The antiderivative of $e^{x^2}$

Question

I am asked to show that show that the antiderivative of $e^{x^2}$ cannot be expressed as $R(e^x)$, where $R$ is a rational function over the reals. Can this be shown without using transcendental extensions, differential algebra etc?

Answer 1

Yes, by comparing the speed of divergence at infinity. Indeed, if $\frac{d}{dx}(R(e^x)) = e^{x^2}$, then for all $x$, $e^{x^2} = e^xR'(e^x) = F(e^x)$ with $F(X) = XR'(X)$ is a rational fraction too. Therefore, if you call $d$ the degree of $F$ (degree of the numerator minus degree of the denominator), then $e^{x^2} = F(e^x) = \mathrm{O}(e^{dx})$ at $+\infty$ hence $x^2 - dx$ is bounded, which is a contradiction.