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?

  • $\begingroup$ 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. $\endgroup$ Jul 20, 2015 at 2:07
  • $\begingroup$ @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. $\endgroup$ Jul 20, 2015 at 2:12
  • $\begingroup$ @Winther I believe I've demonstrated that the premises of Fubini-Tonelli fail. $\endgroup$ Jul 20, 2015 at 2:13
  • $\begingroup$ @CameronWilliams Yeah. I just noticed, I only focused on the $\infty$-end of the integral. $\endgroup$
    – Winther
    Jul 20, 2015 at 2:13
  • $\begingroup$ @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. $\endgroup$ Jul 20, 2015 at 2:14

1 Answer 1


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

  • $\begingroup$ Should it be $2+t^2$ in the numerator? $\endgroup$ Jul 20, 2015 at 2:36
  • $\begingroup$ @PhilipHoskins Thanks. I fixed it. $\endgroup$ Jul 20, 2015 at 2:56
  • $\begingroup$ Cool. Thanks for taking the time to answer my question! $\endgroup$ Jul 20, 2015 at 3:14
  • $\begingroup$ 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? $\endgroup$
    – DaveNine
    Jul 20, 2015 at 3:15
  • $\begingroup$ @DaveNine Yes, and I used Phil Hoskin's argument in showing that. $\endgroup$ Jul 20, 2015 at 3:18

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