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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!

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    $\begingroup$ Are you working in dimension $2$? Also, it should be $\mathcal F(k,\omega)$ in $(3)$, right? What exactly are these indefinite integrals? The antiderivative up to a constant? $\endgroup$
    – LL 3.14
    Commented Sep 28, 2021 at 17:16
  • $\begingroup$ Thank you for pointing these out. I have included them within my post. $\endgroup$
    – sined
    Commented Sep 28, 2021 at 17:19
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    $\begingroup$ Thank you, it is a bit more clear. You did not answered to my question about indefinite integrals? In particular in $(3)$, the integral without bounds becomes an integral from $0$ to $\infty$. $\endgroup$
    – LL 3.14
    Commented Sep 28, 2021 at 17:25
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    $\begingroup$ @LL3.14 I have included now all the limits of the integrals. Thanks! $\endgroup$
    – sined
    Commented Sep 28, 2021 at 17:40

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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.

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  • $\begingroup$ Thank you! Do you know whether we could exclude $\omega=0$ from our integral, e.g., by making the integral $\lim_{\delta \rightarrow 0} \int_{\delta}^{\infty}...$, similarly to the CPV? Would this result in an answer that does not diverge at $\omega\rightarrow 0$? $\endgroup$
    – sined
    Commented Sep 28, 2021 at 20:09
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    $\begingroup$ Your integral is an integral in the $k$ variable, not in $\omega$. So for every $\omega ≠ 0$, your integral is indeed finite, and you should be able to prove that it is bounded above by someting like $|\ln(|\omega|)|$ when $\omega\to 0$. In particular, it means that $\cal G$ is defined almost everywhere and is a (Lebesgue) integrable function. You cannot change it to become another function ... or perhaps I do not get what you are looking for? $\endgroup$
    – LL 3.14
    Commented Sep 28, 2021 at 21:27
  • $\begingroup$ The problem that I am having is the following: $G(t)$, defined by Eq. (2), is well-defined at $t=0$. However, if we think about $G(t)$ as $G(t=0)=\int_{-\infty}^{\infty} d\omega S(\omega)$ we get the divergence at $\omega\rightarrow 0$, which would result in an infinit $G(t=0)$, no? Am I missing something here? $\endgroup$
    – sined
    Commented Sep 29, 2021 at 17:47
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    $\begingroup$ No, even if $\cal G$ is infinite at $\omega=0$, it is still an integrable function. This is because $$ \int_0^1 |\ln \omega|\,\mathrm d ω = [\omega \,(1+\ln \omega)]_0^1 = 1 < \infty $$ and $$ \int_1^\infty \frac{\mathrm d \omega}{|\omega|^2} = [1/\omega]_0^1 = 1 < \infty $$ ... a function can be integrable and infinite. Then $G(0) = ∫ {\cal G} < \infty$ is indeed a valid formula $\endgroup$
    – LL 3.14
    Commented Sep 29, 2021 at 21:35

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