I have been wondering about this for a while, but with no real luck in figuring it out. The famous "Sophomore's dream" identity refers to two similar integrals, one of which is

$$\int_{0}^{1} x^{-x}\ dx = \sum_{n=1}^{\infty} n^{-n}$$

which has a pleasing symmetry between integrand and summand. However, I also notice that $x \mapsto x^{-x}$ is a rapidly-converging-to-zero function of $x$, and hence it seems like it might be more "natural" to then wonder about the integral

$$\int_{0}^{\infty} x^{-x}\ dx$$.

Numerical integration suggests it is approximately 1.99545596. Yet what I would like to find is some other representation that is not directly an integral, whether it be a series, or even a finite expression using any already-established mathematical functions and/or constants.

And it doesn't seem very clear at all how to do this. Obviously, with upper limit $\infty$, Taylor expansion of the integrand is of no use since it will only ever have finite radius of convergence. The other line of attack that I thus think of is to try and express $x^{-x}$ as a series of some form of decaying functions that are simpler to integrate. We do have that

$$x^{-x} = e^{-x \ln x}$$

but this is of no use: it doesn't yield any series in terms of $e^{-x}$-like terms. We have the interesting substitution $x = e^{W(u)}$, $dx = \frac{1}{1 + W(u)}\ du$ (equiv. to $u = x \ln x$) with the Lambert W-function, which gives

$$\int_{0}^{\infty} x^{-x}\ dx = \int_{0}^{\infty} e^{-u} \frac{du}{1 + W(u)}$$

but this doesn't help for series expansions.

What is known about this integral?

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    $\begingroup$ I don't think you're going to be able to get an answer to this. However, there are some partial results in this direction. Check out Jean Jacquelin's paper on the sophomore's dream function here: math.eretrandre.org/tetrationforum/attachment.php?aid=788 $\endgroup$ Jun 20, 2019 at 8:35
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    $\begingroup$ This looks hard.. But atleast we only have to find $\int_1^\infty x^{-x} dx$ since the first part is known. However expanding into power series here is painful. $\endgroup$
    – Zacky
    Jun 20, 2019 at 9:07
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    $\begingroup$ Well if there is a nice expression of this integral as a series, it's probably quite complicated. The inverse symbolic calculator reveals nothing. $\endgroup$
    – YiFan
    Jun 26, 2019 at 10:34
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    $\begingroup$ See this paper : fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . $\endgroup$
    – JJacquelin
    Oct 27, 2019 at 17:57

2 Answers 2


Although it is not exactly for what the OP is asking for, but still may be interesting. Ramanujan provides curious expansion $$ \int_{0}^\infty x^{-x}dx \sim \sum_{n \in \mathbb{Z}} n^{-n} $$ (be careful, because the sum is divergent). See Theorem 2 in Berndt, Evans http://math.ucsd.edu/~revans/Elegant.pdf.


Please see this post as I am thinking it would be better to see some of the representations I have found with graphical proof:

in this link.

This answer has the following as an example answer, although the participants in my question and answer have been unable to find a form that does not require a limit in the answer. Please see my answer for additional expressions. Here is proof of the result.

$$\int_{\Bbb R^+}x^{-x}dx \mathop=^{\text{lim}}\mathrm{\sum_{n=1}^{\infty}n^{-n}\frac{Γ(n,-\infty n)}{Γ(n)}=\lim_{x\to \infty}\sum_{n\ge1}\frac{Q(n,-nx)}{n^n}=\lim_{x\to\infty} 500e^x-\frac{125e^{2x}(4x^2-2x)}{2x} -\frac{250e^{3x}(-27x^3+18x^2-6x)}{81x}+…=1.99545595…}$$

You can go to the link to see the derivation process of using the maclaurin series for exp(y) and integrating using the incomplete regularized gamma function. Please correct me and give me feedback.


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