The fourier transform is given as below :
$$F(\omega) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} f(t) e^{-i \omega t} \mathrm{d}t$$
Now $$F(\omega) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} f(t) e^{-i \omega t} \mathrm{d}t$$ can be transformed as
$$F(\omega) = 0.5 \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} k s(k) e^{-i \omega k^2} \mathrm{d}k$$
where $ k = \sqrt{t} $ and $ s(k) = f(t^2) $
The exponential term inside the integral looks like a gaussian distribution although it is a complex function. Is there any connecting probabilistic interpretation along these lines of the fourier transform or series ? Under such an idea a fourier transform would be an expectation of a gaussianish distribution.
Looking forward to insights.
The characteristic function of a real-valued random variable $X$ is given by
$$\varphi_X (t):=E[e^{itX}],$$
which is an object that completely characterizes the distribution of $X$. If $X$ admits a density function $f$, then $\varphi_X(-t)$ is the Fourier transform of its density function (modulo any differences in a normalizing constant or sign conventions).