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


  • 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 $g_1(x)=x$.
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    Nice question (+1) For more than $15$ $x's$, we can safely replace the interval to $[0,3]$ without making a significant error , probably even smaller than $[0,y]$ with some $y<2$. But whether we can calculate those integrals numerically, is another story. The function just gets too big. – Peter Feb 15 '22 at 18:04
  • The sequence of values seems to become almost stationary although the power tower oscillates for small values between two accumulation points depending on the number of entries. How can this be explained ? – Peter Feb 15 '22 at 18:31
  • Have you tried Pade approximation? There are some experts on that on this site who can probably bash this. Nevertheless, it's quite fascinating that the odd number of tetrations diverge. – dezdichado Feb 17 '22 at 18:02
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    @dezdichado I experimented with it for a bit and found that it is sometimes quite tricky to control for parts of the approximant where it becomes a lower bound, or near singularities. Probably this is the way to go (there is about $0.1$ margin for error), and I am particularly curious why the sequence is monotonically increasing - [these positive values](https://cdn.discordapp.com/attachments/929707996672561182/943832040493690911/cons.png) are the consecutive differences. – TheSimpliFire Feb 17 '22 at 18:24
  • Why not finish the “one method of attack”? – Tyma Gaidash Feb 18 '22 at 02:25
  • If we split it into $(0, \mathrm{e}^{-\mathrm{e}})$, $(\mathrm{e}^{-\mathrm{e}}, 1)$ and $(1, b)$ and $(b, \infty)$ ($b$ is chosen later), I think the difficult part is $(0, \mathrm{e}^{-\mathrm{e}})$. The easy part is $\int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{1}{^{2n}x} \mathrm{d} x \le \int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{-\ln x}{W(-\ln x)} \mathrm{d} x = 1 - \mathrm{e}^{1 - \mathrm{e}} + \int_1^\mathrm{e} y^{-y}\mathrm{d} y$. – River Li Feb 19 '22 at 13:04
  • @RiverLi If one does prove inequalities for $(b,\infty)$ and $(0,e^{-e})$, how would you combine them all into one inequality and answer the question; maybe add/subtract the inequalities if it is possible? – Tyma Gaidash Feb 19 '22 at 13:24
  • @TymaGaidash I mean find good estimation of the four parts, and see if the sum less than 2. According to rough bounds, the sum can be near 2. So perhaps we need good estimation. – River Li Feb 19 '22 at 13:30
  • By the way, for $0 < x < \mathrm{e}^{-\mathrm{e}}$ and any positive integer $n$, we have $^{2n}x \ge \mathrm{e}^{-1}$. Perhaps there are better bounds such as the form $^{2n}x \ge 1 - c\sqrt x$ or $^{2n} x \ge 1 + c x\ln x$. – River Li Feb 19 '22 at 13:52

4 Answers4


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

River Li
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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.

