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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.

Semiclassical
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Veteran
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1 Answers1

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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

$$ \tag{5}I(k_1,k_2)~=~\frac{1}{2\pi}I_1(k^{\prime}_1)~I_2(k_2) ~=~\frac{-1}{2\pi} \frac{1}{k_1+k_2+i0^+}\frac{1}{k_2+i0^+}. $$

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Qmechanic
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