One common way of finding the propagator of the diffusion equation is to Fourier transform only the spatial variables and then solve the resulting ODE (as it is done, e.g., in this video). However, I was wondering if performing both spatial and temporal transformations could also be justified, as it usually makes analytical progress simpler especially in the case of (linearly) coupled diffusion PDEs (i.e. working with algebraic equations is simpler than convolutions).
The following is the derivation of 1-D diffusion kernel ( $G(x,t)= \int \frac{dq}{2\pi}\frac{d\omega}{2\pi} e^{-i\omega t + i q x} \tilde{G}(q,\omega)$ ): \begin{align} \frac{\partial G}{\partial t} &= D \frac{\partial^2 G}{\partial x^2} + \delta(x)\delta(t) \rightarrow (-i\omega + D q^2) \tilde{G} = 1,\\ G(x,t)&= \int \frac{dq}{2\pi}\frac{d\omega}{2\pi} e^{-i\omega t + i q x} \tilde{G}(q,\omega) = \int \frac{dq}{2\pi}\frac{d\omega}{2\pi} \frac{e^{-i\omega t + i q x}}{-i\omega+Dq^2}. \end{align} We first perform the frequency integration in the complex plane. Noting that there is one pole in the lower half plane at $\omega = -i D q^2$, the integration is zero if $t<0$ (where the contour should be closed in the upper half plane), and for $t>0$ we get: \begin{align} G(x,t) = \Theta(t) \int \frac{dq}{2\pi} e^{-Dq^2 t + i q x} = \Theta(t) \frac{e^{-\frac{x^2}{4 D t}}}{{\sqrt{4\pi D t}}}, \end{align} where the Heaviside $\Theta$ enforces $G(x,t<0)=0$.
What I am wondering is whether defining $\tilde{G}$ in the first place is justified, as from the final solution it appears that $\int_{-\infty}^\infty G \, dt$ diverges, indicating one might need to use a Laplace transform instead; if it is not justified, then what is going on that this approach still gives the correct final result?