TITLE: Derivative of Integral (in Fourier transform)
QUESTION [6 upvotes]: I've taken some analysis, but somehow Fourier transforms were never brought up until they were assumed to be familiar.  Fun.  Anyway, in a class example (showing the integral of a Gaussian is again a Gaussian), the professor did the following step, which confused me.  By our definition we have
$$\hat{f}(\gamma):=\int f(t)e^{-2\pi i t\gamma}dt$$
from which we concluded
$$(\hat{f})'(\gamma)=-2\pi i \int t f(t) e^{-2\pi i t\gamma}dt$$
it would appear that he was using a "theorem," that if $f(\gamma)=\int g(\gamma,t)dt$ then $f'(\gamma)=\int (\dfrac{d}{d\gamma} g(\gamma,t))dt$, but I'm not aware of any such theorem.  Is this true?  If so, is it supposed to be obvious, or does it have a name?  If not true, is there a simple argument as to why it's true here?  (our $f$ was $f(t)=e^{-rt^2}$ for some fixed real $r>0$).

REPLY [2 votes]: IT is not at all obvious. The result follows from what is called the Dominated Convergence Theorem.
It is usually tought in measure theory course. Perhaps you have not taken measure theory yet.
Some good references for Fourier analysis are introduction to harmonic analysis by Katznelson, an introduction to Fourier analysis by Pinsky.
They make heavy use of measure theory. For measure theory, you can see real and complex analysis by Rudin. Measure theory by Halmos is also good.
For a light introduction to Fourier analysis avoiding measure theory, see A first course in harmonic analysis by Anton Dietmar.