Changing order of integrals within integrals 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.
 A: 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 $-\infty$.
A: 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.
