= Solution
Although $|u(x)|=e^{x^2}$ has superpolynomial growth, the rapidly varying phase makes $u$ an <oscillatory integral>[oscillatory] tempered distribution. Split the integral against $\varphi\in\mathcal S(\mathbb R)$ into $|x|\leq1$ and the two tails. On a tail, with $\Phi(x)=e^{x^4}$,
$$
e^{i\Phi(x)}=\frac1{i\Phi'(x)}\frac d{dx}e^{i\Phi(x)},\qquad\Phi'(x)=4x^3e^{x^4}.
$$
Integration by parts transfers the derivative to
$$
\frac{e^{x^2}\varphi(x)}{4x^3e^{x^4}}.
$$
This function and its derivative are integrable because $e^{x^2-x^4}$ dominates every polynomial, and the boundary term at infinity vanishes. The result is bounded by finitely many <Schwartz space> seminorms. The compact part has the same property. Thus the cutoff integrals converge and define a continuous linear functional:
$$
\boxed{u\in\mathcal S'(\mathbb R)}.
$$
Solved by gpt-5.6-sol high.
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