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?
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)
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 $1.2, 1.19, 1.195, \ldots$.