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Is there an explicit formula for the Fourier transform of the generalized function of 2 variables $$\frac{1}{x+y^2+i0}?$$

Remark. Equivalent question: consider the Schroedinger equation one the line $$i\frac{\partial}{\partial t}\Psi(x,t)=\frac{\partial^2}{\partial x^2}\Psi(x,t).$$

Find the kernel $K(x,t)$ such that for $t>0$ $$\Psi(x,t)=\int_{-\infty}^\infty K(x-y,t)\Psi(x,0)dy.$$

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    $\begingroup$ The answer to the second question is the standard free particle particle propagator from quantum mechanics: $K(x,t) = (C_1/t^a) e^{iC_2 x^2/t}$ (exercise: fix the right constants $a, C_1, C_2$). $K(x,t)$ is not exactly the Fourier transform from the first question, they are related, as you say. I don't know if that's enough for you. $\endgroup$ – Igor Khavkine Aug 15 '18 at 22:30
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I think$^\ast$ the Fourier transform only vanishes for $u>0$, for $u<0$ instead $$ \frac{1}{2\pi}\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\,\frac{e^{\mathrm{i}(ux+vy)}}{x+y^2+\mathrm{i}0^+}=-ie^{i\pi/4}\sqrt{\pi }(- u)^{-1/2}\exp\left({\frac{i v^2}{4 u}}\right)\;\;\text{for}\;\; u<0. $$ You can check it out on Wolfram Alpha.

$^\ast$ This remark refers to a version of the other answer before the latest edit, where zero for all $u$ was obtained.

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The Fourier transform for $1/(x+y^2+\mathrm{i}\epsilon), \epsilon\in\mathbb{R}$ is formally compute as \begin{align} \int_{\mathbb{R}^2}\frac{e^{2\pi\mathrm{i}(ux+vy)}}{x+y^2+\mathrm{i}\epsilon}\,dx\,dy&=\int_{\mathbb{R}}e^{-2\pi\mathrm{i}(uy^2-vy)}\,dy\int_{\mathbb{R}}\frac{e^{2\pi\mathrm{i}ux}}{x+\mathrm{i}\epsilon}\,dx\\ &=\frac{\mathrm{i}e^{\frac{\pi\mathrm{i}v^2}{2u}}}{\sqrt{{2\mathrm{i}u}}}\int_{\mathbb{R}}\frac{x\sin(2\pi ux)-\epsilon\cos(2\pi ux)}{x^2+\epsilon^2}\,dx\quad(u\neq 0 )\\ &=\frac{\mathrm{i}e^{\frac{\pi\mathrm{i}v^2}{2u}}}{\sqrt{{2\mathrm{i}u}}}\left(\pi{\rm sign}(u)-\pi+o_{u,\epsilon}(1)\right)= \frac{\pi\mathrm{i}e^{\frac{\pi\mathrm{i}v^2}{2u}}}{\sqrt{{2\mathrm{i}u}}}({\rm sign}(u)-1).~as~\epsilon \rightarrow 0. \end{align}

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