I'm trying to piece together how to get to derive the Fourier transform from the Fourier series, reading my lecturer's notes.
Now, if $f(x)$ happens to not be periodic, we can still give a complete description of $f(x)$ in terms of sines and cosines, but we'd need to sum over a continuum of $k$-values, instead of just integer multiples of a fundamental harmonic. We can think of a non-periodic function as the limiting case of a periodic one, where the wavelength $\lambda$ tends to infinity, meaning the $k_{n}$'s become more and more closely spaced together until they form a continuum. In the limit $k \to \infty$, the wavenumber $k$ becomes a continuous variable and the summation goes over to an integral: $$f(x) = \sum_{n=-\infty}^{+\infty}C_n \ e^{ik_n x} \to \int_{-\infty}^{+\infty}C(k)\ e^{ikx}\ dk$$
In my mind, it makes sense that a periodic function ceases to be periodic if its wavelength is infinite, I thinik that's obvious. However, that does not make it obvious to me that the wavenumber is continuous. As the wavelength gets larger and larger, the wavenumber gets smaller and smaller, however it is always some value. If I took a graph of a sine wave from $-\pi$ to $\pi$ and blew it up so its wavelength got more and more massive, its wavenumber would always have some single value. Is this because we're taking it in the case of an infinite wavelength? I could see how this could maybe make $k$ continuous, but I couldn't justify it well for myself.
Secondly, why are we using the case of a non-periodic function to make the transition from infinite summation to integral for the Fourier transform derivation? What if we're making the derivation for a periodic function? Could we still make this step?