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