real analysis - Exchanging the order of integration in $ int_0^infty int_{-infty}^infty sin(x^2)x e^{-t^2 x^2} dt dx$?

For context, this gives one way to evaluate the Fresnel sine integral at infinity. The problem I'm running into is

$$ \int_0^\infty \left[ \int_{-\infty}^\infty \vert\sin(x^2)x e^{-t^2 x^2}\vert dt \right] dx$$ $$= \sqrt{\pi} \int_0^\infty \vert \sin(x^2) \vert dx$$
$$ = \infty,$$

so Fubini does not apply. However, naively switching the order does give the right result. Is there a nice way to justify exchanging the order?

Answer

By Fubini, we have

$$ \int_0^R \int_{-\infty}^\infty \sin(x^2)x e^{-t^2 x^2} \,dt\, dx
= \int_{-\infty}^\infty \int_0^R \sin(x^2)x e^{-t^2 x^2} \,dx \, dt = \int_{-\infty}^\infty f(R,t) \, dt$$
where
$$ f(R,t) = \int_0^R \sin(x^2)x e^{-t^2 x^2} \,dx = \frac{1-(\cos(R^2) + t^2 \sin(R^2))e^{-t^2 R^2}}{2(1+t^4)} .$$
(This follows from integrating by parts twice, and solving the resulting equation.) Hence
$$ |f(R,t)| \le \frac{2+t^2}{2(1+t^4)} ,$$
which is integrable. Hence by the Lebesgue dominated convergence theorem, we have
$$
\lim_{R\to \infty} \int_{-\infty}^\infty f(R,t) \, dt = \int_{-\infty}^\infty \lim_{R\to \infty} f(R,t) \, dt .$$
Hence

$$ \int_0^\infty \int_{-\infty}^\infty \sin(x^2)x e^{-t^2 x^2} \,dt\, dx
= \int_{-\infty}^\infty \int_0^\infty \sin(x^2)x e^{-t^2 x^2} \, dx \, dt$$