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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 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 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 at 16:13

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