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I fear asking 'how/why is this possible' might lead me down a dangerous path, but is there an intermediate step here that may make it clearer to me why this is legal?

Taken from a proof of Parseval's Theorem:

$$ \int_{-\infty}^{\infty}g(t)\cdot\left[\dfrac{1}{2\pi}\int_{-\infty}^{\infty}G^\ast(\omega)e^{j\omega t}\cdot d\omega\right]\cdot dt $$ $$ = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}G^\ast(\omega)\left[\int_{-\infty}^{\infty}g(t)e^{j\omega t}\cdot dt\right]\cdot d\omega $$

I'm told it's "simpler because it's indefinite", but I don't really follow how it's done in any case.

Would be greatful if someone could explain, or tell me what the process is actually called so that I can search better than "change order of integrals" - 'order' is a troublesome word in searches for math help, for obvious reasons.

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  • $\begingroup$ Have you encountered Fubini's Theorem? $\endgroup$
    – Clement C.
    Apr 8, 2014 at 16:41
  • $\begingroup$ @ClementC No, I haven't. $\endgroup$
    – OJFord
    Apr 8, 2014 at 16:42
  • $\begingroup$ (You may want to have a look at the link, then) $\endgroup$
    – Clement C.
    Apr 8, 2014 at 16:43
  • $\begingroup$ Thanks, that certainly does help. But how is the $e^{j\omega t}$ term shifted between them in this example, taking a different parameter as constant/running in each case? $\endgroup$
    – OJFord
    Apr 8, 2014 at 16:49
  • $\begingroup$ Yes: roughly, $\int_x\int_t dxdt\ f(x,t) = \int_t\int_x dxdt\ f(x,t)$. $\endgroup$
    – Clement C.
    Apr 8, 2014 at 16:57

2 Answers 2

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Look for "Fubini's theorem" and "Tonelli's theorem".

One can define a "double integral" $\displaystyle\iint_{[a,b]\times[c,d]} f(x,y)\,d(x,y)$ without defining integrals with respect to $x$ or $y$ separately. This is a well defined finite number if and only if $\displaystyle\iint_R |f(x,y)|\,d(x,y)$ (the integral of the absolute value) is $<\infty$.

The "iterated integrals" are $\displaystyle\int_a^b \left(\int_c^d f(x,y)\, dy\right)\,dx$ and $\displaystyle\int_c^d \left(\int_a^b f(x,y)\, dx\right)\,dy$. In some cases these are defined when the double integral is not, and then the may be unequal.

Fubini's theorem says that the two iterated integrals are equal to the double integral (and hence to each other) whenever the double integral is defined, i.e. the double integral of the absolute value is finite.

Tonelli's theorem says that the two iterated integrals are equal to the double integral (and hence to each other) if $f(x,y)\ge 0$ for all $(x,y)\in[a,b]\times[c,d]$ (regardless of whether that value is finite or infinite).

Consequently the two iterated integrals can be unequal only if the double integral takes the indeteriminate form $\infty-\infty$, i.e. the integral of $f$ over the region where it is non-negative is $+\infty$ and the integral of $f$ over the region where it is negative is $-\infty$.

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The result is known as Fubini's theorem:

$$\int_A\left(\int_Bf(x,y)\mathrm{d}y\right)\mathrm{d}x=\int_B\left(\int_Af(x,y)\mathrm{d}x\right)\mathrm{d}y=\int_{A\times B}f(x,y)\mathrm{d}(x,y)\\ \text{Iff $f$ is measurable ($f(x,y)$ is $A\times B$ integrable) and }\int_{A\times B}f(x,y)\mathrm{d}(x,y)<\infty$$

This is used many times in many places. For example, trying to evaluate (taken from this site) $$\int^1_0\int^1_xe^{y^2}\mathrm{d}y\mathrm{d}x$$ Directly would be a problem (the reason being that there's no elementary anti-derivative of $e^{y^2}$).

As for the $e^{j\omega t}$ term, see that it's not being moved - you're moving the $g(t)$ in and changing $\mathrm{d}\omega \mathrm{d}t$ to $\mathrm{d}t\mathrm{d}\omega$. This is because $e^{j\omega t}$ is a function of $\omega$ and $t$ and so cannot be moved out of any integral.

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  • $\begingroup$ Thank you, I still don't see how the $e^{j\omega t}$ term in the example in my OP can be 'moved' (to stay in the 'inner' integral) though? $\endgroup$
    – OJFord
    Apr 8, 2014 at 16:52
  • $\begingroup$ @OllieFord It's not being moved - you're moving the $g(t)$ in and changing $\mathrm{d}\omega \mathrm{d}t$ to $\mathrm{d}t\mathrm{d}\omega$. See that $e^{j\omega t}$ is a function of $\omega$ and $t$ and so cannot be moved. $\endgroup$
    – user122283
    Apr 8, 2014 at 16:54
  • $\begingroup$ Well it's 'moved' in the sense that it's being integrated wrt a different variable first. $\endgroup$
    – OJFord
    Apr 8, 2014 at 16:56
  • $\begingroup$ @OllieFord Not exactly. As I've stated above, Fubini's theorem allows us to write $$\dfrac{1}{2\pi}\int^\infty_{-\infty}g(t)\int^\infty_{-\infty}\left[G^{*}(\omega)e^{j\omega t}\mathrm{d}\omega\right]\mathrm{d}t=\\ \dfrac{1}{2\pi}\int^\infty_{-\infty}G^{*}(\omega)\int^\infty_{-\infty}\left[g(t)e^{j\omega t}\mathrm{d}t\right]\mathrm{d}\omega$$As long as we "reverse" the functions, too. Note that the limits of integration are the same, and so you don't have to change them. $\endgroup$
    – user122283
    Apr 8, 2014 at 16:59
  • $\begingroup$ It seems potentially misleading not to be completely explicit about what "integrable" means. It means the integral of the absolute value is finite. The two iterated integrals do indeed fail to be equal to each other in some cases where the double integral of the absolute value is infinite. $\endgroup$ Apr 8, 2014 at 17:02

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