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Integrating x^x

  1. Jul 5, 2007 #1
    [tex]\int x^xdx[/tex]

    After calculus today, our professor casually asked us to integrate the above problem. Most of us stated DNE (does not exist), or possibly x+1 / x+1 . He stated that both of these solutions were incorrect and than challenged us to write a paper on the subject which seemed ridiculous to me but here I am.

    So a few questions...

    Can this be done by Taylor Series expansion??

    [tex]\int x^x dx=x + \frac{\left( -1 + 2\,\log (x) \right) \, x^2}{4} + \frac{\left( 2 - 6\,\log (x) + 9\,{\log (x)}^2 \right) \,x^3}{54} + \frac{\left( -3 + 12\,\log (x) - 24\,{\log (x)}^2 + 32\,{\log (x)}^3 \right) \,x^4}{768} + \frac{\left( 24 - 120\,\log (x) + 300\,{\log (x)}^2 - 500\,{\log (x)}^3 + 625\,{\log (x)}^4 \right) \,x^5}{75000} + \frac{\left( -5 + 30\,\log (x) - 90\,{\log (x)}^2 + 180\,{\log (x)}^3 - 270\,{\log (x)}^4 + 324\,{\log (x)}^5 \right) \,x^6}{233280} + {O(x^7)[/tex]

    or is the answer simply dF/dx is x^x ???
    Last edited: Jul 5, 2007
  2. jcsd
  3. Jul 5, 2007 #2
    I've seen this many times, the last time someone jokingly asked me to solve it, to which the answer was solve you xxxxxxx :smile:

    I asked a friend about this very thing and he said in fact the only way it can be done is with an infinite Taylor Expansion.

    So there really is no "solution", and all maths programs I've tried wont solve it.

    Nor will pen and paper.

    [itex]\int e^xdx=e^x+C[/itex]

    Same deal here basically, the solution is either non "renormalizable" or itself.
    Last edited: Jul 5, 2007
  4. Jul 5, 2007 #3


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    There is no integral in terms of elementary functions. However, since that is a continuous function, it certainly does have an integral. That's why "does not exist" is incorrect.

    I have no idea why anyone in your class would have thought the integral would be x+ 1/x+ 1. It should be easy to see that the derivative of that is not xx.
  5. Jul 5, 2007 #4
    I think the way to show that it can't be integrated in terms of an elementary function is using the Risch Algorithm (see here). As for what the algorithm actually is, I have no idea!
  6. Jul 5, 2007 #5
    I read the wiki: I know I'm wrong in the head, according to that, if it is correct. If there is no elementary solution they use computer algorithms to get heuristic solutions which give approximations without proof. In other words only rough answers and there is no suggestion that they will remain infinitely good approximations. Which neatly explains why my maths program tells me to get bent :smile:

    well actually it gives the solution

    [itex]\int x^xdx\rightarrow \int x^xdx[/itex]

    Which is the same thing :smile:
    Last edited: Jul 5, 2007
  7. Jul 5, 2007 #6
    Your first approach was good. Taylor Series. Integrate term-by-term. Then leave it as such. That's all you should do.
  8. Jul 5, 2007 #7
    [tex]\int_0^1 x^x dx = \sum_{n=1}^{\infty} (-1)^{n+1} n^{-n}[/tex]


    You cannot use a Taylor series for [tex]x^x[/tex] centered at 0 because the function is not defined on an open interval containing 0.

    Consider the complex valued funtion [tex]f(z) = z^z = e^{z\log z}[/tex] this function is holomorphic on [tex]\mathbb{C} - (-\infty,0][/tex] so as along as we stay away from the branching of [tex]\log z[/tex] we can use the complex version of Taylor's theorem. For instance if we are on the neightborhood [tex]N(1,1) = \{z\in \mathbb{C} : |z-1|<1\}[/tex] then this function is holomorphic on this open disk and hence has a Taylor series expansion. So you need to change you center of series, for instance, for x=1 if you really want a Taylor series.
    Last edited: Jul 5, 2007
  9. Jul 5, 2007 #8
    My understanding of this article is that the Risch algorithm is a decision procedure used to determine the function of an indefinite integral. I don't gather that there actually is an algorithim that proves it CANNOT be integrated in terms of elementary functions. If there is, can someone post it?
  10. Jul 5, 2007 #9
    The mathematician Liouville showed that [tex]\int f(x) \exp [g(x)] \ dx[/tex] where [tex]g(x)[/tex] is non-constant is elementary for rational functions [tex]f(x),g(x)[/tex] if and only if there exists a rational function [tex]h(x)[/tex] which solves the ordinary differential equation: [tex]f(x) = h'(x) + h(x)g(x)[/tex].

    I believe this can be modified to show [tex]\int x^x \ dx[/tex] is not an elementary function.
  11. Apr 19, 2011 #10
    I'm interested in how this modification can be done. I've had a go at it without much luck.
  12. Apr 20, 2011 #11
  13. Apr 24, 2011 #12
    Great link. Thanks for that. I guess all that's left is to prove that it is indeed a non-elementary integral.
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