Is each of $\int_0^\infty\frac{dx}{x^x},\int_0^\infty\frac{dx}{x^{x^{x^x}}},\int_0^\infty\frac{dx}{x^{x^{x^{x^{x^x}}}}},\cdots$ less than $2$?

Question

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

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

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

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 $g_1(x)=x$.

Answer 1

I think this can be done (he says incautiously). Once upon a time I reproduced Euler's steps with this iteration; perhaps I can do it again. The overall behaviour is, if I remember correctly, not too bad. On the interval $\exp(\exp(-1))< x < \infty$ the iteration $a_{n+1} = x^{a_n}$ diverges, monotonically growing. On the interval $\exp(-\exp(1)) \leq x \leq \exp(\exp(-1))$ the iteration converges (again I think monotonically). On the interval $0 < x < \exp(-\exp(1))$ the even-numbered iterates converge (monotonically) to one answer, while the odd-numbered iterates converge to another (the iteration has a two-cycle, but more than that the convergence is monotonic in each subsequence).

So I expect that one could prove the claim by splitting the interval up into those three and considering each one separately.

Now to go and try to put my money where my mouth is.

Answer 2

Some thoughts:

(It is not a rigorous proof. The integrals are calculated using Maple. For a rigorous proof, we need analytical upper bounds for the integrals. )

Let $$I(n) = \int_0^\infty \frac{1}{^{2n}x}\mathrm{d} x.$$

One can prove that $I(1) < 2$ and $I(2) < 2$.

In the following, assume that $n \ge 3$.

We have $$I_1(n) := \int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{1}{^{2n}x} \mathrm{d} x \le \int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{-\ln x}{W(-\ln x)} \mathrm{d} x < 1.495. $$

We have $$I_2(n) := \int_1^{5/3} \frac{1}{^{2n}x}\mathrm{d} x \le \int_1^{5/3} \frac{1}{^{6}x}\mathrm{d} x < 0.385.$$

Since $^4 x > 5$ for all $x \ge 5/3$, we have $$I_3(n) := \int_{5/3}^\infty \frac{1}{^{2n}x}\mathrm{d} x \le \int_{5/3}^\infty \frac{1}{x^{x^5}}\mathrm{d} x < 0.00005.$$

Since $^{2n}x \ge \mathrm{e}^{-1}$ for all $0 < x < \mathrm{e}^{-\mathrm{e}}$, we have $$I_4(n) := \int_{\frac35\mathrm{e}^{-\mathrm{e}}}^{\mathrm{e}^{-\mathrm{e}}} \frac{1}{^{2n}x}\mathrm{d} x \le \int_{\frac35\mathrm{e}^{-\mathrm{e}}}^{\mathrm{e}^{-\mathrm{e}}} \frac{1}{x^{x^{1/\mathrm{e}}}}\mathrm{d} x < 0.0715.$$

One can use Mathematical Induction to prove that $^{2n} x \ge \frac34 $ for all $0 < x < \frac35 \mathrm{e}^{-\mathrm{e}}$. We have $$I_5(n) := \int_0^{\frac35 \mathrm{e}^{-\mathrm{e}}} \frac{1}{^{2n}x}\mathrm{d} x \le \int_0^{\frac35 \mathrm{e}^{-\mathrm{e}}} \frac{1}{x^{x^{3/4}}}\mathrm{d} x < 0.0485.$$

Thus, $I(n) = I_1(n) + I_2(n) + I_3(n) + I_4(n) + I_5(n) < 2$.

Answer 3

In the region $0<x<1$, $\frac{1}{^{2n}x}$ is an increasing sequence. If $y\leq e$ and $x<1$, then $$\frac{1}{y}\geq\frac{1}{e}\quad\Rightarrow \quad x^{1/y}\leq x^{1/e}\quad\Rightarrow \quad x^{x^{1/y}}\geq x^{x^{1/e}}\geq\frac{1}{e}\quad\Rightarrow \quad\frac{1}{x^{x^{1/y}}}\leq e$$ so the sequence is bounded above by $e$. By the monotone convergence theorem, the sequence must converge to some $y\leq e$, and such a $y$ must satisfy $$\frac{1}{y}=x^{x^{1/y}}$$ When, $x\geq e^{-e}$, then the solution is given simply by $x=\frac{1}{^2y}$. However, when $x<e^{-e}$, we must use the paramaterization Thomas Browning gave: $$(x,y)=(t^{(t^{-t/(1-t)})/(1-t)},t^{-t/(1-t)}) \quad 0<t<1$$ It follows that $$\int_0^1\frac{1}{^{2n}x}dx<1+\int_1^e\frac{1}{^2x}dx-\int_0^1t^{(t^{-t/(1-t)})/(1-t)}\frac{d}{dt}t^{-t/(1-t)}dt<1+0.6734-0.0904=1.583$$ For $n\geq 3$, we get $$\int_1^\infty\frac{1}{^{2n}x}dx\leq \int_1^\infty\frac{1}{^{6}x}dx=\int_0^1\frac{(1/x)^2}{^{6}(1/x)}dx<0.3838$$ Hence, $$\int_0^\infty\frac{1}{^{2n}x}dx \leq 1.583+0.3838=1.9668<2$$ Since we already know $$\int_0^\infty \frac{1}{^2x}<2$$ we need only check $$\int_0^\infty \frac{1}{^4x}dx=\int_0^1 \frac{1}{^4x}dx+\int_0^1\frac{(1/x)^2}{^{4}(1/x)}dx<1.4026+0.4324=1.835<2$$ which completes the proof.

Comment. There isn't a lot elegant going on here integral-wise. The bounds can be calculated using Reimann sums with sufficiently small meshes ($\delta\approx10^{-6}$) so as to bound the error term. By extending the error term given here to many to one functions, we find that the computed integrals here have an error term given by $$k\delta(\max f-\min f)\quad\quad\quad k=\max\{|f^{-1}(y)|:y\in\mathbb{R}\}$$ where $f$ is the integrand, and the maxima and minima are over the range of integration. For the functions above $f\geq 0$, $k\leq 2$ and bounds on the maxima are easy to compute.

Answer 4

Some thought :

From here https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0 :

We have :

$$ f(x)=\frac{-(W(-\ln(x)))}{(\ln(x))}= x^{x^{·^{·^·}}}$$ for $e^{-e}<x<1$

We have :

$$x^{x}> ^{4}x >\cdots>f(x)$$

Now it seems we have on $[\frac{2}{5},1)=I$ :

$$f''(x)>0$$

So for $a,b,c\in I$ we have :

$$\int_{a}^{b}\frac{1}{f'\left(c\right)\left(x-c\right)+f\left(c\right)}dx\geq\int_{a}^{b}\frac{1}{f\left(x\right)}dx$$

So we have an arbitrary accuracy on this interval .

Hope it helps .

Edit :

Some hint to show the convexity of $f(x)$ on $(I)$ :

We have :

$$f'(x)=\frac{e^{-2W\left(-\ln\ x\right)}}{x\left(W\left(-\ln\ x\right)+1\right)}$$

We substitute :

$$y=W\left(-\ln\ x\right)$$

We get :

$$f'\left(e^{-ye^{y}}\right)=\frac{e^{-2y}}{\left(y+1\right)e^{-ye^{y}}}$$

Now we are interested by :

$$g(y)=\frac{e^{-2y}}{\left(y+1\right)e^{-ye^{y}}}$$

$$g'(y)=\frac{e^{\left((e^{y}-2)y\right)}\left(e^{y}(y+1)^{2}-2y-3)\right)}{\left(y+1\right)^{2}}$$

We deduce that the function $g(y)$ is increasing on $(0.55,1)$

Now conclude is easy !

Alternative approach 25/02/2022 :

I use an integral analogue of the Kantorovitch inequality but before a fact :

Let $0<x\leq 1$ then we have :

$$2-^{2n}x\leq \frac{1}{^{2n}x}$$

Now the theorem :

[Schweitzer,P.] If $g,1/g\in L([a,b])$ with $0<m\leq f\leq M$ then :

$$\int_{a}^{b}g\left(x\right)dx\cdot\int_{a}^{b}\frac{1}{g\left(x\right)}dx-\frac{\left(M+m\right)^{2}}{4Mm}\left(b-a\right)^{2}\leq 0$$

Now we can use the lemma and the theorem and find an upper bound but before all we need to find a lower bound for $^{2n}x$ .I think we can use Mathematical induction as the nice idea @RiverLi .

Edit 11/04/2022: Using this link About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$ and defining :

$$f\left(x\right)=x^{-x^{x^{x^{x^{x}}}}}$$

We have :

$$\int_{0}^{\frac{5}{3}}x^{-x^{\frac{16}{27}}}dx+\int_{\frac{5}{3}}^{\infty}f\left(x\right)dx<2$$

Edit 27/04/2022 :

Using the previous edit and the Vasc's paper about three open exponential inequalities combining with a reversed Cauchy Schwartz inequality for integrals and Jensen's inequality for integrals (applying to the logarithm) we can find a bound for the integral of :

$$x^{x^{16/27}}$$ on $(0,1)$

It seems we can also use the inequality on $[1,1.5)$ :

$$^{6}x\ge (2-x^{1.55})^{-0.5}$$