The formula for computing the coefficients of Fourier series is: $$ C_{k} = \frac{1}{2L} \int_{-L}^{L} f(x) e^{-i2 \pi \omega_{k} x} \: dx$$ where $\omega_{k} = \frac{k}{2L}$.
I understand that we can compute them efficiently via algorithm/computer. The Discrete Fourier Transform can be obtained as follows, by the fact that $f(x)$ is $2L$-periodic.
\begin{align*}C_{k} &= \frac{1}{2L} \int_{-L}^{L} f(x) e^{-i2 \pi \omega_{k} x} \: dx\\ &= \frac{1}{2L} \int_{0}^{2L} f(x) e^{-i2 \pi \omega_{k} x} \: dx \end{align*} then we approximate the integral by discretization, let $x_{n} = n \Delta x, \:\:\: n=0,1,2,...,N-1$ where $(N-1) \Delta x = 2L$. So $\Delta x = \frac{2L}{N-1}$. Which means $x_{n} = \frac{2nL}{N-1}$.
\begin{align*} C_{k} &\approx \frac{1}{2L} \sum_{n=0}^{N-1} f(x_{n}) e^{-i 2 \pi \frac{k}{2L} x_{n}} \: \Delta x\\ &= \frac{1}{2L} \sum_{n=0}^{N-1} f(x_{n}) e^{-i 2 \pi \frac{k}{2L} \left(n \frac{2L}{N-1} \right) } \: \left( \frac{2L}{N-1} \right)\\ &= \frac{1}{N-1} \sum_{n=0}^{N-1} f_{n} e^{-i 2 \pi \frac{k n}{N-1} } \end{align*} But in DFT we just multiply it by $N-1$ (DFT computes $\sum_{n=0}^{N-1} f_{n} e^{-i 2 \pi \frac{k n}{N-1} }$)
(Q) But what is the common form's $ \hat{F} = \int_{-\infty}^{\infty} f(x) e^{-2i \pi \omega x} \: dx$ origin??
I see that based on this video at minute 11:26 (https://www.youtube.com/watch?v=_eq8rUr1QIg&t=383s), we take the limit of $L$ to $\infty$, and since $\omega_{k} = \frac{k}{2L}$ then we can regard $\Delta\omega = \frac{1}{2L}$. Taking $L$ to $\infty$ implies $\Delta\omega = 0$. Is this the origin? or is there other approach? no textbook ever mentioned this (as far as I know).