The Fourier transform of a damped cosine and the units of the result If I take a simple transient voltage signal of the form
$$f(t) = V_p e^{-t/\tau} \cos(\omega_0 t)$$
and take the Fourier transform in the normal way
$$F(\omega) = \frac{V_p}{\sqrt{2 \pi}} \int^{+\infty}_{0}  e^{-t/\tau} \cos(\omega_0 t)  e^{-i \omega t} \ \ dt $$
giving the result 
$$F(\omega) = \frac{V_p}{2\sqrt{w\pi}} \left( \frac{1}{1/\tau - i (\omega_{0} - \omega)} + \frac{1}{1/\tau + i (\omega_{0} + \omega)} \right)$$
Seeing as I am interested in a Lorentzian like function, I ignore the second term in the bracket -- and then take the absolute-squared value of what is left giving 
$$\left|F(\omega)\right|^{2} = \frac{V^{2}_{p}}{2 \pi}\frac{1}{(2/\tau)^{2} + 4(\omega - \omega_{0})^{2}}$$ which when making the approximation of $2/\tau = \Delta\omega$ the linewidth gives a quantity in units of $[V^{2}/Hz^{2}]$. Consistent with a power spectrum density.

This is the crux of my question as if I consider what really happens with a measurement say on a signal/spectrum analyser, the time transient voltage is dissipated over the input impedance of the signal/spectrum analyser, which will have a thermal Johnson-Nyqist noise of $[V/\sqrt{Hz}]$ -- and define my noise floor. 

I feel that my result should somehow have the same units, that a spectral lineshape, originating from the FT of a time transient voltage signal should have either units of $[V/\sqrt{Hz}]$ or $[V^{2}/Hz]$. How can I rectify my inconsistency with units as in the end I wish to write something like
$$PSD = \sqrt{\left|F(\omega)\right|^{2} + e_{n}^{2}}$$
Where $e_{n}$ (units of $[V/\sqrt{Hz}]$) is the noise floor of my spectrum.



Some additional thought. I believe this problem can be rectified by the definition of the Power Spectrum density
$$S(\omega) = \lim_{T \rightarrow \infty} \frac{1}{T} |F(\omega)|^{2}$$
this factor of $1/T$ takes care of the dimensionality problem -- however I am unsure how to apply this equation as it seems to be in the limit just sends the function to zero because of the $1/T$ factor.
if $v(t)$ is an electromotive force (a voltage expressed in units of volt) that is a function of time (let's say expressed in units of second), then the Fourier Transform (or the Laplace Transform):

$$ \mathscr{F} \Big \{ v(t) \Big \} \triangleq V(i \omega) = \int\limits_{-\infty}^{\infty} v(t) e^{-i \omega t} \, \mathrm{d}t  $$

will have units of volt-second.  And $\omega$ will be in radian/second.