{x^x}}}}},\cdots$ was less than $2$ and received answers on bounding the tetration integrals by a smaller-order power tower, but the numerical results were attained computationally.

While I don't know whether a simple, analytical proof exists to prove the upper bound of $2$, I was especially curious as to why the integrals from $1/x^{x^{x^x}}$ onwards seemed to form a monotonically increasing sequence

Is this observation true? That is, if $f_0(x)=x^x$ and $f_{k+1}(x)=x^{x^{f_k(x)}}$, then $$\int_0^\infty{dx\over f_k(x)}<\int_0^\infty{dx\over f_{k+1}(x)}$$ for all $k\ge1$ which can be interpreted as the area gained over $(0,1)$ is greater than that lost over $(1,\infty)$.


Wolfram Cloud gives these consecutive differences. It appears that these differences decay on the order of $x^{-\log x}$, and decrease monotonically in most instances as well.

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    $\begingroup$ > which can be interpreted as the area gained over (0,1) is greater than that lost over (1,∞) $$$$ Well, if we say $M_k=\int_1^\infty{dx\over f_k(x)}$ and $m_k=\int_0^1{dx\over f_k(x)}$, we have $0<M_k \to 0$ and $m_k\to\infty$, moreover $m_k$ grows rate is above linear, so after a certain point $M_{k}-M_{k+1}<m_{k+1}-m_k$, which is exactly "$\int_0^\infty{dx\over f_k(x)}$ is eventually monotonic increasing". It just looks like $k=1$ is that "certain point" $\endgroup$
    – Holo
    Feb 26, 2022 at 16:23
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    $\begingroup$ @Holo $m_k$ does not tend to infinity (its growth rate is below linear). We have $1/f_k(0)=1$ rather than $+\infty$ (as is the case for odd tetrations) and $\max1/f_k(x)=e$. In fact, $m_{k+1}-m_k\to0$ also, so it is difficult to compare them at first sight. $\endgroup$ Feb 28, 2022 at 15:42
  • $\begingroup$ oh right, I forgot how tetretion works for x<1. a transformation between the m_k to the M_k looks possible, but I'm not sure what it would be $\endgroup$
    – Holo
    Mar 1, 2022 at 16:13

1 Answer 1

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Let $h_1(z)=z$, and let $h_{n+1}(z)=z^{h_n(z)}$ so that $h_{2n}(z)$ is the even power tower you are interested in. In this partial answer, I'll show that:

Proposition 1: We have that $$\lim_{n\rightarrow \infty} \int_0^\infty \frac{1}{h_{2n}(x)} dx = C_h=1.91873106\dots $$ where $C_h$ may be expressed exactly as $$C_h=e^{1−e}+\int_{-1/e}^e\frac{e^{-u}u}{W(u)}du-\int_1^{e}e^{uW_{-1}[-\log u/u]} du$$ where $W$ and $W_{-1}$ are branches of the W-Lambert Function.

Thank you Fred Hucht for pointing out in the comments how to explicitely state the integral involving $W_{-1}$.

I also outline an approach to prove monotonicity.


Proof of Proposition 1

The first thing to note is that $e^{1/e}$ is an important transition point for the integrand. Indeed, the integral should be split into three relevant ranges: $$(1)\ \ \ \ \ \ \ \ \ \ I_1=(0,e^{-e}),\ \ \ I_2=(e^{-e},e^{1/e}), \ \ \ I_3= (e^{1/e},\infty).$$

On $I_3$, the power tower becomes vanishingly small very fast. On $I_2$, it converges, and on $I_1$, we will need to introduce a new function to discuss convergence in the case of $n$ even.

Lemma 1: For any $x\in I_2$, $$\lim_{n\rightarrow\infty}h_n(x)=\frac{W(-\log x)}{-\log x}$$ where $W$ is the W-Lambert Function. Alternatively, this equals $u$ where $u$ is the unique value such that $u^{1/u}=x$.

Proof: See Exponentials Reiterated.

At first glance, it is surprising that $1.4^{1.4^{1.4^{\dots}}}$ converges, but it does indeed converge.

To extend this Lemma to $I_1$, we need to restrict to $n$ even, and define the following inverse function:

Definition 1: For $x \in (0,e^{1/e})$, let $L(x)$ denote the unique point $u$ such that $\frac{1}{e}\leq u \leq e$ and $x^{x^u}=u$.

With this definition in mind, we have the following Lemma:

Lemma 2: For any $x \in (0,e^{1/e})$, $$\lim_{n\rightarrow \infty} h_{2n}(x) = L(x).$$ For $0<x<1$, this limit is approached from above, and for $1<x<e^{1/e}$ this limit is approached from below.

Proof: See Exponentials Reiterated.

In particular, due to the monotonicity on the different ranges, Lemma 2 implies uniform convergence, and hence $$\lim_{n\rightarrow \infty}\int_0^{e^{1/e}} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx.$$

Lastly, we note that the integral over $I_3$ tends to zero very quickly. Indeed:

Lemma 3: We have that $$\lim_{n\rightarrow \infty}\int_{I_3} \frac{1}{h_{2n}(x)}dx =0.$$

Proof: For $\epsilon>0$, there exists $N$ such that for $n>N$, $h_{2n}(e^{1/e}+\epsilon)>\frac{1}{\epsilon}$. Hence we can split into ranges $(e^{1/e},e^{1/e}+\epsilon)$, $(e^{1/e}+\epsilon,2)$, and $(2,\infty)$, and obtain a bound of $C\epsilon$ for some fixed constant $C$.

All together, Lemma's $1$, $2$, and $3$ imply the following Proposition, which implies Proposition $1$:

Proposition 2: We have that $$\lim_{n\rightarrow \infty}\int_0^{\infty} \frac{1}{h_{2n}(x)}dx=\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx.$$ $$ = \int_{0}^{e^{-e}}\frac{1}{L(x)}dx + \int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx.$$

There are a few ways to clean up this second integral, such as $$\int_{e^{-e}}^{e^{1/e}}\frac{-\log x}{W(-\log x)}dx = \int_{\frac{1}{e}}^{e}\frac{1}{x}x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)dx$$ $$=\int_{-1}^{1}e^{-2u}e^{e^{-u}u}(1-u)du.$$

Fred Hucht pointed out in the comments that the integral $\int_{0}^{e^{1/e}}\frac{1}{L(x)}dx$ can be expressed in terms of $W_{-1}(x)$, a different branch of the W-Lambert function, in the following way:

$$\int_{0}^{e^{-e}}\frac{1}{L(x)}dx = e^{1−e}-\int_1^{e}e^{uW_{-1}[-\log u/u]} du.$$ This implies the exact expression: $$\lim_{n\rightarrow \infty}\int_0^{\infty} \frac{1}{h_{2n}(x)}dx = e^{1−e}+\int_{-1/e}^e\frac{e^{-u}u}{W(u)}du-\int_1^{e}e^{uW_{-1}[-\log u/u]} du.$$

Monotonicity

For $0<x<1$, $h_{2n}(x)$ is monotonically decreasing as $n$ increases. For $1<x<e^{1/e}$, it is monotonically increasing. I believe that these two effects can be combined together in the following way. Consider $$\int_{I_2} \frac{1}{h_{2n}(x)}dx=\int_{-\frac{1}{e}}^{e}\frac{e^{-u}}{h_{2n}(e^{-u})}du.$$ Splitting at $x=1$, which corresponds to $u=0$, this decomposes into $$\int_{0}^{e}\frac{e^{-u}}{h_{2n}(e^{-u})}du+\int_{0}^{\frac{1}{e}}\frac{e^{u}}{h_{2n}(e^{u})}du=\int_{0}^{1}\frac{e^{-eu}}{h_{2n}(e^{-eu})}edu+\int_{0}^{1}\frac{e^{u/e}}{h_{2n}(e^{u/e})}\frac{1}{e}du$$

$$=\int_{0}^{1}\left(\frac{e^{1-eu}}{h_{2n}(e^{-eu})}+\frac{e^{u/e-1}}{h_{2n}(e^{u/e})}\right)du.$$

Claim 1: The function $$\frac{e^{1-ex}}{h_{2n}(e^{-ex})}+\frac{e^{x/e-1}}{h_{2n}(e^{x/e})}dx$$ is monotonically increasing in $n$ for $0<x<1$ and for $n>3$.

Monotonicity would then follow from a combination of:

  1. Stronger quantitative bounds on the integral over $I_3$.
  2. A proof of Claim 1.
  3. The fact that $h_{2n}(x)$ converges to $L(x)$ from above on $I_1$.

The convergence from above on $I_1$ should outweigh the error on $I_3$.

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    $\begingroup$ First note some typos: You mixed up $L(x)$ and $1/L(x)$ a bit in your answer, and a $u$ is missing in the third integral after Prop. 2. Now my point: By integrating the inverse function of $1/L$ we find \begin{align}\int_{0}^{e^{-e}}\frac 1 {L(x)} \,\mathrm dx = e^{1-e}-\int_{1}^{e} e^{u\, W_{-1}[-\log(u)/u]} \, \mathrm du,\end{align} with $W_{k}(z) = \mathrm{ProductLog}[k,z]$. The limiting value from Prop. 1 now is $$1.91873106231887960413475821627\ldots\,.$$ $\endgroup$
    – Fred Hucht
    Jan 13, 2023 at 18:47
  • $\begingroup$ @FredHucht: That's great, thank you! $\endgroup$ Jan 13, 2023 at 19:08
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    $\begingroup$ The $u$ in the third integral after Prop. 2 is still missing, should be $e^{e^{-u}u}$. Furthermore, this integral can be simplified by a partial integration to simply give $$ e^{1/e-1} - e^{1-e} + \int_{1/e}^e\mathrm dx \, x^{-x}. $$ $\endgroup$
    – Fred Hucht
    Jan 13, 2023 at 21:52

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