A few years ago I asked about the inequality Prove that $\int_0^\infty\frac1{x^x}\, dx<2$. As I came back to revisit it, I found that each of the following tetration integrals $$\int_0^\infty\frac{dx}{x^x},\int_0^\infty\frac{dx}{x^{x^{x^x}}},\int_0^\infty\frac{dx}{x^{x^{x^{x^{x^x}}}}},\cdots$$ appeared to be bounded above by $2$. In the plot below, each index denotes half the number of tetrations.

Obviously, one method of attack is to show that if $f_1(x)=x^x$ and $f_{k+1}(x)=x^{x^{f_k(x)}}$, then

$\int_0^\infty dx/f_{k+1}(x)>\int_0^\infty dx/f_k(x)$ for each $k>1$, and

$\lim_{k\to\infty}\int_0^\infty dx/f_k(x)<2$.

**Comments.**

- The first step in the method above means that the area gained in the interval $(0,1)$ is greater than the area lost in $(1,\infty)$. It appears that the consecutive differences (plotted below as
`%97`

) $$\int_0^\infty\frac{dx}{f_{k+1}(x)}-\int_0^\infty\frac{dx}{f_k(x)}$$ decay on the order of $k^{-\log k}$, and decrease monotonically in most instances as well.

_{ I have now cross-posted this problem on MathOverflow. }

- For the second step, the limiting case turns out to be very easy to prove. We have $$\lim_{k\to\infty}\int_0^\infty\frac{dx}{f_k(x)}<\lim_{k\to\infty}\int_0^\infty\frac{dx}{g_k(x)}=-\int_0^\infty t\cdot\frac d{dt}\frac1{t^t}\,dt=\int_0^\infty\frac1{x^x}\,dx<2$$ where $g_{k+1}(x)=x^{g_k(x)}$ and $g_1(x)=x$.