Given the integral:

$$f(x) = \int_{-\infty}^{+\infty} \left[\int_{-a/2}^{a/2} \cos\left(\frac{\pi y}{a}\right) e^{-iky}\,{\rm d}y\right] \ e^{i\left(kx - \frac{\hbar t}{2m}k^2\right)}\,{\rm d}k$$

How would I prove or disprove, that I can change the order of integration. If I evaluate the integral as it is now it is very difficult. When changing the order of integration it is almost trivial.

Am I correct in assuming that Fubbini's theorem doesn't apply as the function is not integrable?

  • $\begingroup$ Why you think the function is not measurable? $\endgroup$
    – Fabian
    Oct 17 '15 at 17:08
  • 1
    $\begingroup$ I think you mean to talk about integrability. The integral would not even be defined for non-measurable functions. $\endgroup$
    – Andrew
    Oct 17 '15 at 17:08
  • $\begingroup$ @Andrew yes I meant integrable $\endgroup$
    – john
    Oct 17 '15 at 17:10
  • $\begingroup$ Entire functions are very well measurable. $\endgroup$ Oct 17 '15 at 17:10
  • $\begingroup$ You are right in that you cannot apply Fubini's theorem here. This is a conditionally convergent integral and the theorems about such things are very few and far between. $\endgroup$ Oct 17 '15 at 17:11

I'm not sure how to justify the change of order, can't memorize, if I'm not totally wrong, it should be something about uniform convergence of the integrand.

I would like to suggest you to turn the complex exponential in the interior integral to $ \sin + i\cos$ , you should arrive at something like $const \cos (something)$ so that the remaining\outer integration will become somewhat similar to that one. Should not be that bad.

  • 3
    $\begingroup$ This should be a comment, not an answer. $\endgroup$ Oct 17 '15 at 17:29

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