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Imagine I have an integral of the following form:

$$I = \int_{-\infty}^{\infty} d\tau_1 \int_{\tau_1}^\infty d\tau_2 \int_{\tau_2}^\infty d\tau_3\ f(\tau_1,\tau_2,\tau_3) \tag{1}$$

Can I always commute the integrals, by changing the integration limits accordingly? And if not, when is it allowed/not allowed? For example:

$$I \overset{?}{=} \int_{-\infty}^\infty d\tau_2 \int_{-\infty}^{\tau_2} d\tau_1 \int_{\tau_2}^\infty d\tau_3\ f(\tau_1,\tau_2,\tau_3) \tag{2}$$

It seems to me that the region over which I integrate is the same, however I did run into discrepancies when numerically integrating $(2)$ vs. $(1)$ in some instances. I am not sure if they are artifacts from the numerical integration, hence the question.

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1 Answer 1

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In general, it is not always possible. Fubini's Theorem gives some conditions under which it is possible to change the order of integration (see, for instance, https://en.wikipedia.org/wiki/Fubini%27s_theorem).

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  • $\begingroup$ Thank you for your answer! Do you know if I can change the order of integration as above in the case where the integral is finite and $f(\tau_1,\tau_2,\tau_3) \geq 0$, i.e. $\left| f(\tau_1,\tau_2,\tau_3) \right| = f(\tau_1,\tau_2,\tau_3)$ (the integrand is real)? Is that a sufficient condition? $\endgroup$
    – Pxx
    Commented May 25, 2020 at 10:51
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    $\begingroup$ As you can find here: en.wikipedia.org/wiki/…, if the integral is finite, and if $f$ is always non negative as in your assumptions, you can change the order. $\endgroup$ Commented May 25, 2020 at 11:03
  • $\begingroup$ Great, thanks a lot! $\endgroup$
    – Pxx
    Commented May 25, 2020 at 11:09

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