The solution of a certain Fokker-Planck equation in the Fourier space $(k,t)$ is,
$$ I(k,t)=\exp \left[ -\frac{k^2}{a^2 + k^2 } \frac{t}{\tau} \right] \tag1 $$
If we consider a parameter $x_0 = a^{-1}$, then we can rewrite this expression as follows,
$$ I(k,t)=\exp \left[ -\frac{k^2 x_0^2}{1 + k^2 x_0^2} \frac{t}{\tau} \right] \tag2 $$
Let us consider the limit, $kx_0 \rightarrow 0$, clearly, we can approximate the above solution as a Gaussian upto leading order in $k$,
$$
I_{lim}(k,t) = \lim_{kx_0\rightarrow 0} I(k,t) = \exp\left[ -k^2 x_0^2\frac{t}{\tau} \right] \tag3
$$
which upon Fourier transform ($k\rightarrow x$) gives the function $G(x,t)$ as Gaussian solution.
$$
G_{lim}(x,t) = \frac{1}{\sqrt{4\pi (x_0^2/\tau)t}} \exp \left[-\frac{x^2}{4(x_0^2/\tau)t} \right] \tag4
$$
Now it is possible to calculate the exact Fourier inverse transform of equation (2) (without considering any limiting behaviour)
(Refer to this answer)
The complete solution $G(x,t)$ is,
$$
G(x,t) = \exp\left(-t/\tau \right) \left[ 2\pi \delta(x)+\frac{1}{x_0}\sqrt{\pi}\sum_{n=1}^\infty \frac{K_{n-1/2}(|x|/x_0)(|x|/x_0)^{n-1/2}}{2^{n-5/2}n!(n-1)!} \right] \tag5
$$
If we get back to find the limiting behaviour from this solution, we first need to modify the limiting expression $kx_0 \rightarrow 0$ to $(x/x_0) \rightarrow \infty$ (using the inverse scaling relationship between Fourier-transform conjugates). Now, can we prove that in $x/x_0 \rightarrow \infty$, the above eq. (5) will yield eq. (4)? ie., can we prove the following? $$ G_{lim}(x,t) = \lim_{x/x_0 \rightarrow \infty} \exp\left(-t/\tau \right) \left[ 2\pi \delta(x)+\frac{1}{x_0}\sqrt{\pi}\sum_{n=1}^\infty \frac{K_{n-1/2}(|x|/x_0)(|x|/x_0)^{n-1/2}}{2^{n-5/2}n!(n-1)!} \right] \\ = \frac{1}{\sqrt{4\pi (x_0^2/\tau)t}} \exp \left[-\frac{x^2}{4( x_0^2/\tau)t} \right] $$
Although, this limit is true indirectly, can we show it directly from the above expansion series?
If such a proof can be shown, it suggests that this series in that limit could give an alternate expression for a Gaussian function.
