How to make sense of the Fourier transform of this distribution

Question

I want to compute the Fourier transform of this distribution:

$$D(f)=\int_{\mathbb{R}} f(t,t^2) \frac{dt}{t}$$

($f$ a Schwartz function on $\mathbb{R}^2$, the integral interpreted with a Cauchy principal value)

Then, with a bit of faith in the good of the world in the step where the integrals are interchanged:

\begin{align} \hat{D}(f) &= D(\hat{f})=\int_\mathbb{R} \hat{f}(t,t^2) \frac{dt}{t} =\int_\mathbb{R} \int_{\mathbb{R}^2} f(x,y) e^{-i(xt+yt^2)} d(x,y) \frac{dt}{t}\\ &\overset{\text{faith}}{=} \int_{\mathbb{R}^2} f(x,y)\int_\mathbb{R} e^{-i(xt+yt^2)} \frac{dt}{t} d(x,y)=\int_{\mathbb{R}^2} f g \end{align}

where (purely symbolically),

$$g(\alpha,\beta)=\int_{\mathbb{R}} \frac{1}{t} e^{-i(\alpha t+\beta t^2)} dt$$

Now I have two problems,

1. How to remove the dependence on faith?

2. Is $g$ (that is, $\hat{D}$) a function, i.e. does that integral above make sense in some way for $(\alpha,\beta)\not=(0,0)$? If yes, is it maybe bounded? Or maybe even smooth?

Update:

• I believe the answer to 1. is to consider truncations, i.e. integrate only over $a\le |t|\le b$ and then always prove things uniformly in those parameters.

• I have shown $\hat{D}$ (i.e. $g$) to be a function and that it is in fact bounded.

• Now I'm missing only the smoothness. As a comment suggests, there may be no smoothness along $\beta=0$.