I have been trying to find an interesting constant over the domain of the infinite tetration of x and have just almost figured out the area with a non integral infinite sum representation. Just one constant is in my way. D denotes the domain. This question is different from this one as it has the full domain such that the imaginary part is $0$ and not for a single point. Here is a demo of my expansion. Here is my source for the product logarithm/W-Lambert function series. My inspiration for the series is here.

Finally here is a graph of the constant. My work is as follows. I used a bit of software to help with the evaluation at the end. Here is data about the generalized incomplete gamma function used here.

enter image description here

$$ \mathrm{G=\int_D x^{x^{x^…}}dx=\int_D {^\infty x}\, dx=\int_D-\frac{W(-ln(x))}{ln(x)}dx= 1.265188689361227081430914184615901039501069191363542653701819999950085943915822836313002058708863484…\implies G+\int_0^{\sqrt[-e]e }\frac {W(-ln(x))}{ln(x)}dx= \int_{\sqrt[-e]e}^{\sqrt[e]e}\frac {W(-ln(x))}{ln(x)}dx= \sum_{n=1}^\infty\frac{n^{n-1}}{n!} \int_{\sqrt[-e]e}^{\sqrt[e]e} ln^{n-1}(x)dx= \sum_{n=1}^\infty\frac{(-n)^{n-1}}{n!}Γ\left(n,-\frac 1e,\frac 1e\right)=\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)= 0.886369135921835965080748…=G-0.378819553439391116350165…} $$

I have also found the amazing result of being able to integrate the $\mathrm{x^{th}}$ root of x using a theorem on the integral of an inverse function. Here is my work.

$$\mathrm{\int_{eW\left(\frac 1e\right)}^e \left(x^\frac 1x=\sqrt[x]x\right)dx+\int_{e^{-\frac 1e}}^{e^\frac 1e} {^\infty x}\,dx=e^{1+\frac1e}-e^{1-\frac1e} W\left(\frac 1e\right)=e^{1-\frac1e}\left(e^\frac2e-W\left(\frac 1e\right)\right)=e^{1+\frac1e}\left(1-e^{-\frac2e}W\left(\frac 1e\right)\right)\implies \int_{eW\left(\frac 1e\right)}^e x^\frac 1xdx= e^{1+\frac1e}-e^{1-\frac1e} W\left(\frac 1e\right)-\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)}$$

In order to find an exact form of G, I need to find the following. The other form uses the following identity here: $$\mathrm{I= \int_0^ {e^{-\frac 1e}} x^{x^{x^…}}dx=\int_0^{e^{-\frac 1e}} {^\infty x}\, dx=\int_0^ {e^{-\frac 1e}} -\frac{W(-ln(x))}{ln(x)}dx=e^{1-\frac1e}W\left(\frac1e\right)-\int_0^ {eW\left(\frac1e\right)} \sqrt[x]x dx=0.378819553439391116350165…}$$

The previous result is proof of the following. I guess this link here is not accurate anymore. I will give an example if wanted of this result. Using another Wikipedia theorem proves that: $$\mathrm{\int x^{\frac1x}dx=x^{\frac1x+1}+\sum_{n=1}^\infty (-1)^nn^{n-2} Q\left(n,-\frac{ln(x)}{x}\right)+C,eW\left(\frac1e\right)\le x\le e}$$

more info on this result

I found this series based on this answer from @mathphile which does not give the right result as the n=0 term diverges and even trying to use the lower integration bound as $e^{-e}$ still gives the wrong result. The user’s answer would have cracked the question.

How do I evaluate this integral? A closed form is wanted, but optional. Please give me any hints as the already used series expansion for this other integral is not in the interval of convergence. This is the main constant that I need to find. Also, please correct me and give me feedback!


These Taylor series expansions were found:

Taylor Series

Unfortunately, it seems like there may be uneven radiuses of convergence which may make us need to have multiple series. Also see nth derivative of xth root of x at x=1 OEIS which has an actual formula. We are so close to the answer.

