Rayleigh energy equation - applied to Fourier Transform of a sinusoid I saw details indicating "total energy of a Fourier-transformable signal equals the total area under the curve of squared amplitude spectrum of this signal."

I have no issue with that. But I have found pretty much no discussions about the consequence of applying the Rayleigh energy equation to the Fourier transform of sinusoidal signal - which of course has infinite energy. There are certainly discussions indicating that the energy equation can only be applied to finite energy signals, and therefore is not applicable to the Fourier transform of a sinusoidal time domain signal. 

So - what is the bottom line? Does this mean that if one attempts to apply the energy equation to the Fourier transform of a sinusoid, then the result will be a finite value or finite number - and it will be a meaningless number, right?

That is - when applied to finite energy signals, the energy equation returns meaningful results. But when applied to infinite energy signals like a sinusoid, then it returns a value, but it will be a meaningless result, right?
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.