Integral in Commutator of Scalar fields

[praharmitra]

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?

 

[eljose79]

Thank you for your question. I am happy to assist you in understanding this integral and how to calculate it.

First, let's break down the integral into smaller parts to make it easier to understand. We have:

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}}

The first part of the integral, \frac{1}{2\pi^2 r}, is just a constant term that will not affect the final result. So let's focus on the integral itself:

\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}}

To solve this integral, we will need to use a technique called integration by parts. This technique is used when we have a product of two functions in our integrand, in this case, p and the rest of the terms.

Integration by parts involves rewriting the integral as follows:

\int\limits_a^b u(x) v'(x) dx = [u(x) v(x)]_a^b - \int\limits_a^b u'(x) v(x) dx

Where u(x) and v(x) are two functions, and u'(x) and v'(x) are their derivatives.

To apply this technique to our integral, let's define u(p) = p and v'(p) = \frac{\sin(p r) \sin \left[(p^2 + m^2)^{1/2} t \right]} { (p^2 + m^2 )^{1/2}}. Then, we can rewrite our integral as:

\int\limits_0^\infty dp u(p) v'(p) = [u(p) v(p)]_0^\infty - \int\limits_0^\infty dp u'(p) v(p)

Now, let's calculate the first part of the integral, [u(p) v(p)]_0^\infty. Since p goes