inequivalence of Fourier transform?

Question

I am working with the following function \begin{equation}\tag{1} {\cal F}\left(\textbf{k},\omega\right)=\frac{1}{i\omega+{\cal D}\left|\textbf{k}\right|^{2}} \end{equation} with $\textbf{k}=(k_x,k_y)$, $F\left(\textbf{k},t\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{i\omega t}}{i\omega+{\cal D}\left|\textbf{k}\right|^{2}}\,\mathrm d\omega = e^{-{\cal{D}\,\textbf{k}^2 |t|}}$ by means of $\frac{1}{2\pi}\int_{\Bbb R} \frac{e^{i\omega t}\,\mathrm d \omega}{i \omega+A}=e^{-A|t|}$. Ideally, I want to calculate the function $G(t)$ or ${\cal G}(\omega)$ with $G(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} {\cal G}(\omega)\,e^{i\omega t} \,\mathrm d\omega$, defined, respectively by \begin{align*} G(t) = \iint_{\Bbb R^2} \frac{k_{x}^{2}}{\textbf{k}^{2}}\,e^{-2|\textbf{k}|d} F(\textbf{k},t)\,\mathrm d\textbf{k} &= \int_{0}^{2\theta} \cos^{2}\theta\,\mathrm d\theta\int_{0}^{\infty} ke^{-2kd}e^{-{\cal D}k^{2}|t|} \,\mathrm d k \\\tag{2} &= \pi\int_{0}^{\infty} k\,e^{-2kd}e^{-{\cal D}k^{2}|t|}\,\mathrm d k \\ &= \pi\, \frac{\sqrt{{\cal D}|t|}-\sqrt{\pi}de^{\frac{d^{2}}{{\cal D}|t|}}\text{erfc}\left(\frac{d}{\sqrt{{\cal D}|t|}}\right)}{2({\cal D}|t|)^{3/2}} \end{align*} and \begin{align*} {\cal G}(\omega) = \iint_{\Bbb R^2}\frac{k_{x}^{2}}{\textbf{k}^{2}}e^{-2|\textbf{k}|d}{\cal F}(\textbf{k},\omega) \,\mathrm d\textbf{k} &= \iint_{\Bbb R^2}\frac{k_{x}^{2}}{\textbf{k}^{2}}\frac{e^{-2|\textbf{k}|d}}{i\omega+{\cal D}\left|\textbf{k}\right|^{2}} \,\mathrm d\textbf{k} \\\tag{3} &= \int_{0}^{2\pi} \cos^{2}\theta \,\mathrm d\theta\int_{0}^{\infty} k\frac{e^{-2kd}}{i\omega+{\cal D}k^{2}}\,\mathrm dk \\ &= \pi\,\int_{0}^{\infty}\frac{{\cal D}k^{3}e^{-2kd}}{\omega^{2}+\left({\cal D}k^{2}\right)^{2}} \,\mathrm dk \end{align*}

My problem is that although $G(t)$ is well behaved, well-defined and continuous, I find $\lim_{\omega\rightarrow 0}{\cal G}(\omega) \rightarrow \infty$, which is very odd. Of course a natural problem appears if we intend to calculate $G(t)$ using Eq. (3) through $G(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty}{\cal G}(\omega)e^{i\omega t} d\omega$, as the mentioned divergence at $\omega\rightarrow 0$ will yield $G(t)\rightarrow \infty$. I have been thinking about the underlying reason for that, but so far I have nothing very concrete. Any thoughts? Thanks!

Answer 1

So actually the function $G$ is continuous but not smooth in $0$ and not integrable for large $|t|$. Using the asymptotic behavior of the $\mathrm{erfc}$ function from https://en.wikipedia.org/wiki/Error_function, one gets $$ \begin{align*}\tag{1} G(t) &\underset{|t|\to\infty}\sim \frac{C}{|t|} \\\tag{2} G(t) &= G(0) + C\,|t| + \underset{|t|\to 0}{O}(|t|^2) \end{align*} $$ This function is in particular not integrable, but still square integrable, so its Fourier transform makes sense in the $L^2$ sense (so as a limit of truncated integrals in the definition of the Fourier transform) or as a tempered distribution.

Now notice that the Fourier transform of $1/{|t|^a}$ is $1/|t|^{1-a}$ when $a$ is not an integer (with suitable definitions of $1/|t|^c$ as distributions when $c>1$) and the Fourier transform of $1/|t|$ is of the form $C_0+C\ln(|\omega|)$ (see e.g. here for $1/|t|$ and here for $|t|$). So, heuristically, since the behavior of the Fourier transform in $0$ gives an idea on the behavior of the function for large values (and reciprocally, the behavior of the Fourier transform for large values gives an idea on the behavior of the function in $0$) one obtains $$ \begin{align*} {\cal G(\omega)} &\underset{|\omega|\to 0}\sim C\ln(|\omega|) \\ {\cal G(\omega)} &\underset{|\omega|\to \infty}\sim \frac{C}{|\omega|^2} \end{align*} $$ which seems to be coherent with your computations.

Notice that since ${\cal F}(G(0)) = G(0)\,\delta_0$, we do not care about the $G(0)$ in my asymptotic formula $(2)$ to get the behavior of $\cal G$ when $\omega\to\infty$ since the Dirac delta "is $0$" for large values of $\omega$.

You should be able to get these asymptotics with your exact formula (cutting the integral appropriately and doing some changes of variable). Another way is to take a smooth and compactly supported function $0\leq\varphi≤1$ and to write $G = \varphi\, G + (1-\varphi)\,G = G_0 + G_{\infty}$ and to use my above reasoning.