Can the sound of infinitely long played music be decomposed in sine (or cosine) forms? Every arbitrary waveform (except non-linear ones) can be decomposed in sine (or cosine) waveforms that spatially extend to infinity. That is if the waveform has a finite spatial extent. But say that I have a waveform that has an infinite spatial extent. Can the same be done for such a waveform? Say for the soundwave of a piece of music that's being played forever, without a beginning and without an end.
A sine (or cosine) waveform extends to infinity, so there is for every sine waveform a corresponding waveform into which it can be decomposed, namely the sine form itself. An arbitrary periodic sound can be decomposed in sine (or cosine) waveforms too (in principle). But what about an arbitrary infinite waveform?
 A: I have to admit that I have problems to understand your intention, therefore I like to put things together:
The Fourier series can describe every (infinitely long) periodic funtion. By the way, it's actually wrong that the Fourier series consists of sin and cos only, but it consists of sin(nx), cos(nx), and the 1-function. That is, because 1-function, sin(nx) and cos(nx) are a basis for all periodic functions. However, so far there are discrete elements (overtones of sin and cos).
Going from discrete steps to continuous coordinates ($n \cdot f \longrightarrow df$, and the sum becomes an integral) yields the Fourier transform, which can describe aperiodic, finite long functions.
What you want is now an indefinitely long, aperidoc function, correct? Such as $f(x) = x$ ? Hm... The integration is from $-\infty $ to $\infty$, but as far as I remember that is an effect of getting from sin and cos to complex numbers $e^{i...}$.
After some Google search I found this very nice site (but it won't help you most likely because it's in German): https://www.mathe-online.at/mathint/fourier/i.html. A very comprehensive overview about fourier series and transform. At the very bottom of that site they claimed that the Fourier transform is only good for functions approaching 0 "sufficiently fast" for $x \longrightarrow \infty$. But there is no proof provided and no info what "sufficiently fast" is...
