Computing the FT of the pullback of the Heaviside distribution I'm very comfortable with the computation for the Fourier transform of the Heaviside function in $\mathbb{R}$, and I'm more or less familiar with pullbacks of distributions (like the delta distribution along the diagonal: $\delta(x-y)$)
Out of sheer curiosity I tried computing the Fourier transform of the distribution $H(x-y)$, however I've been having trouble getting a clean formula in the end.
Obviously if $u(x,y) = H(x-y)$ were this distribution, then
$$\begin{align} \langle\widehat{u},\phi \rangle &:= \langle u, \widehat{\phi} \rangle\\ &= \iint_{\mathbb{R}^2} H(\xi-\eta)\widehat{\phi}(\xi,\eta) \,d\xi d\eta \\ &= \frac{1}{2}\iint_{\mathbb{R}^2} H(s)\widehat{\phi}\left( \tfrac{1}{2}(t+s), \tfrac{1}{2}(t-s) \right) dsdt\\ &= \frac{1}{2}\int_{-\infty}^{\infty}\int_{0}^{\infty}\widehat{\phi}\left( \tfrac{1}{2}(t+s), \tfrac{1}{2}(t-s) \right) dsdt\\ &=\frac{1}{2} \int_{-\infty}^{\infty}\int_{0}^{\infty}\left\{ \iint_{\mathbb{R}^2} \phi(x,y) \, e^{-i\left( \tfrac{1}{2}(t+s)x+\tfrac{1}{2}(t-s)y \right)} dxdy\right\}dsdt \end{align}$$
Here is where I'm getting stuck. I've tried throwing in a cutoff $e^{-\varepsilon s}$ to help with the $ds$ integral, but I'm not exactly sure what to do about the $dt$ integral. I've even tried throwing in a second cutoff, $e^{-\varepsilon |t|}$, but the expressions I get in terms of $x$, $y$, and $\varepsilon$ are quite complicated and I'm not sure how to precisely evaluate the distributional limit as $\varepsilon \to 0^+$:
$$ \begin{align} \widehat{u}(x,y) &= \lim_{\varepsilon \to 0^+} \frac{-8i\varepsilon}{(x-y-2i\varepsilon)((x+y)^2 + 4\varepsilon^2)}\\ &= \lim_{\varepsilon \to 0^+} \frac{1}{(x+y)^2 + 4\varepsilon^2} \left( \frac{16\varepsilon^2}{(x-y)^2 + 4\varepsilon^2} + i \frac{8\varepsilon(y-x)}{(x-y)^2 + 4\varepsilon^2} \right) \\ &= \lim_{\varepsilon \to 0^+} \left( \frac{8\varepsilon}{\left((x+y)^2 + 4\varepsilon^2 \right) \left( (x-y)^2 + 4\varepsilon^2 \right)} \right) \left( 2\varepsilon + i(y-x) \right) \end{align}$$
Any suggestions?
edit: Since $H(x)$ has a nice transform, and since pullbacks like $\delta(x-y)$ can be expressed nicely as line integrals in the plane, I feel like there should be some closed formula for this distribution.
I even tried plugging it into Mathematica, but I get something rather odd:
$$ \widehat{u} = 2\pi^2 \delta(x)\delta(x+y)-2\pi i \frac{\delta(x+y)}{x} $$
This is especially troubling to me since Mathematica seems to be multiplying distributions together, which I know is not allowed in most cases. In particular, it looks like Mathematica thinks the imaginary part involves the product the Cauchy principal value and the delta distribution, which I know for sure is not defined.
 A: Up to a rotation, you want to compute the Fourier transform of $v(x, y):=H(y)$. This is 
$$
\hat{v}(\xi, \eta)=\iint_{\mathbb R^2} H(y)e^{-ix\xi}e^{-iy\eta}\, dxdy = 2\pi \delta(\xi) \hat{H}(\eta), $$ 
where $\hat{H}$ is the 1-dimensional Fourier transform of $H$. 
REMARK. In this formula, we multiply the distributions $(\xi, \eta)\mapsto \delta(\xi)$ and $(\xi, \eta)\mapsto \hat{H}(\eta)$. This looks suspicious, because the product of distributions is not well-defined in general. However, as these lecture notes of Christian Brouder, Nguyen Viet Dang, Frédéric Hélein explain, such a product is well-defined when the singular sets are transversal, and it is the present case. (I think that §2.1.3 of the notes is all we really need here, and nothing more).
To apply this result to the original function $u(x, y)=H(x-y)$, we observe that 
$$
u(x, y)=v(x+y, x-y), $$
and we recall the following general property of the Fourier transform; for every nonsingular matrix $A$, 
$$
\mathcal F(v(A\cdot))(\xi, \eta) = \frac{1}{\lvert \det A\rvert} \mathcal F(v)\left(A^{-T}\begin{bmatrix} \xi\\\eta\end{bmatrix}\right).$$
In our case, 
$$
A=\begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix}, $$ 
with determinant $-2$, so the end result is 
$$
\pi\delta(\tfrac{\xi+\eta}{2})\hat{H}(\tfrac{\xi-\eta}{2}).$$
