Fourier transform of a DC signal is an impulse at the origin.
Now if the spectrum of a signal is the plot of the amplitudes of the respective frequencies of each harmonic, then the Fourier transform shouldn't be an impulse but a signal of the height of the DC value at the origin. Doesn't the impulse imply that the DC signal is also of infinite amplitude?
Dirac is a distribution and not a function. That means it can only be defined by an integral : $$<\delta, \phi> = \int{\phi\cdot d\delta} = \phi(0)$$
Now if we take the Fourier transform of a Dirac, in a distribution sense : $$ s(x) = \delta $$ $$ \begin{split} S(f) &= \int^{+\infty}_{-\infty}{s(x)\times e^{-2\pi ifx}dx}\\ &=\int^{+\infty}_{-\infty}{e^{-2\pi ifx}d\delta}\\ &=e^{-2\pi if\times0} = 1 \end{split} $$
Which means that reciprocally, $TF(1) = \delta$.
Considering that Fourier Transform is linear, $TF(C) = C\times \delta$
The other think to take in account is that a DC signal in the pure mathematical sense doesn't physically exist : it would mean that the signal has existed for an inifinite period of time and will continue to do so until the end of infinity.
First, there are several mistakes in the question. "Spectrum", in its most general term refers to the Fourier transform itself.
The Parseval theorem states that the total signal energy is related to the integral of the magnitude of the Fourier transform. From that, the magnitude of the Fourier transform is related to the energy spectral density. Hence to derive the concentrated energy around a certain frequency, one needs to integrate the energy spectral density around that particular frequency. It is NOT just by looking at the Fourier magnitude at that particular frequency.
Although the Dirac delta is infinite at the origin, its integral is finite which is the quantity we look at for this purpose.
Notice that I intentionally prevented a long (rigorous) answer which requires several lines of equation.
