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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?

terran
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FireWolf15G8
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  • Compare https://math.stackexchange.com/q/155/42969, https://math.stackexchange.com/q/453234/42969 – Martin R Jan 30 '24 at 12:49
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    If you let $R(x)=\frac{P(x)}{Q(x)}$, then you can use the quotient and chain rules to show that $\frac{d}{dx}R\left(e^x\right)$ is also a rational function of $e^x$. So now you need to show that $e^{x^2}$ is not a rational function of $e^x$. – Chris Lewis Jan 30 '24 at 14:06
  • @ChrisLewis i was not sure whether i was allowed to take the derivative or not, thank you! – FireWolf15G8 Jan 30 '24 at 15:15

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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.

Cactus
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