Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 7 like - 0 dislike
1140 views

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

asked Mar 6, 2014 in Mathematics by gaugi (70 points) [ revision history ]
retagged Mar 25, 2014 by dimension10
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
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
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
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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...