TITLE: A new constant?
QUESTION [16 upvotes]: 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?

REPLY [13 votes]: 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)