  • 2
    $\begingroup$ How about expanding $\sqrt[x]x$ as $\sum_{n=0}^{\infty} \frac{1}{n!} \frac{\ln^n(x)}{x^n} $, exchanging sum and integral and expressing $\int_0^{e\text{W}(1/e)} \frac{\ln^n(x)}{x^n} dx $ in terms of the incomplete gamma function? $\endgroup$
    – Vishu
    Jun 12, 2021 at 14:58
  • 1
    $\begingroup$ But for instance this integral diverges. And can you elaborate on what you mean by the wrong value of $I$? $\endgroup$
    – Vishu
    Jun 13, 2021 at 12:48
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Jun 13, 2021 at 13:33
  • 1
    $\begingroup$ @Tavish Infinite tetration area answer. Since you were part of the bounty before, maybe the new answer will be of interest. Do you have any feedback? $\endgroup$ Aug 29, 2021 at 21:56
  • 2
    $\begingroup$ @Tyma Gaidash. I am trying to write an answer but I am stuck. So I am here to ask for your help. In your answer in the last integral, if we substitute $u=\frac{1}{x}$, we will have $\int\frac{1}{u^2}u^{-u}du$. Now using the infinite series of $u^{-u}$ we will have $u^{n-2}\ln^n(u)$ inside the sum and the integral. Since $n$ starts with $0$, we have to separate $n=1$ case form the integral which would produce the integral $\int_{e}^{\infty}\frac{\ln(u)}{u}du$ which diverges. I would like to hear your advice. $\endgroup$ Sep 23, 2021 at 16:46

2 Answers 2


Note that the following method may not work for $0\le x\le e^{-e}$, but should work for convergence interval of the infinite tetration function. This interval is ${e^{-e}}\le x\le {\sqrt[e]e}, \frac1e\le y\le e$:

enter image description here

Although there exist a few techniques for finding a series representation, or related method, to evaluate, this one seems easiest. Graphical visualization of the result. Unfortunately, the integral from 0 to $e^{-e}$ seems to be out of the interval of convergence which is the lower convergence bound of the infinite tetration function. Here is the area under the converging original infinite tetration of x. Let this constant be denoted as $\mathrm G^*$. This leaves the only integral to be found being the following which uses the inverse function theorem mentioned in the question:

$$\mathrm{\int_{e^{-e}}^{\sqrt[-e]e}\,^\infty x\,dx=e^{1-\frac1e}W\left(\frac 1e\right)-e^{-1-e}-\int_{\frac1e}^{e W\left(\frac 1e\right)}\sqrt[x]x\, dx}$$

Let the focus be on the main integral for simplicity. The series representation uses nth derivative formulas from A008405 from the OEIS and $\mathrm{S_{x}^{(y)}}$ as the Stirling Numbers of the First Kind:

$$\mathrm{\int_{\frac1e}^{e W\left(\frac 1e\right)}\sqrt[x]x\, dx= \int_{\frac1e}^{e W\left(\frac 1e\right)}\sum_{n=0}^\infty\frac{\frac{d^n}{dx^n}\left(x^\frac1x\right)_{x=1}}{n!}(x-1)^n\,dx= \sum_{n=0}^\infty\frac{1}{n!}\sum_{m=0}^n S^{(m)}_n\sum_{k=0}^m (-k)^{m-k}\binom mk \int_{\frac1e}^{e W\left(\frac 1e\right)}(x-1)^n\,dx= \sum_{n=0}^\infty\frac{1}{(n+1)!}\sum_{m=0}^n S^{(m)}_n\sum_{k=0}^m (-k)^{m-k}\binom mk \left[\left(e W\left(\frac 1e\right)-1\right)^{n+1}-\left(\frac1e-1\right)^{n+1}\right]= \left(1-\frac1e\right)\sum_{n=0}^\infty\frac{1}{(n+1)!}\sum_{m=0}^n S^{(m)}_n\sum_{k=0}^m (-1)^{k+m+n}k^{m-k}\binom mk \left(1-\frac1e\right)^n-\left(1-eW\left(\frac1e\right)\right)\sum_{n=0}^\infty\frac{1}{(n+1)!}\sum_{m=0}^n S^{(m)}_n\sum_{k=0}^m (-1)^{k+m+n}k^{m-k}\binom mk \left(1-eW\left(\frac1e\right)\right)^n}=.35776216…$$

Note that the sum can be simplified by splitting it, manipulating the alternating part of the sum, and using special functions. Also note that the interval of convergence may be larger for the OEIS expansion. This makes it possible to split the original integral, for example, from $\mathrm{[e^{-e},1]\ to\ [1,e]}$ to get a simpler answer for the area under the infinite tetration function than the answer here. The technique would use the 2 different sum representations seen in this question and answer.

