On the internet I found an evaluation of the integral $\displaystyle \int_0^1 x^{-x} \,\mathrm{d}x$ which results in $$\sum_{x=1}^\infty \frac{1}{x^x}.$$ Seeing this graphically I found that the sum does seem to converge to approximately $1.291$. So as it converges shouldn't we be able to find its exact value? Can someone please tell me what this value is or how I may be able to find it?


While searching about this I came across Sophomore's Dream and this question which is quite interesting but none of them can quite give a closed value (in terms of known constants like $\pi$, $\phi$ or e). So does $\displaystyle \sum_{x=1}^\infty \frac{1}{x^x} $ or $\displaystyle \int_0^1 x^{-x} \,\mathrm{d}x$ not have such a value? Is it irrational? So many questions.

  • 1
    $\begingroup$ The series you wrote seems to be missing an $n$ somewhere. $\endgroup$
    – Marra
    May 15, 2018 at 10:41
  • $\begingroup$ Is this meant to be $\sum_{n=1}^{\infty}{\frac{1}{n^n}}$? $\endgroup$ May 15, 2018 at 10:45
  • $\begingroup$ Yes, extremely sorry about that. $\endgroup$ May 15, 2018 at 10:59
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    $\begingroup$ There is no known "closed form" of the type you ask about. $\endgroup$
    – GEdgar
    May 15, 2018 at 11:23
  • $\begingroup$ See scribd.com/document/34977341/Sophomore-s-Dream-Function $\endgroup$ May 15, 2018 at 13:10

1 Answer 1


The most you can say/prove is that the integral and the series are equal (easy to prove), and that the integral has no elementary anti-derivative (hard to prove, but there's one if you know some Galois theory). There's no known closed form for this value. It's not known whether it's rational/irrational.


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