My question specifically deals with certain real indefinite integrals such as $$\int e^{-x^2} {dx} \ \ \text{and} \ \ \int \sqrt{1+x^3} {dx}$$ Books and articles online have only ever said that these cannot be expressed in terms of elementary functions. I was wondering how this could be proved? I know this is a naive way of thinking, but it seems to me like these are just unsolved problems, not unsolvable ones.
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The three-word answer to your question is: Differential Galois Theory. The two-word answer to your questtion is: It's complicated. I'll leave it to someone who actually knows this stuff to give better answers. – Eric Stucky Jun 24 '15 at 00:23
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See http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral?rq=1 – Plutoro Jun 24 '15 at 00:24
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No, these are theorems. See the Risch algorithm. The basic idea is that it is possible to show that if there is an elementary function with the given integrand as derivative, it must have a certain general form.
If nothing of that form works, there is no elementary antiderivative.
Robert Israel
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The trick is to make precise the meaning of "elementary": essentially, these are functions, which are expressible as finite combinations of polynomials, exponentials and logarithms. It is then possible to show (by algebraically tedious disposition of cases, though not necessarily invoking much of differential Galois theory - see e.g. Rosenthal's paper on the Liouville-Ostrowski theorem) that functions admitting elementary derivatives can always be written as the sum of a simple derivative and a linear combination of logarithmic derivatives. One consequence of this is the notable criterion that a (real or complex) function of the form $x\mapsto f(x)e^{g(x)}$, where $f, g$ are rational functions, admits an elementary antiderivative in the above sense if and only if the differential equation $y'+g'y=f$ admits a rational solution. The problem of showing that $e^{x^2}$ and the lot have no elementary indefinite integrals is then reduced to simple algebra. In any case, this isn't an unsolved problem and there is not much mystery to it once you've seen the material.
Damian Reding
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