Question: I was experimenting in [Wolfram Alpha](http://www.wolframalpha.com/) 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 [$\pi$](http://en.wikipedia.org/wiki/Pi), [$e$](http://en.wikipedia.org/wiki/E_%28mathematical_constant%29), [$\gamma$](http://en.wikipedia.org/wiki/Euler-Mascheroni_constant), etc, or is it a whole new number?

Answer: 
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.  

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**[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]

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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](http://www.wolframalpha.com/input/?i=integral+of+x%5Ex+from+0+to+1.19491) 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 $1.2, 1.19, 1.195, \ldots$.