# Changing to polar coordinates in an exponential

1. Jul 6, 2015

### physichu

Hello :)
I dont get this integral (Peskin & Schroeder P. 27 )

$\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}{1 \over {{E_{\bf{p}}}}}{e^{i{\bf{p}} \cdot {\bf{r}}}}} = {{2\pi } \over {{{\left( {2\pi } \right)}^3}}}\int\limits_0^\infty {dp{{{p^2}} \over {2{E_{\bf{p}}}}}{{{e^{ipr}} - {e^{ - ipr}}} \over {ipr}}}$.

Actualy I dont get the next stage iethr :(

$= {{ - i} \over {2{{\left( {2\pi } \right)}^2}r}}\int\limits_{ - \infty }^\infty {dp{{p{e^{ipr}}} \over {\sqrt {{p^2} + {m^2}} }}}$.

2. Jul 6, 2015

### Dr. Courtney

Between the first two, it appears as if they have integrated over the angular coordinates in a spherical coordinate sytem, leaving only the radial integral. There needs to be some justification about spherical symmetry to simply yield a factor of 2 pi by integrating over the angular coordinates.

Between the second and third expressions, they seem to be using E = p^2/2m and simplifying.

3. Jul 6, 2015

### physichu

where do the other exponential and the $ipr$ denominator come from?

Last edited by a moderator: Jul 6, 2015