# A new constant?

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?

• Can you show more digits? Just out of curiosity :) – Ataraxia Jun 4 '13 at 22:36
• @ZettaSuro I'd like to have more :( That is why I am asking this queestion! – Anonymous Pi Jun 4 '13 at 22:40
• 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 $k$? – Patrick Da Silva Jun 4 '13 at 22:42
• 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... – Gottfried Helms Jun 4 '13 at 22:46
• @anonymousPi: ;-), well I've no (better) algebraic expression for that number - and what's what I understood your question was about... – Gottfried Helms Jun 4 '13 at 22:53

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)

• 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. – Mark McClure Jun 5 '13 at 1:02
• @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.) – Gottfried Helms Jun 5 '13 at 6:40
• @GottfriedHelms Told you it would be a good idea to answer! By the way, I upvoted. – Anonymous Pi Jun 5 '13 at 12:21

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 $1.2, 1.19, 1.195, \ldots$.

• Maybe I wrote it wrong... – Anonymous Pi Jun 4 '13 at 22:41
• 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: $\int_0^{1.19491} x^x dx -1 = 3.7014x10^{-6}$ – Gottfried Helms Jun 4 '13 at 23:24
• – Mark McClure Jun 5 '13 at 0:32
• I didn't downvote, by the way! I definitely see doing the bisection as worthwhile. – Mark McClure Jun 5 '13 at 0:52