Putting this all together gives the area under the infinite tetration curve, the version not defined by the W-Lambert function, as:

$$\mathrm{G^*=\int_{e^{-e}}^{e^{-\frac1e}}\,^\infty x\,dx+ \int_{e^{-\frac1e}}^{e^{\frac1e}}\,^\infty x\,dx=\int_{e^{-e}}^{e^{\frac1e}} x^{x^…} dx= e^{1-\frac1e}W\left(\frac 1e\right)-e^{-e-1}-\sum_{n=1}^\infty\left[\frac{\left[\left(e W\left(\frac 1e\right)-1\right)^n-\left(\frac1e-1\right)^n\right]}{n!}\sum_{m=0}^{n-1} S^{(m)}_{n-1}\sum_{k=0}^m (-k)^{m-k}\binom mk +\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)\right]= e^{1-\frac1e}W\left(\frac 1e\right)-e^{-1-e}-\sum_{n=1}^\infty \left[\frac{1}{n!}\left(\left[\left(e W\left(\frac 1e\right)-1\right)^n-\left(\frac1e-1\right)^n\right]\left(\frac{d^{n-1}}{dx^{n-1}}x^\frac1x\right)_{x=1}+(-n)^{n-1}Γ\left(n,-\frac 1e, \frac 1e\right)\right)\right]= 1.244131300651…}$$

Using similar techniques one can split the integral on the aforementioned interval of $[e^{-e},1] ∪ [1,e] $ as these match the convergence intervals. It uses the Lower Regularized Incomplete Gamma function denoted P(a,z):

$$\mathrm{G^*=1-e^{-e-1}+\sum_{n=0}^\infty \frac{\left(\frac{d^n}{dx^n} x^\frac1x\right)_{x=1}\left(\frac1e-1\right)^{n+1}}{n!}+\sum_{n=1}^\infty (-1)^n n^{n-2}Q\left(n,-\frac1e,0\right)= 1-e^{-e-1}-\sum_{n=1}^\infty (-1)^n\left[n^{n-2}P\left(n,-\frac1e\right)-\sum_{m=0}^{n-1}\sum_{k=0}^m \binom mk\frac{S_{n-1}^{(m)} \left(1-\frac1e\right)^n}{ (-k)^{k-m} n!}\right]= 1.244131300651437208…}$$

As the main sum has been partly shown to converge, our next goal is to find $$\mathrm{-\int_0^{e^{-e}}\frac{W(-ln(x))}{ln(x))} =e^{-e-1}-\int_0^{\frac 1e} x^\frac1x dx=.02105…}$$ in order to complete the integral in the question. However, that is for another time. Simplification would be appreciated. Please correct me and give me feedback!

  • 1
    $\begingroup$ If you can find the integral at the bottom of the question, I will accept the answer. If this sum does work, then please tell me. $\endgroup$ Aug 29, 2021 at 12:17
  • 1
    $\begingroup$ @Rounak I could then possibly use the geometric series if the convergence is well on this, but the $\int W(-\ln(x))dx$ would still be hard with a good convergence. You can also try to answer the integral of $x^\frac1x$. $\endgroup$ Sep 15, 2021 at 11:46
  • 1
    $\begingroup$ Give me some time, if I am able to get anywhere with that integral, then I will tell you. By the way, if you see that the geometric series isn't working then you can just apply the Lagrange inversion theorem which would obviously be much more complicated but still an infinite sum. The inverse of $\textstyle\displaystyle{\frac{1}{1-W(-\ln(u))}}$ is $\textstyle\displaystyle{e^{\left(\frac{1}{u}-1\right)e^{1-\frac{1}{u}}}}$ $\endgroup$ Sep 15, 2021 at 11:59
  • 1
    $\begingroup$ @RounakSarkar Also see Inverse Function integration. There are also many other inversion theorems. Also note there are bell polynomial coefficients for the Lagrange Inversion Theorem. $\endgroup$ Sep 15, 2021 at 12:01
  • 1
    $\begingroup$ Wait, there are other inversion theorems!!. Holy moly. Thanks $\endgroup$ Sep 15, 2021 at 12:04