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  • Maple says $\int_0^1 \frac{1}{^{20} x} \mathrm{d} x \approx 1.550803841$ which is different from your result $\int_0^1\frac{1}{^{2n}x}dx < 1.547$. Would you please calculate $\int_0^1 \frac{1}{^{20} x} \mathrm{d} x$ to see which is true? – River Li Feb 22 '22 at 04:50
  • This answer adds both $[0,1]$ to $[1,\infty)$ intervals easily which is a much better strategy than my deleted answer. – Tyma Gaidash Feb 22 '22 at 05:23
  • @RiverLi oops I accidentally switched the 7 and 3 in 0.6734. I will fix that – Jacob Feb 22 '22 at 06:35
  • Another question: You said "When, $x\geq e^{-e}$, then the solution is given simply by $y=\frac{1}{^2x}$". When $e^{-e} < x < 1$, we know that $x^x > ^4 x > ^6 x > \cdots \to \frac{W(-\ln x)}{-\ln x}$, and $\frac{1}{x^x} < \frac{1}{^4 x} < \cdots \to \frac{-\ln x}{W(-\ln x)}$, right? – River Li Feb 22 '22 at 06:52
  • @RiverLi yes those are two equivalent ways of stating it. – Jacob Feb 22 '22 at 21:37
  • @Jacob Can you give details for the equivalence? – River Li Feb 23 '22 at 00:34
  • @Jacob Can you check the integral $\int_0^1 t^{(t^{-t/(1-t)})/(1-t)}\frac{d}{dt}t^{-1/(1-t)}dt = - 0.0904$? Maple says it is $-\infty$. My calculation: First, $\frac{d}{dt}t^{-1/(1-t)} = t^{-1/(1-t)}(\frac{-\ln t}{(1 - t)^2} - \frac{1}{t(1 - t)})$. Second, $\int_0^1 t^{(t^{-t/(1-t)})/(1-t)} \cdot t^{-1/(1-t)}(\frac{-\ln t}{(1 - t)^2} - \frac{1}{t(1 - t)}) dt = -\infty$. – River Li Feb 23 '22 at 01:07
  • @RiverLi That's strange, I don't know how that slipped through. The paramaterization should have $y=t^{-t/(1-t)}$, not $y=t^{-1/(1-t)}$. I've edited accordingly. – Jacob Feb 23 '22 at 03:27
  • @Jacob It is not very clear to me. Perhaps, you may give more details about how to get $\int_0^1\frac{1}{^{2n}x}dx<1+\int_1^e\frac{1}{^2x}dx-\int_0^1t^{(t^{-t/(1-t)})/(1-t)}\frac{d}{dt}t^{-t/(1-t)}dt$. Did you split $(0, 1)$ into $(0, e^{-e})$ and $(e^{-e}, 1)$? For $x > e^{-e}$, what is your limit of sequence $\frac{1}{^{2n} x}$? It should be $\frac{-\ln x}{W(-\ln x)}$. I did not see it, so you may do something different. – River Li Feb 23 '22 at 03:42
  • @RiverLi for $x>e^{-e}$, $$y=\frac{-\ln x}{W(-\ln x)}\quad \text{iff}\quad x=\frac{1}{y^{-y}}$$ – Jacob Feb 24 '22 at 22:12
  • @Jacob I think you mean $x = \frac{1}{y^y}$ and $e^{-e} < x < 1$. Yes, it is true. But I don't understand " 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 $y=\frac{1}{^2x}$". "by $y = \frac{1}{^2 x}$" do you mean "by $x = \frac{1}{^2 y}$"? – River Li Feb 25 '22 at 00:14
  • Actually, in my answer, I have used the substitution $x = y^{-y}$ to get $\int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{1}{^{2n}x} \mathrm{d} x \le \int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{-\ln x}{W(-\ln x)} \mathrm{d} x = 1 - \mathrm{e}^{1 - \mathrm{e}} + \int_1^\mathrm{e} y^{-y}\mathrm{d} y$. – River Li Feb 25 '22 at 00:20
  • @Jacob Your approach is very nice (though there are 3 typos, and some details are missing). (+1) According to your approach, $\int_0^{e^{-e}} \frac{1}{^{2n} x} \mathrm{d} x \le \int_0^1 y(t) x'(t) \mathrm{d} t$. Also, $\int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{1}{^{2n}x} \mathrm{d} x \le \int_{\mathrm{e}^{-\mathrm{e}}}^1 \frac{-\ln x}{W(-\ln x)} \mathrm{d} x = 1 - \mathrm{e}^{1 - \mathrm{e}} + \int_1^\mathrm{e} y^{-y}\mathrm{d} y$. Adding them up, we get your result. – River Li Feb 25 '22 at 01:00

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 $[\frac{2}{5},1)=I$ :


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


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 :


Now we are interested by :



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 :


We have :


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 $[1,1.5)$ :

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

Erik Satie
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  • It works on $(0,0.95]$ with https://math.stackexchange.com/questions/879832/6-cauchy-schwarz-for-integrals?rq=1] – Erik Satie Apr 28 '22 at 11:14

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.

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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 16 '22 at 14:48
  • Aha! The claim is not true, essentially for the reason that I hinted at. Integrate 1/x^(x^x) on the interval 1.0e-8 up to exp(-exp(1)), and one gets a value of more than 14. Yes that's numerical integration, but the function is monotonically decreasing there and a right-hand Riemann sum will establish a clear inequality. – Rob Corless Feb 16 '22 at 15:36
  • More, 1/x^(x^x) = 1/x - (ln x)^2 + higher order terms as x->0 so the integral is actually divergent. – Rob Corless Feb 16 '22 at 15:41
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    @RobCorless Your answer may be better posted as a comment (at least for now). Note that I intentionally considered only even tetrations ($1/x^x$, $1/x^{x^{x^x}}$ etc) which converge to $\approx1.918$ rather than odd numbers of tetrations like $1/x^{x^x}$, which diverge. – TheSimpliFire Feb 16 '22 at 17:08
  • You are quite right. I missed that. – Rob Corless Feb 16 '22 at 21:45
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    Hi @RobCorless , welcome. In general, something should be written as an answer if it actually solves the problem or at least the crux of the problem i.e., a good hint. Anything else, such as an idea where the person writing isn't positive that it will work, such as this, should instead be posted as a comment . This is to make it easier for people navigating this site. – Mike Feb 17 '22 at 18:15