Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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|improve this question

1 Answer 1

up vote 2 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|improve this answer
Thanks a lot! That really helped. –  Prahar Aug 8 '12 at 1:28

Your Answer


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.