11

It happens quite often that something like an integral or the solution to a certain equation cannot be expressed in terms of elementary functions. For example, the solution to the equation $\cos x=x$ or $xe^x=\alpha$, or the integral of $x^x$ with respect to $x$.

But how do we actually prove that there is no elementary formula that equals those numbers? Does the proof require metalogic/theory of formal languages?

I'm not sure about how to tag the question, so feel free to edit the tags with the most appropriate ones and then delete this line.

Lee Mosher
  • 135,265
Elvis
  • 1,543

1 Answers1

7

Using Liouville's Theorem we can show that $e^{-x^2}$ does not have an elementary antiderivative.

First assume that $\int e^{-x^2}dx=F(x)$ is an elementary antiderivative, i.e. $F(x)$ is a finite combination of elementary functions (polynomials, exponentials, logarithms, trigonometric functions, etc), such that $$F'(x) = e^{-x^2}.$$

Liouville’s Theorem says that if $F(x)$ exists and is elementary, then it must have the form $$F(x) = R(x) + \sum_i c_i \ln(g_i(x)),\tag{1}$$ where $R(x)$ is a rational function, $c_i$ are constants, and $g_i(x)$ are algebraic functions. Taking the derivative, $$R'(x) + \sum_i c_i \frac{g_i'(x)}{g_i(x)}=F'(x)=e^{-x^2}.\tag{2}$$

But $e^{-x^2}$ is transcendental by the Lindemann–Weierstrass theorem, i.e. it does not satisfy any algebraic equation with coefficients in $\mathbb{Q}(x)$ as required by (2). Hence, (1) cannot possibly be expressed as a combination of rational functions and logarithmic terms.

Therefore, $\int e^{-x^2}dx$ cannot be expressed in terms of elementary functions.

For your other question to prove that an equation does not have elementary solution, I would probably try to prove that e.g. $xe^x$ or $\cos(x)-x$ is transcendental in that it does not satisfy any algebraic equation with coefficients in $\mathbb{Q}(x)$, again probably making use of the Lindemann–Weierstrass theorem.

pshmath0
  • 11,208