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 $t>0$ $$\Psi(x,t)=\int_{-\infty}^\infty K(x-y,t)\Psi(x,0)dy.$$
The Fourier transform vanishes for $u>0$, for $u<0$ instead
$$
I(u,v)=\frac{1}{2\pi}\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\,\frac{e^{\mathrm{i}(ux+vy)}}{x+y^2+\mathrm{i}0^+}$$
$$\qquad\qquad=-ie^{i\pi/4}\sqrt{\pi }(- u)^{-1/2}\exp\left({\frac{i v^2}{4 u}}\right)\;\;\text{for}\;\; u<0.
$$
For $u=0$ there is a delta function, $I(0,v)=-i\pi\delta(|v|)$. Note that
$$\int_{-\infty}^\infty I(u,v)\,dv=\begin{cases}
0&\text{if}\;\;u>0,\\
-\pi i&\text{if}\;\;u=0,\\
-2\pi i&\text{if}\;\;u<0.
\end{cases}$$
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}
Chanced on this and after much rumination I have decided to point out the following flaws in the postings. Firstly, I assume that when you write $$\frac 1 {x+y^2+i0}$$ you mean a limit of the function $$\frac 1 {x+y^2+i\epsilon}$$ as $\epsilon$ goes to zero. In fact, since the value of the limit depends on whether it is taken over positive or negative values, you probably mean $$\lim_{\epsilon \to 0+}\frac 1 {x+y^2+i\epsilon}.$$
However, the expression $\frac 1 {x+y^2}$ can also be regarded as a distribution in a natural way which is related to the concept of the principal value of a function at a singularity. This is due to the fact that $\log|x+y^2|$ is locally integrable and so can be interpreted as a distribution. We can define the above expression simply to be $D_x \log|x+y^2|$ (distributional derivative).
The crucial point is that despite the trap set by the notation, these two distributions are distinct--they differ by a delta like function supported on the singular set $x+y^2=0$ and of course this part is crucial in physical applications.
Now there is school of thought on this site that believes that one solves research problems of this type simply by bashing something into a computer. Since it is no skin off my nose, I can only wish them good luck in their enterprise but feel obliged to point out that they are missing the (in my opinion crucial) term in the FT which arises from the singular part.
I might add that there is nothing formal in the approach sketched here--the details can be made perfectly rigorous using simple elementary methods at the level of freshman analysis. These tools were worked out in the 50´s of the previous century.
