Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, in the calculation of $ D(t,r) = \left[ \phi(x) , \phi(y) \right] $, where $ t= x^0 - y^0,~ \vec{r} = \vec{x} - \vec{y} $ you need to calculate the following integral $$ D(t,r) = \frac{1}{2\pi^2 r} \int\limits_0^\infty dp \frac{ p \sin(p r) \sin \left[(p^2 + m^2)^{1/2} t \right]} { (p^2 + m^2 )^{1/2}} $$ For $m=0$, the integral is simple. We get $$ D(t,r) = \frac{1}{4\pi r} \left[ \delta(t - r) - \delta(t + r) \right] $$ I even know what the answer for $ m \neq 0 $. I have no idea how to calculate it though. Any help?

share|cite|improve this question
up vote 3 down vote accepted

Using Gradshteyn and Ryzhik (seventh edition) 3.876 (1) $$\int_0^\infty \frac{\sin{(p \sqrt{x^2+a^2})}}{\sqrt{x^2+a^2}} \cos(b~x)dx=\frac{\pi}{2} J_0(a\sqrt{p^2-b^2}) ~~[0<b<p]\\ = 0~~[0<p<b] $$ Differential of both sides with respect to $b$ will give the integral you want to calculate.

share|cite|improve this answer
    
Thanks a lot! That really helped. – Prahar Aug 8 '12 at 1:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.