- #1
Tio Barnabe
I should evaluate ##\int d^3 p \ \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}## over all ##\mathbb{R}^3##. How can I do this in spherical coordinates? Since ##\vec{p}## is a position vector in ##\mathbb{R}^3##, our ##\vec{r}## of the spherical coordinates would be just equal to ##\vec{p}##, correct? (The same would be said of ##\vec{x}##, which we could call an ##\vec{r'}##.)
So the integral above would be re-written as ##\int dp \ d\varphi \ d\theta \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. I suppose that the correct range for our "spherical variables" is ##p \in (-\infty, \infty)## and ##\varphi, \theta \in [0, 2 \pi]##. Now the integral would then become ##4 \pi \int dp \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. How do I evaluate it?
EDIT: I don't need to give the final result, just to evaluate the last integral a little further.
So the integral above would be re-written as ##\int dp \ d\varphi \ d\theta \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. I suppose that the correct range for our "spherical variables" is ##p \in (-\infty, \infty)## and ##\varphi, \theta \in [0, 2 \pi]##. Now the integral would then become ##4 \pi \int dp \ p^2 \exp(i \vec{p} \cdot \vec{x}) / \sqrt{|p| + m^2}##. How do I evaluate it?
EDIT: I don't need to give the final result, just to evaluate the last integral a little further.
Last edited by a moderator: