# Is $\int_0^\infty{dx\over x^{x^{x^x}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^x}}}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^{x^{x^x}}}}}}}<\cdots$ true?

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. • > 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"
– Holo
Feb 26, 2022 at 16:23
• @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. Feb 28, 2022 at 15:42
• 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
– Holo
Mar 1, 2022 at 16:13

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 $$0, this limit is approached from above, and for $$1 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 $$0, $$h_{2n}(x)$$ is monotonically decreasing as $$n$$ increases. For $$1, 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 $$0 and for $$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 $$I_3$$.

• 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 $$1.91873106231887960413475821627\ldots\,.$$ Jan 13 at 18:47
• @FredHucht: That's great, thank you! Jan 13 at 19:08
• 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 $$e^{1/e-1} - e^{1-e} + \int_{1/e}^e\mathrm dx \, x^{-x}.$$ Jan 13 at 21:52