Fourier transform of distribution without physicist's $\delta$-function Say I want to find the Fourier transform of the following distribution:

The solution I have uses the physicist's definition of the Dirac $\delta$-function $\delta(x) = \int_\mathbb{R} e^{-i x y} dy$ which is mathematically ill-defined:

I cannot figure out how to solve this problem in a mathematically rigorous way, using only the allowed operations on distributions, though it should be possible to do...
Any help is appreciated.
 A: First - I will apply Foubini theorem a lot of times, since test functions are very integrable=)
Second - I will systematically omit all normalisation constants in Fourier transform, as well as integration limits.
Third - I will denote by $\phi_y(x) = \phi(x,y)$. $\forall y\quad \phi_y\in D(\Bbb R^m)$.
Fourth - the corresponding Fourier variables to $(x,y)$ will be $(\xi,\eta)$.
By definition we can write
$$\langle \hat U, \phi(x,y)\rangle  = \left\langle U, \int\phi(x,y) \exp(ix\xi + i y\eta) dx\,dy\right\rangle$$
$$=\int \left(\int\phi(x,y) \exp(ix\xi + i y\eta) dx\,dy\right)\big|_{\eta=0}d\xi$$
Since everything is smooth, we can pass the value of $\eta=0$ inside the integral:
$$=\int \left(\int\phi(x,y) \exp(ix\xi) dx\,dy\right) d\xi$$
$$=\int dy\int d\xi \left(\int\phi_y(x) \exp(ix\xi) dx \right) $$
$$=\int dy \left\langle 1,   \hat \phi_y(\xi)\right\rangle$$
(in the last integral $\hat \phi_y(\xi)$ is a FT with respect to $x$, and $1$ is a distribution acting on the variable $\xi$)
$$=\int dy \left\langle \hat 1,  \phi_y(x)\right\rangle$$
$$=\int dy \left\langle \delta_0,  \phi_y(x)\right\rangle$$
$$=\int     \phi_y(0) dy = \int \phi(0,y)dy.$$
In other words, you initial distriubtions is, in fact, a tensor product $1\otimes \delta_0$; you apply FT to this distribution, and you will get 
$$F[1\otimes \delta_0] = F[1]\otimes F[\delta_0] = \delta_0\otimes 1,$$ which coreponds to out previous result.
