This is in standard texts. For example,
in J.D. Jackson, Classical electrodynamics, third edition, Eq. 14.67
\begin{equation}
\frac{d^2I}{d\omega d\Omega} = \frac{e^2\omega^2}{4\pi^2 c}
\left | \int_{-\infty}^\infty \hat n \times (\hat n \times \vec \beta)
e^{i\omega(t-\hat n \cdot \vec r(t)/c)} dt \right |^2
\end{equation}
as the intensity distribution as a function of frequency and direction.
Here $\vec r(t)$ is the particle trajectory,
$c\vec \beta = \frac{d\vec r(t)}{dt}$,
and $\hat n = \hat x \sin\theta \cos\phi+\hat y \sin\theta\sin\phi
+\hat z \cos\theta$ is the unit vector in the direction of radiation. Jackson's result is in Gaussian units.
Jackson discusses when this is valid. Essentially when the radiation
is localized in space and time (for example, no radiation before the big bang.
Your example of a constant acceleration is a well known
badly behaved trajectory.)
A standard way to derive the result is to start with an arbitrary
localized in space and time current and charge density. Plug these into
the Lorenz gauge retarded coulomb's law equations for the vector and
scalar potentials. Fourier transform the potentials in time to get their
frequency components, take the large distance limit to separate the
radiation fields which go like $r^{-1}$. The result is that the
radiation fields are proportional to the component of the space-time Fourier transform of the
current density evaluated at $\vec k = \frac{\hat n\omega}{c}$, and $\omega$ which is
transverse to the radiation direction.
Calculating the time integral of the
Poynting vector then gives the intensity (i.e.
the energy radiated per unit frequency and solid angle).
You can see that the integral in Jackson's result is proportional to
the Fourier transform of the current density at this $\vec k$ value.
That is
\begin{equation}
\begin{split}
\vec J(\vec k,\omega) &= \int d^3r' dt' ec\vec \beta
\delta^3(\vec r'-\vec r(t')) e^{-i \vec k\cdot \vec r'+i\omega t' }
\\
&= ec \int dt' \vec \beta(t') e^{-i\vec k \cdot \vec r(t') +i\omega t'}
\end{split}
\end{equation}
with $\vec k = \hat n \frac{\omega}{c}$. Taking $- \hat n \times (\hat n \times \vec J(\vec k,\omega))$ gives the transverse component.
