Monday, 24 April 2017

real analysis - How can you prove that a function has no closed form integral?



I've come across statements in the past along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations:




  • addition/subtraction

  • multiplication/division

  • raising to powers and roots


  • trigonometric functions

  • exponential functions

  • logarithmic functions



, which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example.



What sort of techniques are used to prove statements like this? What is this branch of mathematics called?







Merged with "How to prove that some functions don't have a primitive" by Ismael:



Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions.



I wonder how we can prove that kind of assertion?


Answer



It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ non-constant, the antiderivative



$$f(x)\exp(g(x)) \, \mathrm dx$$




can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution to the differential equation:



$$f = h' + hg$$



$e^{x^2}$ is another classic example of such a function with no elementary antiderivative.



I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes: http://www.sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf



Liouville's original paper:





Liouville, J. "Suite du Mémoire sur la classification des Transcendantes, et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients." J. Math. Pure Appl. 3, 523-546, 1838.




Michael Spivak's book on Calculus also has a section with a discussion of this.


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