# Integration of complex exponentials

1. Oct 18, 2015

### spaghetti3451

I am trying to understand how to go from the first line to the next:

$\frac{1}{(2\pi)^{3}}\int d^{3}p\ e^{-it\sqrt{{\bf{p}}^{2}+m^{2}}}\ e^{i {\bf{p}}({\bf{x}}-{\bf{x}}_{0})} = \frac{1}{2\pi^{2}|{\bf{x}}-{\bf{x}}_{0}|} \int_{0}^{\infty} dp\ p\ sin(p|{\bf{x}}-{\bf{x}}_{0}|)\ e^{-it\sqrt{{p}^{2}+m^{2}}}$.

As an initial guess, I'm thinking that the vector $\bf{p}$ is decomposed in spherical polar coordinates, evidenced by the facts that $\frac{1}{(2\pi)^{3}}$ has been multiplied by $4\pi$ and that the volume element is afterwards $dp\ p$ with limits from $0$ from $\infty$. Also, I'm thinking that firstly the exponential $e^{i {\bf{p}}({\bf{x}}-{\bf{x}}_{0})}$ is broken using Euler's relation into its sine and cosine components.

I wonder what's happened to the cosine component, the factor of $i$ in the sine component, how the polar and azimuthal angles have been separated out from the vector expression $-it\sqrt{{\bf{p}}^{2}+m^{2}}$ and how the factor of $|{\bf{x}}-{\bf{x}}_{0}|$ in the denominator outside the integral appears.

2. Oct 18, 2015

### andrewkirk

I can help with this bit:
The expression ${\bf{p}}^{2}$ is more correctly written $\|\bf{p}\|^2$, which is equal to $p^2$. Hence
$-it\sqrt{{\bf{p}}^{2}+m^{2}}$ can simply be replaced by $-it\sqrt{{p}^{2}+m^{2}}$

I think this will be something to do with the following:
Convert the integral into
$$\int_0^\infty \left[\int_{N_p} f(p,\phi,\theta) dA + \int_{S_p} f(p,\phi,\theta) dA\right]\,dp$$
where $N_p$ and $S_p$ are respectively the Northern and Southern hemispheres of the sphere of radius $p$.
Then think about what cancels out between the two hemispheric integrals when you match antipodal points.