1
$\begingroup$

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?

Edit:

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.

$\endgroup$
5
  • 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 $\sum_{n=1}^{\infty}{\frac{1}{n^n}}$? $\endgroup$
    – Rhys Hughes
    May 15, 2018 at 10:45
  • $\begingroup$ Yes, extremely sorry about that. $\endgroup$
    – Raheeb Hassan
    May 15, 2018 at 10:59
  • 2
    $\begingroup$ 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 $\endgroup$
    – cgiovanardi
    May 15, 2018 at 13:10

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.