Multiple Fourier Integrals involving Heaviside Theta Function

Solution 1:

Products of distributions is a well-known delicate topic, see e.g. this mathoverflow post. Unless if each distribution factor depend on different variables, such as, e.g. $\delta^3(\vec{r})=\delta(x)\delta(y)\delta(x)$. So the safest thing is to write OP's integral in two manifestly independent variables. E.g.

$$ I(a_1,a_2) ~:=~ \iint_{\mathbb{R}^2} \! \mathrm{d}x^1 \mathrm{d}x^2~ e^{i (a_1 x^1+a_2 x^2)} \theta(x^1-x^2) $$ $$\tag{1}~=~ \iint_{\mathbb{R}^2} \! \mathrm{d}y^1 \mathrm{d}y^2 ~ e^{i (b_1 y^1 + b_2 y^2)} \theta(y^1)~=~I_1(b_1)~I_2(b_2) .$$

Here we have defined a linear coordinate transformation (with unit determinant to avoid a Jacobian factor for simplicity)

$$ \tag{2} \begin{bmatrix}y^1 \\ y^2 \end{bmatrix} ~=~\begin{bmatrix} \frac{1}{2} &-\frac{1}{2} \\ 1-t & t \end{bmatrix} \begin{bmatrix}x^1 \\ x^2 \end{bmatrix} $$

in such a way that the new variable $y^1$ becomes the argument of the Heaviside step function. Here $t\in\mathbb{R}$ is a free parameter. It is interesting to trace how the integral (1) stays independent of the value of this free parameter $t$. The inverse transformation is

$$ \tag{3} \begin{bmatrix}x^1 \\ x^2 \end{bmatrix} ~=~\begin{bmatrix} t&\frac{1}{2} \\ t-1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix}y^1 \\ y^2 \end{bmatrix}. $$

In order for the argument $a_1 x^1+a_2 x^2=b_1 y^1 + b_2 y^2$ of the exponential (1) to stay form invariant, we define new variables

$$ \tag{4} \begin{bmatrix}b_1 \\ b_2 \end{bmatrix} ~=~\begin{bmatrix} t& t-1 \\ \frac{1}{2}& \frac{1}{2} \end{bmatrix} \begin{bmatrix}a_1 \\ a_2 \end{bmatrix}. $$

The first integral factor is

$$\tag{5} I_1(b_1)~:= \int_{\mathbb{R}} \! \mathrm{d}y^1~e^{i b_1 y^1}\theta(y^1) ~=~\frac{i}{b_1+i0^+}~=~P\frac{i}{b_1}+\pi\delta(b_1),$$

cf. the Sokhotski–Plemelj formula. The second integral factor is

$$\tag{6} I_2(b_2)~:= \int_{\mathbb{R}} \! \mathrm{d}y^2~e^{i b_2 y^2} ~=~2\pi ~\delta(b_2).$$

While the two factors (5) and (6) clearly depend on $t$, their product [=OP's integral (1)] is independent of $t$:

$$ I(a_1,a_2)~=~I_1(b_1)~I_2(b_2)~=~\frac{2\pi i}{a_1+i0^+}\delta(a_1+a_2)$$ $$\tag{7} ~=~2\pi i ~P\frac{1}{a_1}\delta(a_1+a_2)+2\pi^2 \delta(a_1)\delta(a_2). $$