Fourier transformation of heaviside functions I need to know the value of following integration.
$\frac{1}{2\pi}\int_{-\infty}^{\infty}dx_{1}dx_{2} e^{i k_{1}x_{1} +ik_{2}x_{2}} \theta(x_{1})\theta(x_{2}-x_{1})$,
where $\theta$ is Heaviside theta function. 
Above integration looks like a two-dimensional Fourier transformation of Heaviside theta functions, $\theta(x_{1})\theta(x_{2}-x_{1})$. To do this, I thought two ways as below. 
1) $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx_{1} e^{ik_{1}x_{1}}  \theta(x_{1})
\Big( \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx_{2} e^{ik_{2}x_{2}}
\theta(x_{2}-x_{1})
\Big)$
or
2) $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx_{2} e^{ik_{2}x_{2}} 
\Big( \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dx_{1} e^{ik_{1}x_{1}}
 \theta(x_{1})\theta(x_{2}-x_{1})
\Big)$.
As far as I have checked in mathematica, above two ways lead slightly different results. I thought they should be same. I don't know why. 
Please help me figure out which one is correct way. 
 A: See this similar Math.SE post for an explanation of the following manipulations:
$$ I(k_1,k_2)~:=~\frac{1}{2\pi}\iint_{\mathbb{R}^2}\!\mathrm{d}x^1\mathrm{d}x^2~ e^{i (k_1x^1 +k_2x^2)} ~\theta(x^1)\theta(x^2-x^1)$$
$$\tag{1}~=~\frac{1}{2\pi}\iint_{\mathbb{R}^2}\!\mathrm{d}x^1\mathrm{d}x^{\prime 2}~ e^{i (k^{\prime}_1x^1 +k_2x^{\prime 2})} ~\theta(x^1)\theta(x^{\prime 2})~=~\frac{1}{2\pi}I_1(k^{\prime}_1)~I_2(k_2),$$
where
$$ \tag{2} x^{\prime 2}~:=~x^2-x^1, \qquad k^{\prime}_1~:=~k_1+k_2, $$
$$ \tag{3} I_1(k^{\prime}_1)
~:=~\int_{\mathbb{R}}\!\mathrm{d}x^1~ e^{i k^{\prime}_1x^1} ~\theta(x^1)
~=~\frac{i}{k^{\prime}_1+i0^+}
~=~P\frac{i}{k^{\prime}_1}+\pi\delta(k^{\prime}_1),$$
$$ \tag{4} I_2(k_2)
~:=~\int_{\mathbb{R}}\!\mathrm{d}x^{\prime 2}~ e^{i k_2 x^{\prime 2}} ~
\theta(x^{\prime 2})
~=~\frac{i}{k_2+i0^+}
~=~P\frac{i}{k_2}+\pi\delta(k_2).$$
Therefore OP's integral (1) becomes
$$ \tag{5}I(k_1,k_2)~=~\frac{1}{2\pi}I_1(k^{\prime}_1)~I_2(k_2)
~=~\frac{-1}{2\pi} \frac{1}{k_1+k_2+i0^+}\frac{1}{k_2+i0^+}. $$
