Proving Integral of x^x Does Not Exist

n1person
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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?
 
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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
 


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


so what we have is:

<br /> \int f(x)e^g^(^x^)dx<br />
<br /> g(x)=0<br />
<br /> f(x)=x^x<br />

so the formula given

<br /> <br /> f(x)=R&#039;(x)+g(x)R(x)<br /> <br />

just goes to

<br /> f(x)=R&#039;(x)<br />

which isn't overly illuminating :(
 


n1person said:
so what we have is:

<br /> \int f(x)e^g^(^x^)dx<br />
<br /> g(x)=0<br />
<br /> f(x)=x^x<br />

so the formula given

<br /> <br /> f(x)=R&#039;(x)+g(x)R(x)<br /> <br />

just goes to

<br /> f(x)=R&#039;(x)<br />

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.
 


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