Proof of the statement about Fourier transform

Question

I have a problem proving the statement about Fourier transform

$$ \begin{align*}f(t) &= \int_{-b}^b S(\nu)e^{2\pi i \nu t}d\nu\\ &= \int_{-b}^b S(\nu)e^{2\pi i \nu t_0}e^{2\pi i \nu (t-t_0)}d\nu\\ &= \int_{-b}^b S(\nu) e^{2\pi i \nu t_0}\sum_{n=0}^\infty \frac{(2\pi i \nu (t-t_0))^n}{n!}d\nu \\ &= \sum_{n=0}^\infty\frac{(2\pi i (t-t_0))^n}{n!}\int_{-b}^b \nu^n S(\nu)e^{2\pi i \nu t_0}d\nu\\ &= \sum_{n=0}^\infty\frac{(t-t_0)^n}{n!} f^{(n)}(t_0)\end{align*} $$

Question: Why can we move the sum out of the integral?

Answer 1

Usually, when one is doing Fourier transform, one has the integral on the real line; if one performs the integral on a compact interval, say $[-b,b]$ as in the post, then one is looking for the Fourier series.

Regarding the exchange of the summation sign and integral sign, one can apply Fubini-Tonelli theorem by considering the counting measure. Of course, one needs assumptions on the function $S(\nu)$.

Answer 2

Maybe I'm being dumb and missing something here, but the only term in the sum that is dependent on the integral is $ v $. It just seems that they took the $ v $ out of the sum and pulled the remaining part of the sum out of the integral, since it no longer depends on $ v $.

Is that it, or am I completely wrong?