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I was experimenting in Wolfram Alpha the answer to the equation $\int_0^k x^x dx=1$ And I got about 1.19... But, What is this number k (and could you calculate it to more decimal places?) And is it constructed out of $\pi$, $e$, $\gamma$, etc, or is it a whole new number?

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  • $\begingroup$ Can you show more digits? Just out of curiosity :) $\endgroup$ – Ataraxia Jun 4 '13 at 22:36
  • $\begingroup$ @ZettaSuro I'd like to have more :( That is why I am asking this queestion! $\endgroup$ – Anonymous Pi Jun 4 '13 at 22:40
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    $\begingroup$ It's just a number... there's plenty of them. Numbers that we give a name to are not just weird transcendental quantities, we remember them because they have applications! By the way, what's $k$? $\endgroup$ – Patrick Da Silva Jun 4 '13 at 22:42
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    $\begingroup$ strange, I get 1.1949070080264606820 using Pari/GP with 200 digits precision and the lower limit non-zero but as small as 1e-120. Change of the lower limit even nearer to zero does not affect the above shown decimal digits... $\endgroup$ – Gottfried Helms Jun 4 '13 at 22:46
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    $\begingroup$ @anonymousPi: ;-), well I've no (better) algebraic expression for that number - and what's what I understood your question was about... $\endgroup$ – Gottfried Helms Jun 4 '13 at 22:53
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Using the Newton-iteration I computed this to about 200 digits using Pari/GP with 200 digits float-precision. The formula to be iterated, say, 10 to 20 times, goes $$ x_{m+1} = x_m - { \int_0^{x_m} t^t dt - 1 \over x_m^{x_m} } \qquad \qquad \text{initializing } x_0=1$$ This gives $x_{20} \sim 1.1949070080264606819835589994757229370314006804 \\ \qquad 736144499162269650773566266768950014200599457247 \\ \qquad 787580258584233234409032116176621553214684894972 \\ \qquad 73271827683782385863978986910763464541103507567 ... $

where further iterations don't affect the shown decimals.


[update] Perhaps it is of interest to find the number $k$ where the integral does not equal $1$ but $k$ itself instead. We get for $$ \int_0^k x^x dx = k \qquad \qquad \to \qquad k \sim 1.54431721079037838813184037292... $$ [/update]
The pari/GP code used was

m=1  \\ initialize 
     \\ iterate the next two commands until err is sufficiently small
err=(intnum(t=1e-160,m,t^t)-1)/(m^m)
m=precision(m-err,200)
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    $\begingroup$ Very cool! One thing: When I compute the integral from $0$ to your $x_{20}$, I get $1.0\ldots 0999\ldots$ with 120 zeros before other stuff arises. Should your t=1e-120 be t=1e-200? I don't know pari/GP. $\endgroup$ – Mark McClure Jun 5 '13 at 1:02
  • $\begingroup$ @Mark: True, sorry, I've been a bit hasty transferring my result to the answer. Using the lower bound with 1e-160 suffices to get 200 digits correct - I confirmed that also using 400 digits internal precision. (I think, for my final result I'd in fact used the lower bound of 1e-200 which is my "standard"/default precision for such tasks, but where I had stepped to 1e-60,1e-80 and then 1e-120 for the lower bound to improve speed of computation.) $\endgroup$ – Gottfried Helms Jun 5 '13 at 6:40
  • $\begingroup$ @GottfriedHelms Told you it would be a good idea to answer! By the way, I upvoted. $\endgroup$ – Anonymous Pi Jun 5 '13 at 12:21
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Wolfram Alpha thinks that $k=1.19491$ exactly. I'm sure that's only a rounding artifact, but funny nevertheless. This was found in about 5 minutes via bisection, i.e. trying $1.2, 1.19, 1.195, \ldots$.

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  • $\begingroup$ Maybe I wrote it wrong... $\endgroup$ – Anonymous Pi Jun 4 '13 at 22:41
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    $\begingroup$ Well, it knows (wolframalpha.com/input/?i=integral_0^{1.19491}++x^x+dx++-1), there is an error from the 6'th digit on, but WA doesn't document this. If you ask the integral minus 1 you get the result: $ \int_0^{1.19491} x^x dx -1 = 3.7014x10^{-6} $ $\endgroup$ – Gottfried Helms Jun 4 '13 at 23:24
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    $\begingroup$ Try integral of x^x from 0 to 1.19491 to 20 digits $\endgroup$ – Mark McClure Jun 5 '13 at 0:32
  • $\begingroup$ I didn't downvote, by the way! I definitely see doing the bisection as worthwhile. $\endgroup$ – Mark McClure Jun 5 '13 at 0:52

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