We can have another approach for your constant $G^{*}$ in your answer:-


Substitute, $u=-\ln(x)$ then $dx=-e^{-u}du$, so



$\textstyle\displaystyle{=-\int\frac{W(u)}{u}du-\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\int u^{n-1}W(u)du}$

$\textstyle\displaystyle{=-\frac{1}{2}W(u)(W(u)+2)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\int u^{n-1}W(u)du}$

Let's do the integral in the sum first,

$\textstyle\displaystyle{\int u^{n-1}W(u)du}$

Substitute, $v=W(u)$ then $du=(v+1)e^vdv$. So,

$\textstyle\displaystyle{\int v^ne^{nv}(v+1)dv}$ $\textstyle\displaystyle{=\int v^{n+1}e^{nv}dv+\int v^ne^{nv}dv}$

Finally substitute $t=-nv$ then $dv=-\frac{1}{n}dt$, so

$\textstyle\displaystyle{\frac{(-1)^{n}}{n^{n+2}}\int t^{n+1}e^{-t}dt+\frac{(-1)^{n+1}}{n^{n+1}}\int t^ne^{-t}dt}$



So finally,



Then the definite integral would be-




We can simplify this using the generalized incomplete function-


And then using the recurrence relation which can be easily obtained through integration by parts-







From here and here we can simplify even further-



I can't simplify more than this.

For the last integral in your answer, I think I can give you a sum using the lagrange inversion theorem-

Let, $f(x)=x^\frac{1}{x}$



Using the Lagrange Inversion theorem we have


Where, $\textstyle\displaystyle{f_n=\lim_{h\rightarrow 1}\frac{d^{n-1}}{dh^{n-1}}\left[\left(\frac{h-1}{e^{-W(-\ln(h))}-1}\right)^n\right]}$

Here I have taken $a=1$

By the ratio test we have the radius of convergence for the sum-





I don't really know how to calculate $f_n$ cause wolfram alpha doesn't give me an answer when $n>3$. So unless I have an easily manageable formula for $f_n$, I don't know what the limit is, so I would ask for help in that.

So, assuming that the bounds of the integral are within the radius of convergence,







If I am able derive more about $f_n$ then I will add it to the answer.

Finally if your derivation in the answer is correct then,




Which looks really messy but that's all I can give you for now.

To be honest I prefer the constant $G^{*}$ better than $G$.




$\textstyle\displaystyle{f_n=\lim_{h\rightarrow 1}\frac{d^{n-1}}{dh^{n-1}}\left[\left(\frac{h-1}{e^{-W(-\ln(h))}-1}\right)^n\right]}$

  • 1
    $\begingroup$ The goal would be to merge both answers to finally get $$\int_0^{e^{\frac1e}} x^{x^…}dx$$, but I would first need to know the interval of convergence of your general indefinite integral so we can determine if the formula can be used on all of $0\le x\le e^{\frac1e} $. If so, then we can possibly use your integral by itself to find the entire area. At the very least, you need to have the sum formula work on $0<x<\frac1e$. $\endgroup$ Sep 24, 2021 at 11:50
  • 1
    $\begingroup$ Your sum for $\int_{e^{-e}}^{e^\frac1e } x^{x^…}dx=1.24413…dx$ gives $2.27071$. Check the link. Also note that Г(x,y,z)=Г(x,y)-Г(x,z) $\endgroup$ Sep 24, 2021 at 12:29
  • 1
    $\begingroup$ @Tyma Gaidash. I can't really find any mistakes in the derivation. Have you found any?? $\endgroup$ Sep 24, 2021 at 13:04
  • 2
    $\begingroup$ @Tyma Gaidash. My common sense tells me that there shouldn't be a finite radius of convergence for the first sum because I derived the sum using the series representation of $e^{-u}$ whose radius of convergence is infinite. So I am really confused. By the way you can use "\unicode{x22F0}" to represent $\unicode{x22F0}$ $\endgroup$ Sep 24, 2021 at 13:43
  • 2
    $\begingroup$ @Tyma Gaidash. We can actually simplify the sum using the generalized incomplete gamma function as you suggested. I will edit to add that. By the way, you may remember that I had put a really complicated divergent limit in a comment under your question. The antiderivative my approach produces is unable to take 0 as an input or even as a limit for that matter, so we have to do a different approach. $\endgroup$ Sep 26, 2021 at 17:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .