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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?

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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.

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Thanks a lot! That really helped. –  Prahar Aug 8 '12 at 1:28
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