In reference to a recent course about Fourier analysis, various misconceptions arose, which I couldn't resolve via available internet resources and the posts on this forum. This post contains a few questions, which are listed below, and are closely related to the Fourier series and Fourier transform.
If the Fourier series can approximate almost any well-behaved (periodic) function/signal, on an interval between -L and L, via the addition of sines and cosines (or complex exponentials for that matter), then what would be the reason to define the Fourier transform? As a matter of fact, the Fourier transform is merely suited for non-periodic functions (unless Delta-Diracs are introduced), and provides a continuous frequency range. However, why would one create a continuous frequency range while the Fourier series has already provided all the required harmonics to compose the required function? It seems no more information than those harmonics is required, hence, why introduce a continuous analog that requires an infinite period?
Secondly, it is often stated that the Fourier transform yields the pertaining weight of a certain frequency given a function, but how can this be the correct 'weight' or 'Fourier coefficient' if the $d\omega/2\pi$ term is omitted. Here I'll demonstrate my train of thought: $$C_n=\frac{1}{2L}\int\limits_{-L}^{L}f(t)e^{-\frac{i\pi nt}{L}}dt$$ $$\omega =\frac{n \pi}{L} \to \Delta \omega = \frac{\pi}{L} \to \frac{\Delta \omega}{2\pi}=\frac{1}{2L}$$ $$C_n=\frac{\Delta \omega}{2\pi}\int\limits_{-\frac{\pi}{\Delta \omega}}^{\frac{\pi}{\Delta \omega}}f(t)e^{-i\omega t}dt$$ Taking the limit of $\lim \limits_{\Delta\omega \to 0}C_n$ should give the continuous version of the so called 'weight', but then the $d\omega$ is used by an integral to sum up an infinite amount of complex exponentials and continuous $C_n$ coefficients, which essentially defines the inverse Fourier transform: $$f(t)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}f(t)e^{-i\omega t}dt d\omega$$ [Fundamentally a continuous exponential Fourier series]
Again, how could the official Fourier transform, $F(\omega)=\int\limits_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$, provide a 'weight' if the required $\frac{d\omega}{2\pi}$ term is dropped off?
- Lastly, I realised that the Discrete Fourier Transform (DFT) is very reminiscent of the exponential Fourier series, and it turns out that the DFT is virtually a numerical way for determining the Fourier coefficients of a measured signal. Consequently, why is the DFT called a transform, even though it doesn't even require the continuous Fourier transform. In fact, the DFT is customarily used by computers, thereby rendering the Fourier Transform obsolete.
To summarize, I don't seem to comprehend the motivation for the Fourier transform compared to the Fourier series.
Thanks in advance.