Equivalent of Fourier transform for Gaussians?
The (continuous) Fourier transform of a Gaussian is a Gaussian:

$$ \mathscr{F}\Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \, dt $$

$$ \mathscr{F}\left\{ e^{-\pi t^2} \right\} = e^{-\pi f^2} $$

$$ \mathscr{F}\left\{ e^{-\pi \alpha t^2} \right\} = \frac{1}{\sqrt{\alpha}} e^{-\frac{\pi}{\alpha} f^2} \qquad \alpha &gt; 0 $$

$$ \mathscr{F}\left\{ e^{-\pi \alpha (t-\tau)^2} \right\} = \frac{1}{\sqrt{\alpha}} e^{-\frac{\pi}{\alpha} f^2} e^{-j 2 \pi f \tau} $$

$$ \mathscr{F}\left\{ \sum_m c_m e^{-\pi \alpha_m (t-\tau_m)^2} \right\} = \sum_m \frac{c_m}{\sqrt{\alpha_m}} e^{-\frac{\pi}{\alpha_m} f^2} e^{-j 2 \pi f \tau_m} $$

note these gaussians have "tails" that go on forever, but gaussians usually die off to very close to zero very rapidly.

assuming no time or frequency aliasing, let $x[n] \triangleq x\big(n/f_\text{s}\big)$ where $f_\text{s}$ is the sample rate.  Then the DTFT is

$$\begin{align}
X\big(e^{j\omega}\big) &amp;\triangleq \sum_{n=-\infty}^{\infty} x[n] \, e^{-j \omega n} \\
&amp;= f_\text{s} \sum_{n=-\infty}^{\infty} x\big(n/f_\text{s}\big) \, e^{-j f_\text{s} \omega (n/f_\text{s})} \frac{1}{f_\text{s}} \\
&amp; \approx f_\text{s} \int\limits_{-\infty}^{+\infty} x(t) \, e^{-j 2 \pi \frac{f_\text{s} \omega}{2 \pi} t} \, dt \\
&amp;= f_\text{s} \ \mathscr{F}\Big\{ x(t) \Big\} \Bigg|_{f=f_\text{s}\frac{\omega}{2 \pi}} \\
\end{align}$$

so if we let $$x(t) = \sum_m c_m e^{-\pi \alpha_m (t-\tau_m)^2}$$

$$\begin{align}
x[n] &amp;= \sum_m c_m e^{-\pi \alpha_m ((n/f_\text{s})-\tau_m)^2} \\
&amp;= \sum_m c_m e^{-\pi (\alpha_m/f_\text{s}^2) (n-f_\text{s}\tau_m)^2} \\
&amp;= \sum_m c_m e^{-\pi \widehat{\alpha_m} (n-\widehat{\tau_m})^2} \\
\end{align}$$

then the DTFT is

$$\begin{align}
X\big(e^{j\omega}\big) &amp; \approx f_\text{s} \ \sum_m \frac{c_m}{\sqrt{\alpha_m}} e^{-\frac{\pi}{\alpha_m} f^2} e^{-j2\pi f\tau_m} \Bigg|_{f=f_\text{s}\frac{\omega}{2 \pi}} \\
&amp;= f_\text{s} \ \sum_m \frac{c_m}{\sqrt{\alpha_m}} e^{-\frac{\pi}{\alpha_m} \left(f_\text{s}\frac{\omega}{2 \pi}\right)^2} e^{-j2\pi \left( f_\text{s}\frac{\omega}{2 \pi} \right)\tau_m} \\
&amp;= f_\text{s} \ \sum_m \frac{c_m}{\sqrt{\alpha_m}} e^{-\pi\frac{f_\text{s}^2}{4 \pi^2 \alpha_m} \omega^2} e^{-j \omega (f_\text{s}\tau_m)} \\
\end{align}$$

Then, assuming you choose your DFT length $N$ to be sufficiently large (you're saying in the millions) to cover all of the gaussians with their specific delays $f_\text{s}\tau_m$ in samples and widths of ca. $\frac{f_\text{s}}{\sqrt{\alpha_m}}$.  Then the $N$-point DFT will sample this DTFT at $N$ equally spaced points around the unit circle.

$$\begin{align}
\mathcal{DFT:} \quad X[k] &amp;= \sum_{n=0}^{N-1} x[n] \, e^{-j 2 \pi \frac{nk}{N}} \\
&amp;= X\big(e^{j\omega}\big) \Bigg|_{\omega = 2\pi\frac{k}{N}} \quad \mathcal{:DTFT}\\
&amp;= f_\text{s} \ \sum_m \frac{c_m}{\sqrt{\alpha_m}} e^{-\pi\frac{f_\text{s}^2}{4 \pi^2 \alpha_m} \omega^2} e^{-j \omega (f_\text{s}\tau_m)} \Bigg|_{\omega = 2\pi\frac{k}{N}}\\
&amp;=  \ \sum_m \frac{c_m}{\sqrt{\alpha_m/f_\text{s}^2}} e^{-\pi\frac{f_\text{s}^2/\alpha_m}{4 \pi^2} \left(2\pi\frac{k}{N}\right)^2} e^{-j 2\pi\frac{k}{N} (f_\text{s}\tau_m)} \\
&amp;= \sum_m \frac{c_m}{\sqrt{\widehat{\alpha_m}}} e^{-\frac{\pi}{\widehat{\alpha_m}} \left(\frac{k}{N}\right)^2} e^{-j 2\pi\frac{k}{N} \widehat{\tau_m}} \\
\end{align} $$

So summing up and I will remove the little hats in the discrete-time domain, if you have a collection of gaussian pulses with heights of $c_m$, peak width parameters $\frac{1}{\sqrt{\alpha_m}}$ (in units of samples), and peak positions at $\tau_m$ (also in units of samples), then your input should look like:

$$ x[n] = \sum_m c_m e^{-\pi \alpha_m (n-\tau_m)^2} $$

and the result of the $N$-point DFT should look like:

$$ X[k] \approx \sum_m \frac{c_m}{\sqrt{\alpha_m}} \ e^{-\pi/(N^2\alpha_m) k^2} \ e^{-j 2\pi(\tau_m/N) k} $$

for $-\tfrac{N}{2} &lt; k &lt; \tfrac{N}{2}$ .  Remember with the DFT, $X[k+N]=X[k]$ for all integer $k$.