Proving Integral of x^x Does Not Exist

[n1person]

Recently I have begun thinking about the function x^x. I am well aware that there is no elementary function to define it's antiderivative, and intuitively it makes sense (I cannot think of an elementary function who's derivative is x^x). However, how would one go about proving this rigorously?

 

[g_edgar]

MARCHISOTO and ZAKERI (1994): "An Invitation to Integration in Finite Terms" , The College Mathematics Journal 25 No. 4 Sept. pp 295 - 308

J. F. Ritt, Integration in finite terms: Liouville's theory of elementary methods, 1948

 

[n1person]

Does anyone happen to have a more accessible resource about this subject? Like something online?

 

[Big-T]

http://www.sosmath.com/calculus/integration/fant/fant.html

 

[n1person]

so what we have is:

$$\int f(x)e^g^(^x^)dx$$
$$g(x)=0$$
$$f(x)=x^x$$

so the formula given

$$<br /> f(x)=R'(x)+g(x)R(x)<br />$$

just goes to

$$f(x)=R'(x)$$

which isn't overly illuminating :(

 

[g_edgar]

so what we have is:

$$\int f(x)e^g^(^x^)dx$$
$$g(x)=0$$
$$f(x)=x^x$$

so the formula given

$$<br /> f(x)=R'(x)+g(x)R(x)<br />$$

just goes to

$$f(x)=R'(x)$$

which isn't overly illuminating :(

No, $x^x$ is not a rational function, so this is not the way to fit the theorem.
In fact, the case of $x^x$ (as explained in the actual references) is a bit more involved than that simplistic web page says.

 

[Big-T]

How about letting $$x^x = e^{x\ln x}$$?