Asymptotics of integral representation of distribution

Question

I initially posted this question at MSE (here), but I have gotten no response, so I figured I would ask it to this community.

Background: I am studying the PDE $$\,\,\,\,\,\,\,\,\,\,\,\,i\partial_t \psi(x,t)=-\Delta\psi-\epsilon^4\Delta^2\psi\\ \psi(x,0)=f(x)$$ for $x\in\mathbb{R}, t\ge 0$, $\epsilon>0$ small, and $f\in\mathcal{S}(\mathbb{R})$. I am particularly interested in the behavior of $\psi$ in the limit as $\epsilon \to 0$. It is alreacy clear the the limit will be quite singular as the order of the PDE changes suddenly as soon as $\epsilon\neq 0$. The Fourier transform in space of $\psi$, denoted $\widetilde{\psi}(k,t)$, satisfies $$\widetilde{\psi}(k,t)=e^{-itk^2}e^{i\epsilon^4 t k^4}\widetilde{f}(k).$$ Let $\psi_0(x,t)$ denote the solution to the PDE when setting $\epsilon=0$. Then, $$\widetilde{\psi}(k,t)=\widetilde{\psi_0}e^{i\epsilon^4 t k^4}.$$ Inverting the Fourier transform, we find that $$\psi(x,t)=\int_{-\infty}^\infty\psi_0(y,t)\,F_\epsilon(x-y,t)\,dy,$$ where $$F_\epsilon(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}e^{i\epsilon^4 tk^4}\,dk$$ We treat $F$ as a tempered distribution since we uniquely care about its action on an element of $\mathcal{S}$, in this case $\psi_0$, which gives us the solution to the desired PDE. As distributions, $F_0(x,t)=\delta_0$. My question for this post is as follows: is it correct, to expand $e^{i\epsilon^4 tk^4}=\sum_{m=0}^\infty (i\epsilon^4 t)^m k^{4m}$, switch the order of integration and integrate to obtain $$F_\epsilon(x,t)=\delta_0+\sum_{m=1}^\infty (i\epsilon^4 t)^m \,\delta^{\,(4m)}_0,$$ where $\delta^{(n)}_0$ is the $n$-th derivative of the dirac mass at $0$? If not, what is the correct way to proceed to find an asymptotic expansion of $F_\epsilon$ as $\epsilon\to 0$? Or is there a better way to proceed to understand the behavior of the solutions to this PDE for small $\epsilon$?

EDIT:

I will add some clarifications as to what I'm seeking with a toy example. Consider the distribution $G_t:\mathcal{S}\to\mathbb{R}, \phi(x)\mapsto \int g_t(x)\phi(x)dx$, where $g_t=\frac{1}{2\pi}\int e^{ikx}e^{-tk^2}\,dk=\frac{1}{2\sqrt{\pi t}}e^{-x^2/4t}.$ We know that $G_t\to\delta_0$ as $t\to 0$ as distributions. However, we can more precise by asking what is the behavior of $G_t-\delta_0$ on $\phi\in\mathcal{S}$ for small but non-zero $t$. To do so, we consider, with $0<\epsilon<1/2$, $$G_t(\phi)=\frac{1}{2\sqrt{\pi}}\int_{|x|\le \,t^{\epsilon-1/2}}\phi(x\sqrt{t})\,e^{-x^2/4}\,dx \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{1}{2\sqrt{\pi}}\int_{|x|\ge \,t^{\epsilon-1/2}}\phi(x\sqrt{t})\,e^{-x^2/4}\,dx. $$ We can Taylor expand $\phi$ in the first integral, which contains the dominant contributions $\phi(0)+\phi''(0)t+\mathcal{O}(t^2)$, while the latter term contains exponentially smaller terms on the order of $ t^{\frac{1}{2}-\epsilon}e^{-t^{2\epsilon}/4t})$. Thus, we have that $G_t=\delta_0+t\delta''_0+\mathcal{O}(t^2).$

Answer 1

The Fourier transform $F_\epsilon(x,t)$ has a closed-form expression in terms of hypergeometric functions, $$F_\epsilon(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}e^{i\epsilon^4 tk^4}\,dk=$$ $$\qquad=(8\pi \epsilon)^{-1}\left[8 e^{i\pi/8}t^{-1/4} \Gamma \left(\tfrac{5}{4}\right) \, _0F_2\left(;\tfrac{1}{2},\tfrac{3}{4};2^{-8}it^{-1}(x/\epsilon)^4\right)-e^{3i\pi/8}t^{-3/4} (x/\epsilon)^2 \Gamma \left(\tfrac{3}{4}\right) \, _0F_2\left(;\tfrac{5}{4},\tfrac{3}{2};2^{-8}it^{-1}(x/\epsilon)^4\right)\right].$$

For small $\epsilon$ one can evaluate the integral by steepest descent, I find $$F_\epsilon(x,t)\rightarrow \Phi_\epsilon(x,t)\equiv \frac{\exp\bigl(-3 (2^{-8}x^4/z)^{1/3}\bigr)}{ \sqrt{3 \pi } (4zx^2)^{1/6}},\;\;z=i\epsilon^4 t.$$ As a check, I note that $\int_{-\infty}^\infty \Phi_\epsilon(x,t)\,dx=1$, while $\lim_{\epsilon\rightarrow 0}\Phi_\epsilon(x,t)=0$ for $x\neq 0$, hence $\Phi_\epsilon(x,t)\rightarrow\delta(x)$ for $\epsilon\rightarrow 0$, as it should be.