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Possible Duplicate:
How can you prove that a function has no closed form integral?

(Forgive my crude lingo)

Why do some integrals seem to be unsolvable; that is to say the indefinite integral cannot be solved symbolically (or is it analytically(?)). For example, those with the form of a radial distribution function using the Lennard-Jones potential:

$$g(r) = \exp(-(r^{-12} - r^{-6}))$$

What prevents me from using my standard Calc II knowledge to come up with a continuous expression for $\int{g(r)dr}$?

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Nick T
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    "There are more things in Heaven and Earth, Horatio, than are dreamt of in your philosophy." There are more far more continuous functions than we can write down using only elementary functions, and antidifferentiation is a procedure that is not "elementary", so it should be no major surprise that it is possible to do something non-elementary to an elementary function and end up with a non-elementary function. Just like no matter how much geometry knowledge you throw at the problem, you still won't be able to duplicate the cube using straightedge and compass; or solve a quintic a radicals. – Arturo Magidin Apr 19 '12 at 16:20
  • @ArturoMagidin Isn't this a duplicate? I swear I've seen a question that amounts to this somewhere else on Math SE. I just don't know where it is, besides the vague mental picture of my mind. – 000 Apr 19 '12 at 16:22
  • @Limitless I would have thought so, but couldn't find anything. – Nick T Apr 19 '12 at 16:23
  • @Limitless: Maybe; there certainly are questions about specific non-elementary integrals. But there is no question similar to this in the List of generalization of common questions, so I don't know off hand if this (or something very similar) has been posted. – Arturo Magidin Apr 19 '12 at 16:25
  • Hmm, I can't seem to find anything either...

    I could've sworn I read something on here that referenced the theorem that developed in order to answer this question.

     – 000 Apr 19 '12 at 16:28
  • @Ihf, I think you found it! – 000 Apr 19 '12 at 16:29
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    It is basically a question of a lack of names. On a planet far far away they only have expressions for what we call the rational functions. So no log in their math, no sin, no exp. Someone there has just discovered the basics of the calculus. She finds, sadly, that $\frac{1}{x}$ and $\frac{1}{1+x^2}$ do not have antiderivatives in terms of "known" (that is, rational) functions. The same is true on that planet even after they discover the algebraic functions, like $\sqrt{1-x^2}$. – André Nicolas Apr 19 '12 at 16:33
  • See this as well. – J. M. ain't a mathematician Apr 19 '12 at 16:38

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