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Integrating 1/(x+ln(x))

  1. Apr 30, 2013 #1
    I've been trying to figure out how to integrate 1/(x+ln(x)) but am not getting anywhere. Mathematica can't do it, and I haven't found it in lists of integrals.

    Does anyone know if this integral exists in closed form?

    Same goes for (x+ln(x))/(1+x+ln(x))

    Thanks!
    zeroseven
     
  2. jcsd
  3. Apr 30, 2013 #2

    micromass

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    If mathematica can't do it, then it means there is no solution in terms of standard integrals. This is because the algorithm employed by mathematica is a very good one. If a solution exist, then the algorithm finds it. And if the algorithm fails, then that implies that a solution doesn't exist. So it goes both ways.
     
  4. Apr 30, 2013 #3
    Thanks, that's helpful and very interesting.

    I have no idea how I would even begin to go about proving that a solution doesn't exist for an integral! Is this something that can only be done by computer algorithms?
     
  5. Apr 30, 2013 #4

    micromass

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    No, it can be done by hand too.

    See http://en.wikipedia.org/wiki/Liouville's_theorem_(differential_algebra) and http://en.wikipedia.org/wiki/Risch_algorithm

    A good book on the topic is Geddes: https://www.amazon.com/Algorithms-Computer-Algebra-Keith-Geddes/dp/0792392590 (despite its name, it's a math book and not a programming book)
     
  6. Apr 30, 2013 #5
    Thanks again, I always wondered how we could be sure that the integral of exp(-x^2) doesn't exist!

    So does this mean that when Mathematica works on an integral, it's actually running a rigorous mathematical proof on whether that integral exists or not? Does it use the Risch algorithm that you just linked to?

    This leads me to a follow-up question.. if these algorithms work for elementary functions, does it extend to non-elementary functions? For example, if I ask Mathematica to
    Integrate[1/(x + ProductLog[x]), x]
    and it can't do it, does this prove that an integral doesn't exist in terms of the function ProductLog?
     
  7. Apr 30, 2013 #6

    micromass

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    It uses an implementation of the Rish algorithm, yes. So if mathematica can't calculate an integral, then it actually proves that it can't be calculated. Of course, the proof is never shown, because it's not very interesting. But in principle, mathematica could be used to prove that integrals can't be calculated.

    But be aware that the practical situation is a bit more complicated. The wiki on Rish algorithm already shows you that there is a certain dependence on initial conditions. Also, the Rish assumes that polynomials can be factored exactly, this might not be true in practice.

    I don't know about what the Rish algorithm does with these. But Liousville's theorem seems to be able answer questions like those. I don't know about mathematica's implementation to say with confidence whether it proves the non-existence of this integral. It is however something you can certainly do in theory (by hand).
     
  8. Apr 30, 2013 #7
    Thanks again micromass.
    That solves my problem (or actually shows that it is unsolvable), and makes me appreciate Mathematica a bit more!

    Cheers,
    zeroseven
     
  9. Apr 30, 2013 #8

    pwsnafu

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    Not in general. For example applying Richardson's theorem we see that "elementary functions + absolute value function" has no equivalent to Risch.
     
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