Integral of Theta Function with $e^{ix}$

Question

I want to evaluate the integral:

$$I=\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2 \ \Theta(x_1-x_2) \ e^{i(ax_1+bx_2)}$$ where $\Theta(x)$ is the Heaviside function.

What I was doing now was taking the relation for $\Theta$: $\Theta (x)=-\frac{1}{2\pi i}\int_{-\infty}^{\infty}d\tau \frac{1}{\tau + i\epsilon} e^{-ix\tau}$ and I got:

$$I= -\frac{1}{2\pi i}\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}d\tau \ \frac{1}{\tau + i\epsilon}e^{-i(\tau-a)x_1}e^{-i(\tau+b)x_2}\\=2\pi i\int_{-\infty}^{\infty}d\tau\ \frac{1}{\tau + i\epsilon}\delta(\tau-a)\delta(\tau+b) =2\pi i\frac{\delta(a+b)}{a + i\epsilon}$$ I didn't know if the integral was convergent and I could simply interchange the integrals, so I tried it in a different form with $X=x_1+x_2$ and $x=x_1-x_2$ : $$I=\frac{1}{2}\int_{-\infty}^{\infty}dX \int_{-\infty}^{\infty}dx \ \Theta(x) \ e^{ia\frac{x+X}{2}+ib\frac{X-x}{2}} \\ =\pi \int_{-\infty}^{\infty}dx \ \Theta(x) e^{-2\pi i\frac{b-a}{4\pi}}\delta(\frac{b+a}{2})$$ With the Fourier Transform of the Heaviside function $\int_{-\infty}^{\infty}dk\ \Theta(k)e^{-2\pi i kx}=\frac{1}{2}(\delta(x)-\frac{i}{\pi k})$ I get $$I=\pi \left(2\pi\delta(b-a)-\frac{4 i}{b-a}\right)\delta(a+b)=2\pi^2\delta(a)\delta(b)+2\pi i\frac{\delta(a+b)}{a}$$ I don't know yet where the $\delta(a)\delta(b)$ should come from in the first method. When I want to check that now and integrate $I$ over $a$ and $b$ I get from the first line: $$\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}db \ I = \int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2 \Theta(x_1-x_2) \delta(x_1)\delta(x_2) \\ = \Theta(0)=\frac{1}{2}$$ and from the second result: $$\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}db \ I=4\pi^2-\int_{-\infty}^{\infty}db \frac{1}{b}=-\infty$$ Where did it go wrong? Is the integral correct?

Thank you in advance.

EDIT: corrected mistake in derivation because of comment.

This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user gaugi

Comments

According to Mathematica, the result is $2 \pi \left(\pi \delta (a) \delta (a+b)+\frac{i \delta (a+b)}{a}\right)$.

This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user DumpsterDoofus

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oh, yes, thank you, I just see that I get that with the second method, too. $\delta(a+b)\delta(a-b)=\delta(a)\delta(b)$. I have to check the first method for that, too.

This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user gaugi

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The two results are in agreement in view of the relation: $\frac{1}{a+i\epsilon} = \frac{1}{a} - i\pi \delta(a)$ with $1/a$ understood in the sense of principal value distribution...

This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user V. Moretti

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Close-voters, would you care to comment on your reasons? Unless you want to migrate this to math or as homework (both of which I'd disagree with), you should provide at least some explanation of what's going on. Voting to reopen.

This post imported from StackExchange Physics at 2014-03-06 21:12 (UCT), posted by SE-user Emilio Pisanty