# 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?

• I think it might help to put this in a bit more context. If this is from a physics course, it might be a simple lack of rigor. If however it is from a math course, there might be something else going on underneath. – Cameron Williams Jul 20 '15 at 2:07
• @Winther There is a pretty steep (exponential) singularity at $t=0$ if you do the integral in the opposite order. I'm not sure Fubini-Tonelli applies here. – Cameron Williams Jul 20 '15 at 2:12
• @Winther I believe I've demonstrated that the premises of Fubini-Tonelli fail. – Philip Hoskins Jul 20 '15 at 2:13
• @CameronWilliams Yeah. I just noticed, I only focused on the $\infty$-end of the integral. – Winther Jul 20 '15 at 2:13
• @Winther It happens. This is a very easily doable Riemann integral but horrific with Lebesgue integration. OP: you may need to interpret this simply as a Riemann integral. The Lebesgue integral fails sometimes with traditional integrals over the whole real line - this is one such case. – Cameron Williams Jul 20 '15 at 2:14

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 $$\int_0^\infty \int_{-\infty}^\infty \sin(x^2)x e^{-t^2 x^2} \,dt\, dx = \int_{-\infty}^\infty \int_0^\infty \sin(x^2)x e^{-t^2 x^2} \, dx \, dt$$
• Should it be $2+t^2$ in the numerator? – Philip Hoskins Jul 20 '15 at 2:36
• Unless I'm wrong, the only reason we can apply Fubini's here in the first step is by knowing that the integrand $g(x,t)=\sin{(x^2)}xe^{-t^2x^2}$ is $L^1$ on $[0,R] \times (-\infty, \infty)$, correct? – DaveNine Jul 20 '15 at 3:15