Rayleigh energy equation - applied to Fourier Transform of a sinusoid
An (infinite) sinusoidal signal does not really have a Fourier transform, since as you say it is not a finite-energy signal. It can be represented as a Fourier series, and there is a useful way to sort of write it as a "pseudo-Fourier transform" with Diract delta distributions, as well. However, if you try to take the magnitude of that spectrum and square it, it is not well-defined, since you would need the squared-magnitude of the Dirac deltas.