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.
Answer
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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