What am I doing wrong?- Differential equations and Integrals

Question

I am trying to find an equation for $\int x^x dx = h$

(I've not been told it's impossible... so I tried to... just to for fun)

So we know that $f^{g(x)}(x) = h(x)$ where $g(x) = k$ (constant)

so: $h'(x) = k \cdot f^{k-1} $

$x^x = h'(x) = k \cdot f^{k-1}(x)$

and thus: $f(x) = (\dfrac{x^x}{c}) ^ {\dfrac{1}{c-1}}$

and if we substitude back "f(x)" we get:
$h(x) = ((\dfrac{x^x}{c}) ^ {\dfrac{1}{c-1}})^{c} = (\dfrac{x^x}{c}) ^ {\dfrac{c} {c-1}}$

and we get that: $\int x^x dx = h(x) = (\dfrac{x^x}{c}) ^ {\dfrac{c}{c-1}}$

but it doesn't work out for let's say $c=2$ or $c=3$ ... any ideas why?
I am new to this field of maths.

Answer 1

As I noted in the comments, this really can't be done using standard integration techniques, but with a series expansion$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

As to what is wrong with your proof,

$f^{g(x)}(x)=h(x)$ where $g(x)=k$ (constant)

This is completely wrong. Why would $g(x)$ be a constant? What does it even mean to raise a function to another function? I don't really know what you've done here, but it is definitely incorrect.

Answer 2

The integral $\int x^x dx$ cannot be expressed with a finite number of elementary functions. It requires a special function :

https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

$$\int x^x dx=\text{Sphd}(1\,;\,x)+\text{constant}$$ $$\int_0^x t^t dt=\text{Sphd}(1\,;\,x)$$ $\text{Sphd}(\alpha,x)$ is a non-standard special function.

Some properties and series expansions are provided in the paper referenced above.